Learn By real time problems
Topic: Systems of Linear Equations and Matrix Rank (Click to Expand)
Question
Consider the matrices $A = \begin{bmatrix} 1 & 3 & -1 & 4 \\ 2 & 6 & -2 & 8 \\ 3 & 9 & -3 & 12 \end{bmatrix}$ and $b = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.
- (a) Find the Echelon form of $A$ and hence find the rank of $A$. $[2M]$
- (b) Find all the solutions of $AX = b$. $[2M]$
- (c) Suppose if the matrix $b = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}$ what can you conclude about the linear system of equations $AX = b$, where $A$ is the same matrix given in (a). $[1M]$
Detailed Step-by-Step Solution
Part (a): Finding the Echelon Form and Rank
The Concept: Think of a matrix as a stack of equations. The Echelon Form is a simplified version where we use "row operations" to create as many zeros as possible at the bottom. This helps us see which equations are unique and which are just copies of others.
- Start with Matrix A: $$A = \begin{bmatrix} 1 & 3 & -1 & 4 \\ 2 & 6 & -2 & 8 \\ 3 & 9 & -3 & 12 \end{bmatrix}$$
- Use Row Operations:
We want to turn the first numbers in Row 2 and Row 3 into zeros using Row 1.
Target Row 2: $R_2 \to R_2 - 2R_1 \implies [0, 0, 0, 0]$
Target Row 3: $R_3 \to R_3 - 3R_1 \implies [0, 0, 0, 0]$ - The Echelon Form: $$Echelon(A) = \begin{bmatrix} 1 & 3 & -1 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
- Finding the Rank:
The Rank is simply the number of non-zero rows (rows that have at least one number that isn't zero) in the final Echelon form.
Conclusion: There is only one non-zero row. Therefore, Rank of A = 1.
Part (b): Finding all solutions of $AX = b$
The Concept: We use an Augmented Matrix, which is just the matrix $A$ with the result $b$ attached as an extra column. We solve for the variables $x_1, x_2, x_3,$ and $x_4$.
1. Setup the Augmented Matrix $[A|b]$: $$\left[ \begin{array}{cccc|c} 1 & 3 & -1 & 4 & 1 \\ 2 & 6 & -2 & 8 & 2 \\ 3 & 9 & -3 & 12 & 3 \end{array} \right]$$
2. Simplify:
Perform $R_2 - 2R_1$ and $R_3 - 3R_1$. Note that the $b$ column also changes:
$$\left[ \begin{array}{cccc|c} 1 & 3 & -1 & 4 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0
\end{array} \right]$$
3. Write the resulting equation:
This leaves us with only one real equation: $x_1 + 3x_2 - x_3 + 4x_4 = 1$.
4. Parameterize (Free Variables):
Because we have 4 variables but only 1 equation, we have "free choice" for 3 of them. Let's name
them $s, t,$ and $u$:
$x_2 = s, \quad x_3 = t, \quad x_4 = u$
Solve for $x_1$: $x_1 = 1 - 3s + t - 4u$
5. General Solution Form: $$X = \begin{bmatrix} 1 - 3s + t - 4u \\ s \\ t \\ u \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + s\begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + u\begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$ Since $s, t,$ and $u$ can be any numbers, there are infinitely many solutions.
Part (c): What if $b = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}$?
The Concept: Consistency means "Is it possible to find a solution?" If we perform our row operations and end up with a statement that is mathematically impossible (like $0 = 5$), the system is inconsistent.
[Image showing consistent vs inconsistent systems of linear equations]1. Setup & Row Operations:
$R_3 \to R_3 - 3R_1$: The $b$ part is $2 - 3(1) = \mathbf{-1}$.
2. Resulting Matrix: $$\left[ \begin{array}{cccc|c} 1 & 3 & -1 & 4 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \mathbf{-1} \end{array} \right]$$
3. Conclusion:
The third row now says: $0x_1 + 0x_2 + 0x_3 + 0x_4 = -1$, which simplifies to $0 =
-1$.
Because $0$ can never equal $-1$, the system is Inconsistent and has No
Solution.
Topic: Introduction to Matrices (Click to Expand)
In mathematics, a matrix (plural: matrices) is a rectangular grid or "array" of numbers, symbols, or expressions arranged in horizontal rows and vertical columns. Think of it as a powerful way to organize data into a table to perform operations on the entire set at once.
1. Structure of a Matrix
A matrix is typically written inside square brackets $[ \ ]$ or parentheses $( \ )$.
- 📏 Dimensions: The size is called its order, written as $m \times n$ ($m$ rows by $n$ columns).
- 🔢 Elements: Each individual number inside the matrix is called an element or entry.
Example of a $2 \times 3$ Matrix:
$$A = \begin{bmatrix} 1 & 5 & 2 \\ 0 & -4 & 7 \end{bmatrix}$$(2 Horizontal Rows | 3 Vertical Columns)
2. Common Types of Matrices
Matrices come in various shapes depending on their entries and dimensions:
3. Why are Matrices important?
Matrices are a fundamental tool in modern technology and science:
Topic: Working with Matrices (Operations) (Click to Expand)
The most important thing to remember is the Address System. Every element has a specific location denoted as $a_{ij}$, where $i$ is the row and $j$ is the column.
1. Matrix Addition and Subtraction
To add or subtract matrices, they must have the same dimensions. You simply add or subtract the numbers that are in the same position.
2. Scalar Multiplication
This is when you multiply a single number (a scalar) by the entire matrix. You multiply every element inside by that number.
3. Matrix Multiplication (The "Dot Product")
This is more complex. To multiply two matrices, the number of columns in the first must equal the number of rows in the second. You follow a "Row by Column" rule.
- Pick the first row of Matrix A.
- Pick the first column of Matrix B.
- Multiply the corresponding parts and add them together to get the first element of the new matrix.
4. Key Matrix "Sheet Codes" (Formulas)
Summary Table for Matrix Operations
| Operation | Requirement | Method |
|---|---|---|
| Addition | Same dimensions | Element-by-element |
| Subtraction | Same dimensions | Element-by-element |
| Multiplication | Col of A = Row of B | Dot product (Row $\times$ Col) |
| Transpose | None | Flip rows and columns |
Topic: Matrix Multiplication & Casio fx-991ES Guide (Click to Expand)
Since you have a Casio fx-991ES, you have a built-in Matrix Mode that can handle up to $3 \times 3$ matrices automatically. This is a lifesaver for checking your work during exams!
1. How to use Matrix Mode on Casio fx-991ES
- Enter Matrix Mode: Press
MODE$\rightarrow$6 (MATRIX). - Define Matrix A: Press
1 (MatA)$\rightarrow$ Choose dimensions $\rightarrow$ Enter numbers, pressing=after each. - Define Matrix B: Press
SHIFT$\rightarrow$4 (MATRIX)$\rightarrow$2 (Data)$\rightarrow$2 (MatB)$\rightarrow$ Choose dimensions $\rightarrow$ Enter numbers. - Calculate: Press
AC$\rightarrow$SHIFT$\rightarrow$4 (MATRIX)$\rightarrow$3 (MatA)$\rightarrow$×$\rightarrow$SHIFT$\rightarrow$4 (MATRIX)$\rightarrow$4 (MatB)$\rightarrow$=.
2. Manual Step-by-Step Multiplication
If you have to show your work, remember the rule: Run along the Row, Dive down the Column.
Let's multiply $A$ and $B$: $$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$$
$(1 \times 5) + (2 \times 7) = \mathbf{19}$
$(1 \times 6) + (2 \times 8) = \mathbf{22}$
$(3 \times 5) + (4 \times 7) = \mathbf{43}$
$(3 \times 6) + (4 \times 8) = \mathbf{50}$
3. Important Matrix "Sheet Codes"
⌨️ Casio:
SHIFT $\rightarrow$ 4 $\rightarrow$
7 (det).📝 Formula for $2 \times 2$: $(ad - bc)$.
⌨️ Casio: Call up
MatA and press the $x^{-1}$ button.📝 Formula: $\frac{1}{\text{det}(A)} \times \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Topic: The "Stack of Equations" (Linear Systems) (Click to Expand)
Thinking of a matrix as a "stack of equations" is the most powerful way to understand why matrices were invented. It is the bridge between basic algebra and advanced computing.
1. From Sentences to Algebra
Imagine you are solving a simple puzzle: "Two apples and three bananas cost $8$ dollars; one apple and five bananas cost $11$ dollars."
The $x$, $y$, and plus signs are placeholders. The Coefficient Matrix above holds the actual information.
2. The Matrix Equation: $Ax = B$
In linear algebra, we represent that entire "stack" with one elegant formula:
$$Ax = B$$
- 📘 $A$ (The Matrix): The coefficients (The "rules" of the system).
- 📗 $x$ (Variable Vector): The unknowns ($x$ and $y$).
- 📙 $B$ (Constant Vector): The answers ($8$ and $11$).
3. Why is this "stacking" useful?
4. Visualizing the Stack as Space
Each row in the matrix "stack" is actually a line (or a plane in 3D). The solution is the intersection point.
Summary Table
| Algebraic View | Matrix View |
|---|---|
| Equation 1 & 2 | Row 1 & 2 of the Matrix |
| Variables ($x, y$) | Column Vector $x$ |
| Substitution/Elimination | Finding the Inverse $A^{-1}$ |
Topic: The Strategy of Row Echelon Form (REF) (Click to Expand)
Creating as many zeros as possible in Row Echelon Form (REF) or Reduced Row Echelon Form (RREF) is a strategic process designed to isolate variables so the system of equations can be solved effortlessly.
1. The "Isolation" Strategy
By creating zeros in the lower-left corner, you are effectively removing variables from the bottom equations so they can be solved individually.
2. Efficiency and Diagnosis
The zeros create a "staircase" structure that allows for Back-Substitution. They also act as a diagnostic tool for the system:
3. Computational "Cost"
Zeros are "cheap" for computers. In massive data sets, like Google’s PageRank, matrices are often "sparse" (mostly zeros). This is the only way calculations can be performed in reasonable time.
4. Transition to Reduced Row Echelon Form (RREF)
In RREF, zeros eliminate every variable except one in each row, providing the final answer directly.
[Image showing the step-by-step transformation from a system of equations to an augmented matrix to RREF]Example of RREF Result:
$$\begin{bmatrix} 1 & 0 & 0 & | & 5 \\ 0 & 1 & 0 & | & -2 \\ 0 & 0 & 1 & | & 3 \end{bmatrix}$$Translated Solution: $x = 5, y = -2, z = 3$
Summary: The Role of Zeros
| Feature | Role of the Zeros |
|---|---|
| Elimination | They "hide" variables so you can focus on one at a time. |
| Structure | They create the "echelon" (staircase) for back-substitution. |
| Diagnosis | They reveal if a system has no solution or infinite solutions. |
| Final Answer | In RREF, they remove all noise, leaving only the solution. |
Topic: Why Row Operations? (The Smoothie & Noise Analogy) (Click to Expand)
It feels like we are just doing "math for the sake of math," but those zeros are actually a way of filtering out noise to reveal the truth. Think of creating zeros as a process of Elimination.
1. Real-World Case: The "Smoothie Recipe" Mystery
Imagine you have three mystery smoothies. You know the ingredients and the total cost, but not the price per gram of each.
Smoothie 2: 2g Blue + 8g Spin + 1g Prot = $3.50
Smoothie 3: 5g Blue + 2g Spin + 10g Prot = $8.00
The Problem: You can't see the price of Spinach because it's "hidden" behind the others. Creating Zeros is like filtering the recipes:
- First Zero: Subtract Smoothie 1 from Smoothie 2 to remove all Blueberries.
- Second Zero: Do the same for Smoothie 3 until the bottom row has zero Blueberries and zero Spinach.
2. Technical Case: Noise Cancellation
Think about Noise-Canceling Headphones. Your mic picks up a "messy" equation: $Result = Music + Noise$.
The headphones create a "negative" version of the noise. In matrix terms, they perform a Row Operation to create a zero in the Noise column. When the Noise becomes zero, all that is left is the Music.
3. The "Why" in Two Points
Summary: The Meaning of the Zero
| In Math | In the Real World |
|---|---|
| Coefficient is 5 | This factor is actively influencing the total. |
| Coefficient is 0 | This factor has been eliminated or filtered out. |
| Echelon Form | A clear "to-do list" where we solve for one thing at a time. |
Topic: Step-by-Step Row Operations (Filtering the Mystery) (Click to Expand)
Let's take those "messy" smoothie recipes and use Row Operations to create zeros. This will "filter out" the ingredients until the prices are revealed.
1. The Setup: Two Messy Recipes
In matrix form, we write the "stack" as an Augmented Matrix:
$$\begin{bmatrix} 1 & 2 & | & 5 \\ 3 & 2 & | & 7 \end{bmatrix}$$2. The "Zero-Maker" Strategy
The Goal: Find the price of Spinach by zeroing out the Blueberries in the second recipe.
Operation: $R_2 \to R_2 - 3R_1$
- 🫐 Blueberries: $3 - (3 \times 1) = \mathbf{0}$ (Filtered!)
- 🍃 Spinach: $2 - (3 \times 2) = \mathbf{-4}$
- 💰 Price: $7 - (3 \times 5) = \mathbf{-8}$
3. The "Filtered" Result
Our matrix now reveals a simplified truth in Row 2:
$$\begin{bmatrix} 1 & 2 & | & 5 \\ 0 & -4 & | & -8 \end{bmatrix}$$Dividing both sides, we find: Spinach = $2/g$.
4. Finalizing: Reduced Row Echelon Form (RREF)
By substituting the Spinach price back into Recipe 1, we find Blueberries = $\$1$. If we reached the final "Identity" form, the matrix would look like this:
Final Prices: Blueberry = $\$1$ | Spinach = $\$2$
Why did we make the zero?
Without that zero, the ingredients were "blocking" our view of the individual prices. The zero isolated the variable, turning a complex mixture into a clear answer.
Topic: Building Your Own "Keys" (Custom Row Operations) (Click to Expand)
Row operations are like custom-made keys. Every lock (matrix) is different, so you have to build a new key (equation) to fit the specific numbers you are trying to "zero out."
1. The Universal "Key" Formula
You always build your equation based on the Target (number to be zeroed) and the Pivot (the tool you use to attack it).
The Zero-Maker Template:
$$R_{\text{target}} \to R_{\text{target}} - \left( \frac{\text{Target Number}}{\text{Pivot Number}} \right) \times R_{\text{pivot}}$$
2. Three Examples of Custom Keys
Target: 7 | Pivot: 1
Key: $R_2 \to R_2 - 7R_1$ (Result: $7 - 7(1) = 0$)
Target: -4 | Pivot: 1
Key: $R_2 \to R_2 + 4R_1$ (Result: $-4 + 4(1) = 0$)
Target: 6 | Pivot: 2
Key: $R_2 \to R_2 - 3R_1$ (Result: $6 - 3(2) = 0$)
3. The "Golden Rules" of Row Operations
[Image illustrating the three allowed row operations: swapping rows, multiplying a row by a constant, and adding/subtracting rows]- Rule 1: Don't change the Pivot row while you are using it.
- Rule 2: Use only Addition or Subtraction between rows.
- Rule 3: Multiply by a non-zero constant if needed to match the target.
Summary Table
| Target Number | Pivot Number | Use this Equation |
|---|---|---|
| $5$ | $1$ | $R_2 \to R_2 - 5R_1$ |
| $-2$ | $1$ | $R_2 \to R_2 + 2R_1$ |
| $10$ | $2$ | $R_2 \to R_2 - 5R_1$ |
| $3$ | $3$ | $R_2 \to R_2 - R_1$ |
Topic: Matrix Rank (The "Information Density" Metric) (Click to Expand)
In simple terms, the Rank of a matrix is the number of non-zero rows remaining after you have converted it into Echelon Form. It represents the number of "unique pieces of information" or "independent equations" in your stack.
1. Finding Rank in Echelon Form
Count the rows that still have at least one non-zero number. If a row turns into all zeros, it was redundant—providing no new data.
Row 1 & 2: Non-zero | Row 3: All zeros $\implies$ Rank = 2
2. The "Health" of a System
3. Real-World Case: GPS Positioning
To find your 3D location ($x, y, z$), your phone connects to satellites. Each signal is an equation.
[Image showing four satellites providing signals to a single point on Earth representing full rank vs two satellites representing rank deficiency]- Case A: High Rank (Open Field): 4 satellites provide unique equations. Rank = 3. You get a precise blue dot.
- Case B: Low Rank (Urban Canyon): Signals bounce off buildings. Two equations become identical. Rank = 2. Your GPS "jumps" because it can't pin down altitude.
4. Real-World Case: Digital Image Compression (JPEG)
An uncompressed photo is a massive matrix. By using Singular Value Decomposition (SVD), we find the "Rank" of the image and "throw away" low-importance rows (redundant sky pixels, etc.).
Extra Insight: Lowering the rank reduces file size while keeping the visual structure intact. This is how a 10MB photo becomes a 500KB JPEG.
Summary: The Meaning of Rank
| Rank Status | Mathematical Meaning | Real-World Result |
|---|---|---|
| Full Rank | Every equation is unique. | Clear, precise solution. |
| Lower Rank | Equations are "repeats." | Infinite solutions or "blurry" data. |
| Rank = 0 | All entries are zero. | No information / Black image. |
Topic: The Standard Form ($AX = b$) (Click to Expand)
In linear algebra, $AX = b$ is the compact way to write a system of linear equations. It is the "sentence structure" that allows us to treat a massive stack of data as a single mathematical object.
1. Breaking Down the Formula
The Rules (Coefficients)
The Mystery (Variables)
The Result (Constants)
2. Packing Equations into Matrices
Instead of writing variables repeatedly, we "pack" the information into a single operation:
Multiplying $A \times X$ uses the "Row by Column" rule to recreate the original equations.
3. The Geometric & Algebraic Power
4. Real-World Applications
| Field | What $A$ represents | What $b$ represents |
|---|---|---|
| Circuits | Resistance of wires | Voltage of the battery |
| Graphics | 3D Rotation/Scaling | Final pixel position |
| Economics | Cost of resources | Total budget available |
Summary Table
| Component | Algebraic Equivalent | Role in the "Story" |
|---|---|---|
| $A$ | Coefficients (numbers) | The Map or the Rules |
| $X$ | Letters ($x, y$) | The Mystery to be solved |
| $b$ | Answers ($8, 10$) | The Target or Result |
Topic: The Augmented Matrix ($[A | b]$) (Click to Expand)
An Augmented Matrix is a simplified way of writing a system of linear equations by combining the Coefficient Matrix ($A$) and the Constant Vector ($b$) into a single table. It is "augmented" because we are literally "attaching" the answers to the right side of the coefficients.
1. The Anatomy of an Augmented Matrix
If you have the equation $AX = b$, the augmented matrix is written as $[A | b]$. We strip away the variables ($x, y$) and the plus signs to focus purely on the data.
$1x - 5y = -2$
- 🟦 The Left Side: Contains the coefficients of the variables.
- 📏 The Vertical Bar: Represents the "equals" ($=$) signs.
- 🟧 The Right Side: Contains the constants (the answers).
2. Why do we use it?
Efficiency is the goal. When you perform a Row Operation, you must apply it to the entire equation. The augmented matrix ensures you don't forget to change the answer side.
3. Interpreting the "Final Story"
After performing Row Operations, the final rows tell you the status of your solution:
| Final Row Pattern | Algebraic Meaning | Type of Solution |
|---|---|---|
| $[ 1 \quad 0 \quad | \quad 5 ]$ | $1x + 0y = 5$ | Unique ($x=5$) |
| $[ 0 \quad 0 \quad | \quad 0 ]$ | $0 = 0$ (True) | Infinite Solutions |
| $[ 0 \quad 0 \quad | \quad 7 ]$ | $0 = 7$ (False) | No Solution |
4. Real-World Analogy & Calculator Tips
[2 1 | 15]. This shorthand allows the manager to
quickly compare multiple rows to solve for individual item prices.
(3 rows for equations, 4 columns to include the "Augmented" answer column).
Topic: Free Variables & Infinite Solutions (Click to Expand)
When you solve a system and end up with fewer "unique equations" (Rank) than variables, you cannot find a single numeric answer. Instead, you get a General Solution expressed in terms of Free Variables.
1. Pivot vs. Free Variables
In Echelon Form, every column corresponds to a variable. The presence of a Pivot (leading 1) determines if that variable is "locked" or "free."
2. Building the General Solution
Imagine a $3$-variable system with only one unique row in RREF:
$$\left[ \begin{array}{ccc|c} 1 & 2 & 5 & 10 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$$- Identify: $x$ has a pivot. $y$ and $z$ are Free.
- Assign: Let $y = s$ and $z = t$.
- Solve: $1x + 2s + 5t = 10 \implies \mathbf{x = 10 - 2s - 5t}$.
General Solution Vector: $$X = \begin{bmatrix} 10 - 2s - 5t \\ s \\ t \end{bmatrix}$$
3. Real-World Case: The Under-Specified Project
Imagine a client says: "The total of Nails ($x$), Screws ($y$), and Bolts ($z$) must be 100." This is one equation for three variables.
4. Visualizing the Solution Geometry
The number of Free Variables determines the "shape" of the solution in space:
[Image showing geometric representations of solutions: a point for zero free variables, a line for one free variable, and a plane for two free variables]- 0 Free Variables: A single Point (Unique Solution).
- 1 Free Variable: A Line of infinite solutions.
- 2 Free Variables: A Plane of infinite solutions.
Extra Concept - Nullity: The number of free variables is formally
called the Nullity of the matrix.
Rank + Nullity = Total Columns.
Summary Table
| Component | Definition | Role in Solution |
|---|---|---|
| Rank | Number of pivot rows. | Tells you how many variables are "locked." |
| Free Variables | Total Variables - Rank. | Tells you the "dimensions" (Line, Plane). |
| Parameters ($s, t$) | Placeholders for free variables. | Formula that generates all possible answers. |
Topic: Expanding to Vector Form (Parametric Equations) (Click to Expand)
In linear algebra, expanding a general solution into Vector Form is the process of separating constant parts from the variables ($s, t, u$). This reveals how each free variable independently changes the solution.
1. The Vector "Stack"
After solving your row equation ($x_1 + 3x_2 - x_3 + 4x_4 = 1$), you get a "stacked" vertical vector $X$ based on your parameters:
2. The "Deconstruction" Process
We split the big vector into four smaller vectors: one for Constants and one for each Free Variable ($s, t, u$).
Look for numbers without a variable attached.
Result: $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$
Extract coefficients of $s$.
Result: $s \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}$
Extract coefficients of $t$.
Result: $t \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$
Extract coefficients of $u$.
Result: $u \begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \end{bmatrix}$
3. The Final Vector Form
Adding them back together gives the complete general solution. This is the "blueprint" for every possible answer to your system:
4. Why This Works (The Geometry)
| Component | Geometric Meaning | Role |
|---|---|---|
| Constant Vector | A specific point. | The "anchor" of the solution. |
| Variable Vectors | Directions/Spans. | Creates the "infinite" nature of the solution. |
Topic: The Extraction Logic (Sorting the "Folders") (Click to Expand)
Extracting the general solution is like taking a single "combined" list and sorting it into separate folders based on their labels ($s$, $t$, or $u$).
1. Visualizing the "Hidden" Variables
To make extraction easier, we imagine every row contains every label ($s, t, u$), even if their value is zero. This aligns the columns for extraction.
2. Sorting into Vector Folders
We "pull" the data out column by column. Each column becomes its own independent vector.
Pull only plain numbers.
$\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$
Factor out $s$ coefficients.
$s \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}$
Factor out $t$ coefficients.
$t \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$
Factor out $u$ coefficients.
$u \begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \end{bmatrix}$
3. The Final "Assembly"
The solution is the sum of these parts. In the "story" of linear algebra, this is called a Linear Combination.
4. Why do we do this?
This format identifies the "Particular Solution" (the starting point) and the "Basis" for the Null Space (the directions you can slide).
| Action | Mathematical Result | Physical Analogy |
|---|---|---|
| Extract Constants | Particular Solution | Your starting "Anchor" point. |
| Extract $s, t, u$ | Basis Vectors | The "Steering" directions. |
Topic: Vector Spaces, Rank, and Subspaces (Detailed Solution)
This problem explores how we can measure the "size" of a matrix's information (Rank), how to find the fundamental building blocks of its data (Basis), and how to prove if a specific subset of data belongs to that space (Subspace).
(a) Find the rank of the matrix $A$. $[3M]$
(b) Let $W$ be the vector space spanned by the columns of $A$. Find a basis and dimension of $W$. $[2M]$
(c) Show that the set $U = \left\{ r \begin{bmatrix} 1 \\ 1 \\ -2 \\ 1 \end{bmatrix} : r \text{ is a real number} \right\}$ is a subspace of $W$. $[2M]$
Part (a): Finding the Rank of Matrix A
The Concept: The Rank tells us how many "independent" rows or columns a matrix has. We find this by simplifying the matrix into Row Echelon Form.
1. Start with Matrix A: $$A = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 2 & -1 & 0 & 1 \\ 0 & -2 & -2 & -4 \\ 0 & 1 & 1 & 2 \end{bmatrix}$$
2. Row Operations:
- $R_2 \to R_2 - 2R_1 = [0, -1, 0, -1]$
- $R_4 \to R_4 + R_2 = [0, 0, 1, 1]$
- $R_3 \to R_3 - 2R_2 = [0, 0, -2, -2]$
- $R_3 \to R_3 + 2R_4 = [0, 0, 0, 0]$
3. Resulting Echelon Form: $$\begin{bmatrix} \mathbf{1} & 0 & 0 & 1 \\ 0 & \mathbf{-1} & 0 & -1 \\ 0 & 0 & \mathbf{1} & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
Conclusion: There are 3 non-zero rows. $\implies$ Rank(A) = 3.
Part (b): Basis and Dimension of W
The Concept: $W$ is the "Column Space." A Basis is the smallest set of columns needed to "build" the entire space. The Dimension is the count of vectors in that basis.
[Image showing the relationship between pivot columns in echelon form and corresponding columns in the original matrix]Pivots are in Col 1, 2, and 3. We use the original columns of $A$:
Basis for W: $\{ \begin{bmatrix} 1 \\ 2 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ -1 \\ -2 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ -2 \\ 1 \end{bmatrix} \}$
Since there are 3 vectors in our basis:
Dimension(W) = 3
Part (c): Showing U is a Subspace of W
The Concept: We must prove the vector in $U$ lives inside $W$ and that $U$ follows the closed-loop rules of a subspace.
[Image illustrating subspace criteria showing zero vector inclusion and closure under addition and scalar multiplication]1. Is the vector in W?
Observe Column 1 + Column 2 of original matrix $A$:
$$\begin{bmatrix} 1 \\ 2 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ -1 \\ -2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -2 \\ 1 \end{bmatrix}$$
Since it is a linear combination of $A$'s columns, it is in $W$.
2. Is U a subspace?
$U$ is defined as all multiples ($r$) of this vector.
- Zero Vector: If $r=0$, the zero vector is in $U$.
- Closed under Addition: $(r_1\mathbf{v}) + (r_2\mathbf{v}) = (r_1 + r_2)\mathbf{v} \in U$.
- Closed under Scaling: $c(r\mathbf{v}) = (cr)\mathbf{v} \in U$.
Conclusion: $U$ is a 1-dimensional subspace of the 3-dimensional space $W$.
Summary Table: Key Findings
| Component | Value | What it tells us |
|---|---|---|
| Rank of A | 3 | A has 3 independent rows/columns of info. |
| Dimension of W | 3 | The column space is a 3D volume in 4D space. |
| Set U | Subspace | $U$ is a "line" passing through the "3D room" $W$. |
Topic: Column Operations and Consistency (Click to Expand Solution)
To help you prepare for your exam, here is a detailed, step-by-step breakdown of the problem regarding interchanging columns and its effect on system consistency.
1. Understanding the Terminology
- 🧪 System of Equations ($AX = b$): Can we combine the columns of matrix $A$ in some specific way to reach the result $b$?
- ✅ Consistency: The system is "consistent" if there is at least one solution (a way to reach $b$).
- 🧱 Columns ($A_i$): Think of these as different "paths" or "building blocks" available to you.
- 🔄 Interchanging: Swapping the order of the building blocks.
2. The Logical Approach
Imagine you have a box of Lego bricks. You want to build a tower of a specific height ($b$).
3. Formal Proof
We look at the Span (the collection of all possible results you can get by combining the columns).
Step 1: Define the Span of A. The consistency of $AX = b$ depends entirely on whether $b$ lies in the span of the columns $\{A_1, A_2, \dots, A_n\}$.
Step 2: Define the Span of C. Matrix $C$ consists of the same set of columns, just in a different order: $\{A_1, \dots, A_j, \dots, A_i, \dots, A_n\}$.
Step 3: Compare the sets. Since the set of vectors in $A$ and $C$ are identical (order doesn't matter in a set), their span is identical: $$\text{Span}(A_1, \dots, A_n) = \text{Span}(C_1, \dots, C_n)$$
Step 4: Conclusion. If $b$ is in the span of $A$, it is automatically in the span of $C$. If it's not in $A$, it's not in $C$.
Consistency remains unchanged because interchanging columns does not change the Column Space (the set of all reachable vectors).
Consistency Property Summary
| Operation | Effect on Column Space | Effect on Consistency |
|---|---|---|
| Interchange Columns | None (Set remains same) | Unchanged |
| Scale a Column | None (Span remains same) | Unchanged |
Topic: Linear Systems and Reduced Row Echelon Form (RREF) (Click to Expand)
$x = \begin{bmatrix} 0 \\ 4 \\ 0 \end{bmatrix} + c_1 \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 5 \\ 1 \end{bmatrix}$
(a) Find $R$ and $k$.
(b) Find, if possible, the pivotal and non pivotal columns of $R$.
1. Understanding the Terminology
To solve this problem, we need to understand what the pieces of the "general solution" represent:
- Variable Vector ($x$): Since the vectors in the solution have three rows, our system has three variables: $x_1$, $x_2$, and $x_3$.
- Particular Solution ($x_p$): This is the constant vector $\begin{bmatrix} 0 \\ 4 \\ 0 \end{bmatrix}$. It is a specific set of values that makes the equation $Rx = k$ true.
- Free Variables ($c_1, c_2$): These indicate that the system has infinitely many solutions because some variables can be anything.
- RREF ($R$): The simplest version of the original matrix. Pivot columns have a leading $1$ and zeros elsewhere.
2. Breaking Down the General Solution
We can write out an equation for each individual variable ($x_1, x_2, x_3$) from the provided solution:
Row 2: $\mathbf{x_2 = 4 + 2c_1 + 5c_2}$
Row 3: $x_3 = 0 + 0c_1 + 1c_2 \implies \mathbf{x_3 = c_2}$
(a) Find $R$ and $k$
In a system reduced to RREF, we express the pivot variables in terms of the free variables.
From our breakdown, $x_1$ and $x_3$ are equal to the parameters $c_1$ and $c_2$. This means $x_1$ and $x_3$ are the free variables. $x_2$ is the pivot variable because its value depends on the others.
Rewrite the equation for $x_2$ so that all variables are on the left and the constant is on the right:
$x_2 = 4 + 2x_1 + 5x_3$
$-2x_1 + 1x_2 - 5x_3 = 4$
$R = \begin{bmatrix} -2 & 1 & -5 \end{bmatrix}$
$k = \begin{bmatrix} 4 \end{bmatrix}$
(b) Pivotal and Non-pivotal Columns
Column 2 (the $x_2$ column). It contains the leading $1$ used to solve the system.
Column 1 (for $x_1$) and Column 3 (for $x_3$). These correspond to the free variables.
Topic: Linear Algebra - Pivots and Gaussian Elimination (Click to Expand Solution)
In the world of mathematics, specifically linear algebra, we often solve systems of equations by arranging them into a matrix. A "pivot" is the first non-zero number in a row of a matrix as we simplify it. For a system of three equations with three variables ($x, y, z$) to have a unique, clean solution, we generally need three pivots (one for each variable). If a pivot becomes zero during our simplification process, the system "fails," meaning it might have no solution or infinitely many solutions.
You are given a system of three linear equations:
$$\alpha x + 2y + 6z = b_1$$ $$\alpha x + \alpha y + 5z = b_2$$ $$\alpha x + \alpha y + \alpha z = b_3$$ Derive the three possible values of $\alpha$ for which the given linear system will fail to have three pivots, giving justification(s).
Detailed Step-by-Step Solution
To find where the pivots fail, we will use Gaussian Elimination. This is just a fancy way of saying we will subtract equations from each other to "eliminate" variables and turn the system into a triangular shape (Row Echelon Form).
Step 1: Write the system as a matrix
We represent the coefficients of $x, y,$ and $z$ in a grid:
$$A = \begin{bmatrix} \alpha & 2 & 6 \\ \alpha & \alpha & 5 \\ \alpha & \alpha & \alpha \end{bmatrix}$$Step 2: Identify the First Pivot
The first pivot is the top-left entry.
Pivot 1: $\alpha$
Condition for failure: If the first pivot is zero, the system immediately fails to have three distinct pivots.
Value 1: $\alpha = 0$
Step 3: Eliminate the $x$ terms in Rows 2 and 3
Assuming $\alpha \neq 0$, we subtract Row 1 ($R_1$) from Row 2 ($R_2$) and Row 3 ($R_3$) to create zeros under the first pivot.
$R_2 \to R_2 - R_1$: $(\alpha - \alpha)x + (\alpha - 2)y + (5 - 6)z = b_2 - b_1$
$R_3 \to R_3 - R_1$: $(\alpha - \alpha)x + (\alpha - 2)y + (\alpha - 6)z = b_3 - b_1$
Our new matrix looks like this:
$$\begin{bmatrix} \alpha & 2 & 6 \\ 0 & \alpha - 2 & -1 \\ 0 & \alpha - 2 & \alpha - 6 \end{bmatrix}$$Step 4: Identify the Second Pivot
The second pivot is the second entry in the second row.
Pivot 2: $\alpha - 2$
Condition for failure: If this value is zero, we cannot proceed normally.
Value 2: $\alpha - 2 = 0 \implies \alpha = 2$
Step 5: Eliminate the $y$ term in Row 3
Assuming $\alpha \neq 2$, we subtract our new Row 2 from Row 3 to get a zero in the second column of the third row.
$R_3 \to R_3 - R_2$: $(0-0)x + ((\alpha - 2) - (\alpha - 2))y + ((\alpha - 6) - (-1))z = (b_3 - b_1) - (b_2 - b_1)$
Simplify the $z$ coefficient: $\alpha - 6 + 1 = \alpha - 5$
Our final matrix (in upper triangular form) is:
$$\begin{bmatrix} \alpha & 2 & 6 \\ 0 & \alpha - 2 & -1 \\ 0 & 0 & \alpha - 5 \end{bmatrix}$$Step 6: Identify the Third Pivot
The third pivot is the last entry in the diagonal.
Pivot 3: $\alpha - 5$
Condition for failure: If this value is zero, we lose our third pivot.
Value 3: $\alpha - 5 = 0 \implies \alpha = 5$
Final Justification
The system fails to have three pivots if any of the diagonal elements in its Row Echelon Form become zero. Based on our derivation:
- $\alpha = 0$: The first pivot is zero.
- $\alpha = 2$: The second pivot becomes zero after the first stage of elimination.
- $\alpha = 5$: The third pivot becomes zero after the final stage of elimination.
The three values are: $\alpha = 0, 2, 5$.
Topic: Linear Algebra - Invertibility and the Vandermonde Matrix (Click to Expand Solution)
In linear algebra, when we have a system of equations written as $\mathbf{Ax} = \mathbf{b}$, we are basically asking: "Is there a combination of the columns of matrix $\mathbf{A}$ that can produce the result $\mathbf{b}$?"
If a matrix is invertible (or "non-singular"), it means the matrix covers all possible directions in space. In such cases, no matter what result $\mathbf{b}$ you want, there is always exactly one solution for $\mathbf{x}$. We call this a consistent system.
If a data analysis led to a real matrix $$\mathbf{A} = \begin{pmatrix} 1 & a_1 & a_1^2 \\ 1 & a_2 & a_2^2 \\ 1 & a_3 & a_3^2 \end{pmatrix}$$ with distinct $a_1, a_2, a_3$, find all possible values of $\mathbf{b} \in \mathbb{R}^3$ such that the system $\mathbf{Ax} = \mathbf{b}$ is consistent where $\mathbf{x} \in \mathbb{R}^3$.
Detailed Step-by-Step Solution
To understand when this system is consistent, we need to look at the "strength" of matrix $\mathbf{A}$.
1. Recognizing the Matrix Type
The matrix $\mathbf{A}$ shown here has a very specific structure. Each row follows a geometric progression ($1, a, a^2$). This is known in mathematics as a Vandermonde Matrix.
2. Checking for Invertibility (The Determinant)
A system $\mathbf{Ax} = \mathbf{b}$ with a square matrix (like our $3 \times 3$ matrix) is consistent for all values of $\mathbf{b}$ if the matrix is invertible. A matrix is invertible if its determinant is not zero.
For a $3 \times 3$ Vandermonde matrix, there is a famous formula for the determinant:
$$\det(\mathbf{A}) = (a_2 - a_1)(a_3 - a_1)(a_3 - a_2)$$3. Applying the "Distinct" Condition
The problem states that $a_1, a_2,$ and $a_3$ are distinct. This is the key piece of information.
If they are distinct, then $a_2 - a_1 \neq 0$, $a_3 - a_1 \neq 0$, and $a_3 - a_2 \neq 0$. Since none of these factors are zero, their product (the determinant) cannot be zero.
$$\text{Because } a_1 \neq a_2 \neq a_3, \text{ then } \det(\mathbf{A}) \neq 0$$
4. What does a non-zero determinant mean?
When the determinant of a square matrix is not zero:
- The matrix is invertible.
- The columns of the matrix are linearly independent (they don't overlap or lie on the same line/plane).
- The matrix spans the entire 3D space ($\mathbb{R}^3$).
5. Conclusion for $\mathbf{b}$
Because the matrix $\mathbf{A}$ is invertible, the system $\mathbf{Ax} = \mathbf{b}$ has a unique solution for every single possible vector $\mathbf{b}$ in 3D space. There are no "forbidden" values for $\mathbf{b}$.
The system is consistent for all vectors $\mathbf{b} \in \mathbb{R}^3$.
Justification: The matrix $\mathbf{A}$ is a Vandermonde matrix with distinct parameters $a_1, a_2, a_3$. Its determinant is non-zero, making the matrix non-singular (invertible). An invertible matrix $\mathbf{A}$ ensures that for any $\mathbf{b}$, there exists a unique solution $\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$.
Topic: Convexity of Solution Sets to Linear Equations (Click to Expand Solution)
This problem asks us to determine if the set of all possible solutions to a linear system is convex. To solve this, let's break down the terminology first.
Consider the set $C = \{x \mid Ax = b\}$ where $A \in \mathbb{R}^{n \times n}$ is a square matrix and $b \in \mathbb{R}^n$. $C$ represents all possible solutions of a linear system $Ax = b$. Prove or disprove that $C$ is a convex set.
Understanding the Terminology
If you are seeing these terms for the first time, here is what they mean:
- Linear System ($Ax = b$): This is a compact way of writing a group of equations. $A$ is a grid of numbers, $x$ is the list of unknowns we want to find, and $b$ is the result.
- Convex Set: A set is "convex" if, for any two points inside the set, the straight line connecting them is also entirely inside the set.
Analogy: Think of a circle or a square; if you pick two spots inside and draw a line between them, that line stays inside. Now think of a "C" shape; you could pick two points (one at the top tip, one at the bottom) where the connecting line goes through the empty space in the middle. The "C" shape is not convex.
The Proof
To prove that $C$ is convex, we must show that if we take any two solutions, $x_1$ and $x_2$, then any point on the line segment between them is also a solution.
1. Identify our starting points:
Assume $x_1$ and $x_2$ are both in set $C$. By the definition of the set, this means:
$$Ax_1 = b$$ $$Ax_2 = b$$
2. Define a point on the line segment:
A point $x_{\theta}$ on the line segment between $x_1$ and $x_2$ can be written as a "weighted average" using a variable $\theta$:
$$x_{\theta} = \theta x_1 + (1 - \theta)x_2$$
where $0 \le \theta \le 1$.
3. Test if $x_{\theta}$ is also in the set $C$:
To be in $C$, $x_{\theta}$ must satisfy the original equation $Ax = b$. Let's plug it in:
$$A(\theta x_1 + (1 - \theta)x_2)$$
Using the distributive property of matrices ($A(B+C) = AB + AC$):
$$A(\theta x_1) + A((1 - \theta)x_2)$$
Since $\theta$ is just a number (a scalar), we can move it to the front:
$$\theta(Ax_1) + (1 - \theta)(Ax_2)$$
4. Substitute our known values:
From step 1, we know $Ax_1 = b$ and $Ax_2 = b$. Let's substitute those in:
$$\theta(b) + (1 - \theta)(b)$$
5. Simplify:
$$\theta b + b - \theta b$$
$$b$$
Since $A x_{\theta} = b$, the point $x_{\theta}$ is also in set $C$. Because this holds true for any $x_1, x_2 \in C$ and any $\theta \in [0, 1]$, the set $C$ is a convex set.
Topic: Vector Bases and Linear Combinations (Click to Expand Solution)
The Question
(a) Is the set $S = \left\{ \begin{bmatrix} 0 \\ 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 0 \\ 3 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 4 \end{bmatrix} \right\}$ a basis for $\mathbb{R}^4$?(b) Write the vector $\mathbf{v} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ as a linear combination of the elements in $S$.
Part (a): Determining if it is a Basis
What is a Basis?
Imagine you are playing a video game where you can move in 4 independent directions (up/down, left/right, forward/backward, and through time). To be able to reach any point in that 4D space ($\mathbb{R}^4$), you need exactly 4 "tools" (vectors) that are:
- Linearly Independent: No vector can be made by combining the others. (None of them are "redundant").
- Spanning: Together, they can reach every possible point in the space.
The Test:
The easiest way to check if 4 vectors form a basis for $\mathbb{R}^4$ is to put them into a square matrix and calculate the Determinant. If the determinant is not zero, the vectors are independent and form a basis.
Let's build the matrix $A$ using these vectors as columns:
$$A = \begin{bmatrix} 0 & 1 & 2 & 0 \\ 1 & 0 & 1 & 0 \\ 2 & 0 & 0 & 0 \\ 0 & 3 & 0 & 4 \end{bmatrix}$$Calculating the Determinant $|A|$:
To make it easy, we expand along the 3rd row because it has the most zeros:
$$|A| = 2 \cdot (-1)^{3+1} \cdot \det \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 4 \end{bmatrix}$$Expanding that $3 \times 3$ matrix along its 2nd row:
$$|A| = 2 \cdot (1) \cdot \left[ 1 \cdot (-1)^{2+2} \cdot \det \begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix} \right]$$ $$|A| = 2 \cdot 1 \cdot (4 - 0) = 8$$Conclusion for (a):
Since the determinant is 8 (which is not 0), the vectors are linearly independent. Because there are exactly 4 independent vectors in a 4D space, yes, the set is a basis for $\mathbb{R}^4$.
Part (b): Writing as a Linear Combination
What is a Linear Combination?
We want to find "weights" (numbers) $c_1, c_2, c_3, c_4$ such that:
$$c_1 \begin{bmatrix} 0 \\ 1 \\ 2 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 3 \end{bmatrix} + c_3 \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + c_4 \begin{bmatrix} 0 \\ 0 \\ 0 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$This gives us a system of 4 equations based on the rows:
(2) $1c_1 + 0c_2 + 1c_3 + 0c_4 = 0 \implies \mathbf{c_1 + c_3 = 0}$
(3) $2c_1 + 0c_2 + 0c_3 + 0c_4 = 0 \implies \mathbf{2c_1 = 0}$
(4) $0c_1 + 3c_2 + 0c_3 + 4c_4 = 0 \implies \mathbf{3c_2 + 4c_4 = 0}$
Solving the system:
- From (3): $2c_1 = 0 \implies \mathbf{c_1 = 0}$
- Substitute $c_1=0$ into (2): $0 + c_3 = 0 \implies \mathbf{c_3 = 0}$
- Substitute $c_3=0$ into (1): $c_2 + 2(0) = 1 \implies \mathbf{c_2 = 1}$
- Substitute $c_2=1$ into (4): $3(1) + 4c_4 = 0 \implies 4c_4 = -3 \implies \mathbf{c_4 = -3/4}$
The linear combination is:
$$0 \begin{bmatrix} 0 \\ 1 \\ 2 \\ 0 \end{bmatrix} + 1 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 3 \end{bmatrix} + 0 \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix} - \frac{3}{4} \begin{bmatrix} 0 \\ 0 \\ 0 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
Topic: Linear Algebra - Rank and Subspaces (Click to Expand Solution)
The problem you've shared deals with a fundamental constraint in linear algebra: you cannot create more "independent directions" than the number of building blocks you start with.
The Question
Given two linearly independent $D \times 1$ vectors $\mathbf{v}_1$ and $\mathbf{v}_2$, where $D > 3$, we intend to create a $D \times 3$ matrix where each column of the matrix is a linear combination of the given vectors $\mathbf{v}_1$ and $\mathbf{v}_2$. If possible, find vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ and values for the coefficients in the linear combination such that the resulting matrix has full column rank. Otherwise, explain with a suitable mathematical argument why this is not possible.Detailed Explanation
To understand why this is or isn't possible, let's break down the terminology first.
1. What is a "Linear Combination"?
Think of a linear combination as a recipe. If you have two vectors, $\mathbf{v}_1$ and $\mathbf{v}_2$, a linear combination is any vector created by scaling them and adding them together:
$$\mathbf{c} = a\mathbf{v}_1 + b\mathbf{v}_2$$No matter what numbers you pick for $a$ and $b$, the resulting vector $\mathbf{c}$ will always lie on the same 2D "flat surface" (a plane) defined by $\mathbf{v}_1$ and $\mathbf{v}_2$.
2. What is "Full Column Rank"?
A $D \times 3$ matrix has 3 columns. To have full column rank, all 3 columns must be linearly independent. This means none of the columns can be created by mixing the other two. In simpler terms, you need 3 unique "directions" in space.
The Answer: Why it is NOT possible
It is not possible to create such a matrix. Here is the mathematical justification:
1. The Span Constraint
The matrix we are building, let's call it $M$, has columns $\mathbf{c}_1, \mathbf{c}_2,$ and $\mathbf{c}_3$. The problem states that every one of these columns must be a linear combination of $\mathbf{v}_1$ and $\mathbf{v}_2$. Mathematically, this means:
$$\text{col}(M) \subseteq \text{span}\{\mathbf{v}_1, \mathbf{v}_2\}$$2. Dimension of the Subspace
The "span" of a set of vectors is the space they fill. Since we have only two linearly independent vectors ($\mathbf{v}_1$ and $\mathbf{v}_2$), the dimension of their span is exactly 2.
3. The Rank Inequality
A fundamental rule in linear algebra is that the rank of a matrix (the number of independent columns) cannot exceed the dimension of the subspace that contains its columns.
- The subspace available to us has a dimension of 2.
- We want the matrix to have a rank of 3 (full column rank for a $D \times 3$ matrix).
Since $3 > 2$, it is impossible. You are trying to find 3 independent directions in a space that only has 2 dimensions. It's like trying to find 3 perpendicular lines on a flat sheet of paper; you can only find 2.
Let $M = [\mathbf{c}_1 \ \mathbf{c}_2 \ \mathbf{c}_3]$. Since each $\mathbf{c}_i \in \text{span}\{\mathbf{v}_1, \mathbf{v}_2\}$, the entire column space of $M$ is a subspace of $\text{span}\{\mathbf{v}_1, \mathbf{v}_2\}$. Therefore:
$$\text{rank}(M) = \dim(\text{col}(M)) \leq \dim(\text{span}\{\mathbf{v}_1, \mathbf{v}_2\}) = 2$$
For the matrix to have full column rank, we require $\text{rank}(M) = 3$. Because $2 < 3$, the matrix can have a maximum rank of 2, and thus cannot have full column rank.
Topic: Inner Products and Matrix-Vector Multiplication (Click to Expand Solution)
Before diving into the solution, it's helpful to understand what an inner product is. Think of it as a way to "multiply" two vectors to get a single number. For this number to be officially called an inner product, it must follow specific rules (like fairness and consistency). In this problem, we are using a specific Matrix $A$ to define that relationship.
Question
Given the matrix: $$A = \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{pmatrix}$$ and the definition $\langle \mathbf{x}, \mathbf{y} \rangle_A = \mathbf{x}^T A \mathbf{y}$ for vectors in $\mathbb{R}^4$:- Prove that $\langle \cdot, \cdot \rangle_A$ is an inner product on $\mathbb{R}^4$.
- Find all possible vectors $\mathbf{y} = [y_1, y_2, y_3, y_4]^T$ such that $\langle \mathbf{x}, \mathbf{y} \rangle_A = 0$, given $\mathbf{x} = [1, 0, 0, 0]^T$.
Detailed Answer
Part i) Proving it is an Inner Product
To prove a function is an inner product, it must satisfy four specific properties. Let’s break them down:
1. Symmetry: $\langle \mathbf{x}, \mathbf{y} \rangle_A = \langle \mathbf{y}, \mathbf{x} \rangle_A$
Since $\langle \mathbf{x}, \mathbf{y} \rangle_A = \mathbf{x}^T A \mathbf{y}$ is a scalar (a single number), it is equal to its own transpose.
$$\mathbf{x}^T A \mathbf{y} = (\mathbf{x}^T A \mathbf{y})^T = \mathbf{y}^T A^T \mathbf{x}$$Looking at matrix $A$, we see it is symmetric ($A = A^T$). Therefore, $\mathbf{y}^T A^T \mathbf{x} = \mathbf{y}^T A \mathbf{x}$, which is $\langle \mathbf{y}, \mathbf{x} \rangle_A$. Property satisfied.
2. Linearity: $\langle a\mathbf{x} + b\mathbf{z}, \mathbf{y} \rangle_A = a\langle \mathbf{x}, \mathbf{y} \rangle_A + b\langle \mathbf{z}, \mathbf{y} \rangle_A$
This follows directly from the rules of matrix distribution:
$$(a\mathbf{x} + b\mathbf{z})^T A \mathbf{y} = (a\mathbf{x}^T + b\mathbf{z}^T) A \mathbf{y} = a\mathbf{x}^T A \mathbf{y} + b\mathbf{z}^T A \mathbf{y}$$Property satisfied.
3. Positivity: $\langle \mathbf{x}, \mathbf{x} \rangle_A \geq 0$
4. Definiteness: $\langle \mathbf{x}, \mathbf{x} \rangle_A = 0$ only if $\mathbf{x} = \mathbf{0}$
These two are often checked together by seeing if the matrix $A$ is Positive Definite. We can check this using Sylvester’s Criterion (checking the determinants of the leading principal minors):
- $M_1 = |2| = 2 > 0$
- $M_2 = \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} = 4 - 1 = 3 > 0$
- $M_3 = \begin{vmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{vmatrix} = 2(4-1) - (-1)(-2-0) = 6 - 2 = 4 > 0$
- $M_4 = \det(A) = 5 > 0$ (calculated via expansion)
Since all leading principal minors are positive, $A$ is positive definite. Thus, $\mathbf{x}^T A \mathbf{x} > 0$ for all $\mathbf{x} \neq \mathbf{0}$. Properties satisfied.
Conclusion: Since all properties hold, $\langle \cdot, \cdot \rangle_A$ is a valid inner product.
Part ii) Finding the specific vector $y$
We are given $\mathbf{x} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ and we need to solve for $\mathbf{y}$ in the equation $\langle \mathbf{x}, \mathbf{y} \rangle_A = 0$.
Step 1: Write out the multiplication
$$\langle \mathbf{x}, \mathbf{y} \rangle_A = \mathbf{x}^T (A \mathbf{y})$$First, let's find $A \mathbf{y}$:
$$A\mathbf{y} = \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{pmatrix} = \begin{pmatrix} 2y_1 - y_2 \\ -y_1 + 2y_2 - y_3 \\ -y_2 + 2y_3 - y_4 \\ -y_3 + 2y_4 \end{pmatrix}$$Step 2: Multiply by $\mathbf{x}^T$
$\mathbf{x}^T$ is a row vector $[1, 0, 0, 0]$. When you multiply a row vector like this by a column vector, it "picks out" only the first element:
$$\begin{bmatrix} 1 & 0 & 0 & 0 \end{bmatrix} \begin{pmatrix} 2y_1 - y_2 \\ -y_1 + 2y_2 - y_3 \\ -y_2 + 2y_3 - y_4 \\ -y_3 + 2y_4 \end{pmatrix} = 1(2y_1 - y_2) + 0 + 0 + 0 = 2y_1 - y_2$$Step 3: Set to zero
The problem states $\langle \mathbf{x}, \mathbf{y} \rangle_A = 0$, so:
$$2y_1 - y_2 = 0 \implies y_2 = 2y_1$$All possible vectors $\mathbf{y}$ are of the form where the second component is exactly twice the first, while the third and fourth components can be anything. In set notation:
$$\mathbf{y} = \begin{bmatrix} y_1 \\ 2y_1 \\ y_3 \\ y_4 \end{bmatrix} \text{ for any } y_1, y_3, y_4 \in \mathbb{R}$$
Topic: Inner Products and Basis Representations (Click to Expand Solution)
In linear algebra, the standard inner product (often called the dot product) measures how two vectors relate in terms of length and angle. However, if we change our "viewpoint" (the basis), the way we calculate this product changes.
The representation of an inner product with respect to a basis is a matrix, often called the Gram Matrix.
Question
Represent the standard inner product in $\mathbb{R}^3$ with respect to the basis:$B = \{v_1, v_2, v_3\} = \{(1, 0, 1), (0, 1, 1), (1, 1, 0)\}$
Also, find the inner product between $u = (-2, 1, 3)$ and $w = (-4, 5, 9)$.
Detailed Solution
1. Understanding the Terminology
- Standard Inner Product: For two vectors $x = (x_1, x_2, x_3)$ and $y = (y_1, y_2, y_3)$, the product is $\langle x, y \rangle = x_1y_1 + x_2y_2 + x_3y_3$.
- Basis: A set of vectors that act as a "coordinate system." Any vector in the space can be built using these.
- Matrix Representation ($G$): To find the matrix of the inner product relative to basis $B$, we calculate the inner product of every pair of basis vectors. The element in the $i$-th row and $j$-th column is $G_{ij} = \langle v_i, v_j \rangle$.
2. Constructing the Matrix ($G$)
We calculate the dot products for all combinations of $v_1, v_2,$ and $v_3$:
$\langle v_1, v_2 \rangle = (1)(0) + (0)(1) + (1)(1) = 1$
$\langle v_1, v_3 \rangle = (1)(1) + (0)(1) + (1)(0) = 1$
$\langle v_2, v_1 \rangle = \langle v_1, v_2 \rangle = 1$
$\langle v_2, v_2 \rangle = (0)(0) + (1)(1) + (1)(1) = 2$
$\langle v_2, v_3 \rangle = (0)(1) + (1)(1) + (1)(0) = 1$
$\langle v_3, v_1 \rangle = \langle v_1, v_3 \rangle = 1$
$\langle v_3, v_2 \rangle = \langle v_2, v_3 \rangle = 1$
$\langle v_3, v_3 \rangle = (1)(1) + (1)(1) + (0)(0) = 2$
Putting these into a matrix:
$$G = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$$
This matrix represents the inner product with respect to the given basis.
3. Finding the Inner Product of $u$ and $w$
The question asks for the inner product between $u = (-2, 1, 3)$ and $w = (-4, 5, 9)$. Using the standard dot product formula:
$\langle u, w \rangle = 8 + 5 + 27$
$\langle u, w \rangle = 40$
Alternative Interpretation: If the question meant for you to find the inner product using the basis coordinates and the matrix $G$ we just found, you would first need to convert $u$ and $w$ into the new basis $B$, then calculate $[u]_B^T G [w]_B$. However, since $u$ and $w$ are already given in standard components, calculating the standard dot product directly is the standard approach.
- Matrix Representation: $\begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$
- Inner Product of $u$ and $w$: $40$
Topic: Linear Algebra - Rank Invariance and Idempotent Matrices (Click to Expand Solution)
Topic 1: Rank Invariance under Multiplicity
(a) Let $A$ and $B$ be two $n \times n$ matrices. A Linear Algebra Professor asked whether the rank of $B$ and $AB$ are the same if $A$ has full rank. What should be the answer? Justify.
Breaking Down the Terminology
- Matrix: A grid of numbers representing a "transformation" or a map.
- Rank: The "information content" of a matrix. Specifically, it's the number of independent dimensions the matrix can reach.
- Full Rank: A square matrix has "full rank" if it is invertible, meaning no information is lost during the transformation.
The Answer
Yes, the rank of $AB$ is the same as the rank of $B$ if $A$ has full rank.
Detailed Justification
- The Property of Full Rank: Since $A$ has full rank, it is an invertible matrix. This means $A^{-1}$ exists such that $A^{-1}A = I$.
- The "Squashing" Rule: Multiplying by a matrix can never increase the rank; it can only stay the same or decrease. Therefore: $$\text{rank}(AB) \leq \text{rank}(B)$$
- The "Undo" Trick: Since $A$ is invertible, we can write $B$ in terms of $AB$: $$B = I \cdot B = (A^{-1}A)B = A^{-1}(AB)$$
- Applying the Rule Again: Multiplying $AB$ by $A^{-1}$ cannot increase its information: $$\text{rank}(B) \leq \text{rank}(AB)$$
Conclusion: Since $\text{rank}(AB) \leq \text{rank}(B)$ and $\text{rank}(B) \leq \text{rank}(AB)$, it must be that $\text{rank}(AB) = \text{rank}(B)$.
Topic 2: Idempotent Matrices and Rank
(b) If $A^2 = A$, $B^2 = B$ and $I - A - B$ is invertible, then help the students to prove $A$ and $B$ are of same rank.
Breaking Down the Terminology
- Idempotent Matrix ($A^2 = A$): Also called a projection. Applying the projection a second time does not change the result.
- Trace ($\text{tr}$): The sum of the numbers on the main diagonal of a matrix.
- Crucial Fact: For any idempotent matrix, the Rank = Trace.
The Answer/Proof
To prove $\text{rank}(A) = \text{rank}(B)$, we demonstrate their traces are equal.
Step 1: Use the Invertibility Condition
Let $M = I - A - B$. Multiply $M$ by $(A - B)$:
$$(I - A - B)(A - B) = IA - IB - A^2 + AB - BA + B^2$$
Step 2: Simplify using $A^2=A$ and $B^2=B$
$$= A - B - A + AB - BA + B = AB - BA$$
Step 3: Use the Trace Property
A key rule is that $\text{tr}(AB) = \text{tr}(BA)$, so $\text{tr}(AB - BA) = 0$.
$$\text{tr}((I - A - B)(A - B)) = 0$$
Step 4: Connect back to Rank
Because $I-A-B$ is invertible:
$$A(I-A-B) = A - A^2 - AB = -AB \implies A = -AB(I-A-B)^{-1}$$
$$(I-A-B)B = B - AB - B^2 = -AB \implies B = (I-A-B)^{-1}(-AB)$$
Conclusion:
Since $A$ and $B$ are related by multiplying the same matrix ($-AB$) by an invertible matrix on different sides, they share the same rank. Combined with the fact that $\text{rank} = \text{tr}$ for idempotent matrices, we conclude $\text{rank}(A) = \text{rank}(B)$.
Topic: Orthogonality and Orthogonal Projections (Click to Expand Solution)
In geometry, "orthogonal" is just a fancy word for perpendicular (at a 90-degree angle). An orthonormal set is a collection of vectors that are all perpendicular to each other and all have a length (norm) of 1.
The Question
If $\{v_1, v_2, \dots, v_n\}$ is an orthonormal set in a vector space $V$ and if $w \in V$, show that the vector: $$u = w - \sum_{i=1}^{n} \langle w, v_i \rangle v_i$$ is orthogonal to each of the vectors $v_j$ (where $j = 1, 2, \dots, n$).Detailed Step-by-Step Answer
To understand this, we need to know what the math symbols mean. Think of this as a recipe:
- The "Inner Product" $\langle a, b \rangle$: This measures how much vector $a$ goes in the direction of vector $b$. If the result is 0, the vectors are perpendicular (orthogonal).
- The Goal: We need to prove that $\langle u, v_j \rangle = 0$ for any vector $v_j$ in our set.
Step 1: Set up the Inner Product
We want to check if $u$ is orthogonal to an arbitrary vector $v_j$ from our set. We write the inner product of $u$ with $v_j$:
$$\langle u, v_j \rangle = \left\langle w - \sum_{i=1}^{n} \langle w, v_i \rangle v_i, v_j \right\rangle$$Step 2: Use Linearity
The inner product is "linear," which means we can distribute the $v_j$ to each part inside the bracket, just like multiplying in basic algebra:
$$\langle u, v_j \rangle = \langle w, v_j \rangle - \left\langle \sum_{i=1}^{n} \langle w, v_i \rangle v_i, v_j \right\rangle$$Now, we can pull the summation symbol and the constants ($\langle w, v_i \rangle$ is just a number) out of the inner product:
$$\langle u, v_j \rangle = \langle w, v_j \rangle - \sum_{i=1}^{n} \langle w, v_i \rangle \langle v_i, v_j \rangle$$Step 3: Apply the "Orthonormal" Property
Here is the "magic" part. We know the set $\{v_1, \dots, v_n\}$ is orthonormal. This gives us two rules for $\langle v_i, v_j \rangle$:
- If $i = j$ (the vectors are the same), $\langle v_i, v_j \rangle = 1$.
- If $i \neq j$ (the vectors are different), $\langle v_i, v_j \rangle = 0$.
Because of this, every single term in that long sum becomes zero, except for the one single case where $i$ matches our chosen $j$.
Step 4: Simplify
The entire sum collapses into just one term:
$$\sum_{i=1}^{n} \langle w, v_i \rangle \langle v_i, v_j \rangle = \langle w, v_j \rangle \cdot (1)$$Now, let's plug that back into our equation from Step 2:
$$\langle u, v_j \rangle = \langle w, v_j \rangle - \langle w, v_j \rangle = 0$$
Conclusion
Since the inner product is 0, the vector $u$ is officially orthogonal to $v_j$. Because we chose $v_j$ as any arbitrary vector in the set, $u$ is orthogonal to every vector in the set $\{v_1, v_2, \dots, v_n\}$.
In plain English: We took vector $w$, calculated its components along all the $v$ directions, and subtracted them. What was left ($u$) has "zero" component in any of those $v$ directions.
Topic: Vector Spaces and Subspaces (Click to Expand Solution)
A set $V$ is a Vector Space if it follows specific rules, most importantly:
- Additive Closure: If you add two things in the set, the result is still in the set.
- Scalar Closure: If you multiply a member by a number, the result is still in the set.
- Zero Element: The "zero" version of that object must be in the set.
Question (a)
Answer: Yes, $V$ is a vector space. Since $V$ is a subset of the known vector space of polynomials $P_5(x)$, we just need to check if it is a "subspace" by testing closure:
- Zero Polynomial: Does the zero polynomial $0(x)$ belong to $V$? Yes, because $0(a) = 0$.
- Addition: If we have two polynomials $p(x)$ and $q(x)$ where $p(a)=0$ and $q(a)=0$, then $(p+q)(a) = p(a) + q(a) = 0 + 0 = 0$.
- Scalar Multiplication: If $p(a)=0$ and $k$ is a number, then $(kp)(a) = k \cdot p(a) = k \cdot 0 = 0$.
Finding a Basis:
Every $p(x) \in V$ has a root at $a$, so by the Factor Theorem, $p(x) = (x - a) \cdot q(x)$ where $q(x)$ is degree $\leq 4$. Using the standard basis for $P_4$, we multiply by $(x-a)$:
Basis: $\{(x-a), x(x-a), x^2(x-a), x^3(x-a), x^4(x-a)\}$
Question (b)
Answer: No, $V$ is not a vector space. A vector space must contain a zero element. In the world of matrices, the "zero" is the zero matrix:
$$\mathbf{0} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$The zero matrix is not invertible (determinant is 0). Missing the zero element disqualifies $V$. It also fails Addition Closure: If $A = I$ and $B = -I$, both are invertible, but $A+B = \mathbf{0}$, which is not in $V$.
Question (c)
Answer: No, $V$ is still not a vector space. Replacing addition with multiplication breaks fundamental structures:
- Commutativity: Vector space addition must be commutative ($u+v = v+u$). Matrix multiplication is generally not commutative ($AB \neq BA$).
- Scalar Distributivity: Matrix multiplication doesn't distribute with scalars in the required way (e.g., $k(A \times B) \neq (kA) \times (kB)$).
- The "Zero" Problem: Multiplying by the scalar $0$ must result in the "additive identity." For multiplication, that identity is $I$. However, $0 \cdot A = \mathbf{0}$, which is not the Identity matrix and is not in the set.
Topic: Linear Algebra - Symmetric Matrices and Inner Products (Click to Expand Solution)
This problem deals with Linear Algebra, specifically focusing on the properties of symmetric matrices, the definition of inner products, and how to calculate the magnitude (norm) of a vector in a custom mathematical space.
The Problem
Consider the matrix $A = \begin{bmatrix} m & 1 & 1 \\ 1 & m & 1 \\ 1 & 1 & m \end{bmatrix}$, where $m \ge 2$ is a positive integer.- Prove that $A$ is a positive definite matrix.
- Using this matrix $A$, find an inner product on $\mathbb{R}^3$.
- Using the inner product defined in (b), find the norm of the vector $\mathbf{v} = \begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix}$.
(a) Proving $A$ is Positive Definite
What is a "Positive Definite" matrix?
Think of a positive definite matrix like a positive number, but for matrices. Just as $x^2$ is always positive for any non-zero number, a matrix is positive definite if, when you "sandwich" it between a vector and its transpose ($x^T Ax$), the result is always a positive number for any non-zero vector $x$.
One common way to check this is Sylvester's Criterion, which says a symmetric matrix is positive definite if all its leading principal minors (the determinants of the top-left square sub-matrices) are positive.
Step-by-Step Proof:
The matrix is $A = \begin{bmatrix} m & 1 & 1 \\ 1 & m & 1 \\ 1 & 1 & m \end{bmatrix}$.
- First Minor ($D_1$): The top-left $1 \times 1$ entry.
$D_1 = |m| = m$. Since $m \ge 2$, $D_1 > 0$. - Second Minor ($D_2$): The determinant of the top-left $2 \times 2$ sub-matrix.
$D_2 = \begin{vmatrix} m & 1 \\ 1 & m \end{vmatrix} = m^2 - 1$. Since $m \ge 2$, the smallest value is $2^2 - 1 = 3$. Thus, $D_2 > 0$. - Third Minor ($D_3$): The determinant of the whole $3 \times 3$ matrix.
$\det(A) = m(m^2 - 1) - 1(m - 1) + 1(1 - m)$
$\det(A) = (m-1)^2(m+2)$
Since $m \ge 2$, $(m-1)^2$ is positive and $(m+2)$ is positive. Thus, $D_3 > 0$.
Conclusion: Since all leading principal minors are positive, $A$ is positive definite.
(b) Finding an Inner Product on $\mathbb{R}^3$
What is an "Inner Product"?
Usually, we use the "Dot Product" to multiply vectors. However, an inner product is a more general way to define how vectors interact. Any positive definite matrix $A$ can be used to define a valid inner product $\langle \mathbf{x}, \mathbf{y} \rangle$ using the formula: $$\langle \mathbf{x}, \mathbf{y} \rangle = \mathbf{x}^T A \mathbf{y}$$
The Answer:
For any two vectors $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ and $\mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}$ in $\mathbb{R}^3$, the inner product is:
$$\langle \mathbf{x}, \mathbf{y} \rangle = \begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix} \begin{bmatrix} m & 1 & 1 \\ 1 & m & 1 \\ 1 & 1 & m \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}$$Expanded, this looks like: $$\langle \mathbf{x}, \mathbf{y} \rangle = m(x_1y_1 + x_2y_2 + x_3y_3) + (x_1y_2 + x_1y_3 + x_2y_1 + x_2y_3 + x_3y_1 + x_3y_2)$$
(c) Finding the Norm of the Vector
What is a "Norm"?
The norm $||\mathbf{v}||$ is the "length" of a vector in this specific inner product space. It is calculated as the square root of the inner product of the vector with itself: $$||\mathbf{v}|| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle}$$
Step-by-Step Calculation:
Given $\mathbf{v} = \begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix}$:
- Calculate $A\mathbf{v}$ first:
$A\mathbf{v} = \begin{bmatrix} m & 1 & 1 \\ 1 & m & 1 \\ 1 & 1 & m \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ -2 \end{bmatrix} = \begin{bmatrix} m - 3 \\ -m - 1 \\ -2m \end{bmatrix}$ - Calculate $\mathbf{v}^T (A\mathbf{v})$:
$\langle \mathbf{v}, \mathbf{v} \rangle = \begin{bmatrix} 1 & -1 & -2 \end{bmatrix} \begin{bmatrix} m - 3 \\ -m - 1 \\ -2m \end{bmatrix}$
$\langle \mathbf{v}, \mathbf{v} \rangle = 1(m-3) + (-1)(-m-1) + (-2)(-2m) = 6m - 2$
$||\mathbf{v}|| = \sqrt{6m - 2}$
Topic: Matrix Diagonalization, Eigenvalues, and Eigenvectors (Click to Expand Solution)
This problem focuses on Linear Algebra, specifically the concepts of Matrix Diagonalization, Eigenvalues, and Eigenvectors.
The Problem
(a) Suppose $A = \begin{bmatrix} -1 & 0 \\ 1 & -1 \\ 2 & -2 \end{bmatrix}$. Find the diagonal matrix $D$ and the invertible matrix $P$ such that $(A^T A)P = PD$.(b) Why for any $m \times n$ matrix $A$, $A^T A$ is always diagonalizable?
(a) Step-by-Step Solution
What are we trying to do?
The equation $(A^T A)P = PD$ is the standard definition for diagonalization.
- $D$ is a diagonal matrix containing the eigenvalues of $(A^T A)$.
- $P$ is a matrix where each column is an eigenvector corresponding to those eigenvalues.
- $A^T$ is the "transpose" of $A$, which means we turn the rows into columns.
Step 1: Calculate $A^T A$
First, we find $A^T$:
$$A^T = \begin{bmatrix} -1 & 1 & 2 \\ 0 & -1 & -2 \end{bmatrix}$$Now, multiply $A^T$ by $A$:
$$A^T A = \begin{bmatrix} -1 & 1 & 2 \\ 0 & -1 & -2 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 1 & -1 \\ 2 & -2 \end{bmatrix} = \begin{bmatrix} 6 & -5 \\ -5 & 5 \end{bmatrix}$$Step 2: Find the Eigenvalues (Matrix $D$)
To find eigenvalues ($\lambda$), we solve the characteristic equation $\det(A^T A - \lambda I) = 0$:
$$\begin{vmatrix} 6 - \lambda & -5 \\ -5 & 5 - \lambda \end{vmatrix} = 0$$ $$(6 - \lambda)(5 - \lambda) - (-5)(-5) = 0$$ $$\lambda^2 - 11\lambda + 5 = 0$$Using the quadratic formula:
$$\lambda = \frac{11 \pm \sqrt{121 - 20}}{2} = \frac{11 \pm \sqrt{101}}{2}$$So, the diagonal matrix $D$ is:
$$D = \begin{bmatrix} \frac{11 + \sqrt{101}}{2} & 0 \\ 0 & \frac{11 - \sqrt{101}}{2} \end{bmatrix}$$Step 3: Find the Eigenvectors (Matrix $P$)
For each eigenvalue $\lambda$, we solve $(A^T A - \lambda I)\mathbf{v} = 0$.
For $\lambda_1 = \frac{11 + \sqrt{101}}{2}$, we get vector $\mathbf{v}_1 = \begin{bmatrix} 10 \\ 1 - \sqrt{101} \end{bmatrix}$.
For $\lambda_2 = \frac{11 - \sqrt{101}}{2}$, we get vector $\mathbf{v}_2 = \begin{bmatrix} 10 \\ 1 + \sqrt{101} \end{bmatrix}$.
The matrix $P$ is:
$$P = \begin{bmatrix} 10 & 10 \\ 1 - \sqrt{101} & 1 + \sqrt{101} \end{bmatrix}$$(b) Why $A^T A$ is Always Diagonalizable
The reason $A^T A$ is always diagonalizable is rooted in the Spectral Theorem.
- Symmetry: For any matrix $A$, the product $A^T A$ is always a symmetric matrix. We can prove this by showing $(A^T A)^T = A^T (A^T)^T = A^T A$.
- The Property: A fundamental theorem in linear algebra states that every real symmetric matrix is orthogonally diagonalizable.
- Eigenvectors: This means it will always have a full set of $n$ linearly independent eigenvectors, which are required to form the invertible matrix $P$.
Topic: Singular Value Decomposition (SVD) and PCA (Click to Expand Solution)
This problem deals with Singular Value Decomposition (SVD) and Principal Component Analysis (PCA). These are techniques used in data science to simplify complex data and find the most important patterns within it.
Part (a): Finding the Singular Value Decomposition (SVD)
The Goal: We want to break down matrix $A$ into three specific parts: $A = U\Sigma V^T$.
- $V$: Tells us about the "directions" in our input data.
- $\Sigma$: Tells us the "strength" or "importance" of those directions.
- $U$: Tells us how the data maps to the output space.
Given: $$A = \begin{bmatrix} -2 & 0 \\ 1 & 1 \\ 1 & -1 \end{bmatrix}$$
Step 1: Find $A^TA$
To find the singular values and the matrix $V$, we first multiply $A$ by its transpose ($A^T$).
$$A^T = \begin{bmatrix} -2 & 1 & 1 \\ 0 & 1 & -1 \end{bmatrix}$$ $$A^TA = \begin{bmatrix} 6 & 0 \\ 0 & 2 \end{bmatrix}$$Step 2: Find Eigenvalues ($\lambda$) and Singular Values ($\sigma$)
Since $A^TA$ is a diagonal matrix, its eigenvalues are simply the numbers on the diagonal:
$$\lambda_1 = 6, \quad \lambda_2 = 2$$The singular values ($\sigma$) are the square roots of these eigenvalues:
$$\sigma_1 = \sqrt{6}, \quad \sigma_2 = \sqrt{2}$$So, our diagonal matrix $\Sigma$ (same size as $A$) is:
$$\Sigma = \begin{bmatrix} \sqrt{6} & 0 \\ 0 & \sqrt{2} \\ 0 & 0 \end{bmatrix}$$Step 3: Find the matrix $V$
$V$ is formed by the eigenvectors of $A^TA$. Since $A^TA$ is already diagonal, the eigenvectors are the standard basis vectors:
$$v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad v_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \implies V = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$Step 4: Find the matrix $U$
We find the columns of $U$ using the formula $u_i = \frac{1}{\sigma_i} A v_i$:
$u_1 = \begin{bmatrix} -2/\sqrt{6} \\ 1/\sqrt{6} \\ 1/\sqrt{6} \end{bmatrix}$, $u_2 = \begin{bmatrix} 0 \\ 1/\sqrt{2} \\ -1/\sqrt{2} \end{bmatrix}$, $u_3 = \begin{bmatrix} 1/\sqrt{3} \\ 1/\sqrt{3} \\ 1/\sqrt{3} \end{bmatrix}$ (orthogonal to $u_1, u_2$).
Final Answer for (a):
$$A = \begin{bmatrix} -2/\sqrt{6} & 0 & 1/\sqrt{3} \\ 1/\sqrt{6} & 1/\sqrt{2} & 1/\sqrt{3} \\ 1/\sqrt{6} & -1/\sqrt{2} & 1/\sqrt{3} \end{bmatrix} \begin{bmatrix} \sqrt{6} & 0 \\ 0 & \sqrt{2} \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$Part (b): Covariance and Variance
The Terminology
- Covariance Matrix: Describes how much variables change together.
- Eigenvalues of Covariance: Represent the amount of "variance" (spread) along directions.
- Direction of Maximum Variance: The direction of the first Principal Component.
1. Eigenvalues of the Covariance Matrix
If $N=3$ (samples), the covariance matrix $C = \frac{1}{N-1} A^TA$.
Eigenvalues of $C$: $\frac{6}{2} = \mathbf{3}$ and $\frac{2}{2} = \mathbf{1}$.
2. Direction of Maximum Variance
The maximum variance occurs along the eigenvector associated with the largest eigenvalue ($\lambda = 6$).
From Part (a), the eigenvector is: $$v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$
The eigenvalues are 3 and 1. The direction along which the variance is maximum is the x-axis (represented by the vector $[1, 0]^T$).
Topic: Linear Algebra - Spectral Decomposition & Nilpotent Matrices (Click to Expand Solution)
This response covers two fundamental topics in Linear Algebra: Spectral Decomposition (building matrices from vectors) and Nilpotent Matrices (matrices that eventually "vanish" when multiplied by themselves).
Topic 1: Matrix Decomposition and Eigensystems
b) A data analyst at an oil exploration company observed that an $m \times m$ matrix $\mathbf{B}$ can be defined as $$\mathbf{B} = \sum_{i=1}^{n} \left( 6^i \cdot \mathbf{p}_i\mathbf{p}_i^T \right)$$ where $\mathbf{p}_i \in \mathbb{R}^m$ and $n < m$, with the property that vectors are orthonormal ($\langle \mathbf{p}_i, \mathbf{p}_j \rangle = 0$ for $i \neq j$ and $\langle \mathbf{p}_i, \mathbf{p}_i \rangle = 1$).
Derive: i) eigenvalues, ii) eigenvectors, iii) trace, and iv) determinant.
Understanding the Terminology
- Matrix ($\mathbf{B}$): A grid representing a linear transformation.
- Outer Product ($\mathbf{p}_i\mathbf{p}_i^T$): A column multiplied by a row, resulting in a square matrix.
- Inner Product ($\langle \mathbf{p}_i, \mathbf{p}_j \rangle$): A measure of overlap; $0$ means orthogonal (perpendicular).
Answer
i) & ii) Deriving Eigenvalues and Eigenvectors:
Multiply $\mathbf{B}$ by a vector $\mathbf{p}_j$:
$$\mathbf{B}\mathbf{p}_j = \sum_{i=1}^{n} 6^i \mathbf{p}_i (\mathbf{p}_i^T \mathbf{p}_j)$$
Since vectors are orthonormal, $\mathbf{p}_i^T \mathbf{p}_j$ is $1$ if $i=j$ and $0$ otherwise.
$$\mathbf{B}\mathbf{p}_j = 6^j \mathbf{p}_j$$
The Eigenvectors are $\mathbf{p}_1, \dots, \mathbf{p}_n$ and the Eigenvalues are $6^1, 6^2, \dots, 6^n$.
iii) Derive the Trace:
The sum of all $m$ eigenvalues. Since $n < m$, the remaining $m-n$ eigenvalues are $0$.
$$\text{Trace}(\mathbf{B}) = \sum_{i=1}^{n} 6^i = \frac{6(6^n - 1)}{5}$$
iv) Derive the Determinant:
The product of all eigenvalues. Since at least one eigenvalue is $0$:
$$\text{Det}(\mathbf{B}) = 0$$
Topic 2: Nilpotent Matrices
c) A matrix $\mathbf{M} \in \mathbb{R}^{r \times r}$ has the property $\mathbf{M}^k = \mathbf{0}$ for some $k < r$.
Derive: i) trace, ii) determinant, iii) all individual eigenvalues. iv) Is $\mathbf{M}$ invertible?
Answer
iii) Derive all individual eigenvalues:
If $\mathbf{M}\mathbf{x} = \lambda\mathbf{x}$, then $\mathbf{M}^k\mathbf{x} = \lambda^k\mathbf{x}$. Since $\mathbf{M}^k = \mathbf{0}$, we have $\mathbf{0} = \lambda^k\mathbf{x}$. As $\mathbf{x} \neq \mathbf{0}$, $\lambda^k = 0$, so $\lambda = 0$.
All eigenvalues are 0.
i) Trace: Sum of eigenvalues = $0 + 0 + \dots = \mathbf{0}$.
ii) Determinant: Product of eigenvalues = $0 \cdot 0 \cdot \dots = \mathbf{0}$.
iv) Is the matrix M invertible?
No. A matrix is invertible only if its determinant is non-zero. Since $\text{Det}(\mathbf{M}) = 0$, it is singular.
Topic: Linear Algebra – Positive Definite Matrices and Cholesky Factorization (Click to Expand Solution)
Question
b) Consider a $n \times n$ matrix $A$ which we can write as $A = L^T L$ where $L$ is a lower-triangular matrix. We have a function $F(x, R, y)$ that returns the value $x^T R y$ where $R$ is an $n \times n$ matrix and $x$ and $y$ are $n \times 1$ vectors respectively. If we are given both the matrix $A$ and $L$, what is the smallest number of calls that need to be made to this function in terms of $n$ in order to decide that the given matrix $A$ is positive definite, and why?Terminology Breakdown
Before solving, let’s define the terms for someone seeing them for the first time:
- Matrix ($n \times n$): A square grid of numbers with $n$ rows and $n$ columns.
- Vector ($n \times 1$): A single column of $n$ numbers.
- Lower-Triangular Matrix ($L$): A square matrix where all the entries above the main diagonal are zero.
- Transpose ($L^T$): A matrix formed by swapping the rows and columns of $L$. If $L$ is lower-triangular, $L^T$ is upper-triangular.
- Positive Definite (PD): A matrix $A$ is positive definite if, for every non-zero vector $v$, the result of $v^T A v$ is always a positive number (greater than zero).
- The Function $F(x, R, y)$: This performs a calculation called a "bilinear form." It multiplies a row vector, a matrix, and a column vector together to result in a single number (a scalar).
Detailed Answer
1. The Condition for $A$ to be Positive Definite
The problem states that $A = L^T L$. In linear algebra, this is a form of the Cholesky Decomposition. For any matrix $A$ that can be written as $L^T L$, we know that $A$ is at least "Positive Semi-Definite." To be strictly Positive Definite, the matrix $A$ must be non-singular (invertible). Because $A = L^T L$, $A$ is non-singular if and only if $L$ is non-singular.
2. How to tell if $L$ is non-singular
For a triangular matrix (like our $L$), there is a very simple rule: A triangular matrix is non-singular if and only if all of its diagonal elements are non-zero. If even one entry on the main diagonal of $L$ (let's call these entries $L_{11}, L_{22}, \dots, L_{nn}$) is zero, then the matrix $L$ is singular, and consequently, $A$ cannot be positive definite.
3. Using the Function $F$ to check the diagonals
We need to check the $n$ diagonal elements of $L$ to ensure none of them are zero. We can use the provided function $F(x, R, y) = x^T R y$ to extract these values. To get the value of the $i$-th diagonal element ($L_{ii}$), we can use a "basis vector" $e_i$. A basis vector $e_i$ is a vector where the $i$-th entry is $1$ and all other entries are $0$.
If we set:
- $R = L$
- $x = e_i$
- $y = e_i$
The function $F(e_i, L, e_i)$ performs the calculation $e_i^T L e_i$, which isolates and returns exactly the value of $L_{ii}$.
4. Determining the Smallest Number of Calls
Since there are $n$ diagonal elements in an $n \times n$ matrix, and we must ensure that every single one of them is non-zero to guarantee that $A$ is positive definite, we must check each one individually.
Call 2: $F(e_2, L, e_2) = L_{22}$
...
Call $n$: $F(e_n, L, e_n) = L_{nn}$
If any of these calls return $0$, we stop and conclude $A$ is not positive definite. To "decide" (confirm) it is positive definite, we must verify all $n$ entries.
The smallest number of calls is $n$.
Why? A matrix $A = L^T L$ is positive definite if and only if the lower-triangular matrix $L$ is non-singular. For a triangular matrix, non-singularity is guaranteed if and only if all $n$ diagonal entries are non-zero. Since each call to the function $F$ can only extract one specific diagonal entry (by using basis vectors), we require $n$ calls to verify all diagonal positions.
Topic: Inner Products and Positive Definite Matrices (Click to Expand Solution)
Question
Let $A = \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{pmatrix}$ and define $\langle \mathbf{x}, \mathbf{y} \rangle_A = \mathbf{x}^T A \mathbf{y}$ where $\mathbf{x}, \mathbf{y} \in \mathbb{R}^4$. Then,i) prove $\langle \cdot, \cdot \rangle_A$ is an inner product on $\mathbb{R}^4$
ii) find all possible $\mathbf{y} = [y_1, y_2, y_3, y_4]^T$ such that $\langle \mathbf{x}, \mathbf{y} \rangle_A = 0$ where $\mathbf{x} = [1, 0, 0, 0]^T$
To solve this problem, we first need to understand what an inner product is. Think of it as a generalized way to multiply two vectors to get a single number (a scalar). For a matrix-based formula like $\mathbf{x}^T A \mathbf{y}$ to be a valid inner product, the matrix $A$ must satisfy two main conditions:
- Symmetry: The matrix must be the same if you flip it over its diagonal.
- Positive Definiteness: For any non-zero vector $\mathbf{x}$, the result of $\mathbf{x}^T A \mathbf{x}$ must always be greater than zero.
Part (i): Proving it is an Inner Product
Step 1: Check for Symmetry
A matrix is symmetric if $A = A^T$ (the rows are the same as the columns). Looking at $A$:
- Row 1 is $[2, -1, 0, 0]$, Column 1 is $[2, -1, 0, 0]$.
- Row 2 is $[-1, 2, -1, 0]$, Column 2 is $[-1, 2, -1, 0]$.
- Row 3 is $[0, -1, 2, -1]$, Column 3 is $[0, -1, 2, -1]$.
- Row 4 is $[0, 0, -1, 2]$, Column 4 is $[0, 0, -1, 2]$.
Since $A = A^T$, the symmetry condition is satisfied.
Step 2: Check for Positive Definiteness (Sylvester’s Criterion)
To prove a symmetric matrix is "positive definite," we calculate the leading principal minors. If all these determinants are positive, the matrix is positive definite.
- First Minor ($D_1$): $D_1 = |2| = 2 > 0$
- Second Minor ($D_2$): $\begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} = 4 - 1 = 3 > 0$
- Third Minor ($D_3$): $\begin{vmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{vmatrix} = 2(4-1) - (-1)(-2-0) = 4 > 0$
- Fourth Minor ($D_4$): $\det(A) = 2(D_3) - (-1)(-1 \times D_2) = 2(4) - 3 = 5 > 0$
Conclusion for (i): Since $A$ is symmetric and all its leading principal minors are positive, $A$ is positive definite. Therefore, $\langle \mathbf{x}, \mathbf{y} \rangle_A = \mathbf{x}^T A \mathbf{y}$ satisfies all the axioms of an inner product.
Part (ii): Finding $\mathbf{y}$ such that $\langle \mathbf{x}, \mathbf{y} \rangle_A = 0$
We are given $\mathbf{x} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$ and we want to find $\mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{pmatrix}$ such that $\mathbf{x}^T A \mathbf{y} = 0$.
Step 1: Calculate $\mathbf{x}^T A$
Since $\mathbf{x}^T$ is a vector with a $1$ in the first position and $0$ everywhere else, multiplying $\mathbf{x}^T A$ simply picks out the first row of $A$:
$$\mathbf{x}^T A = \begin{pmatrix} 2 & -1 & 0 & 0 \end{pmatrix}$$Step 2: Solve $(\mathbf{x}^T A) \mathbf{y} = 0$
$$\begin{pmatrix} 2 & -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{pmatrix} = 0 \implies 2y_1 - y_2 = 0$$Step 3: State the solution
From $2y_1 - y_2 = 0$, we get $y_2 = 2y_1$. Note that $y_3$ and $y_4$ do not appear, meaning they can be any real number.
All possible vectors $\mathbf{y}$ are of the form:
$$\mathbf{y} = \begin{bmatrix} y_1 \\ 2y_1 \\ y_3 \\ y_4 \end{bmatrix} \text{ for any } y_1, y_3, y_4 \in \mathbb{R}$$
Topic: Linear Algebra - Eigenvalues and Matrix Properties (Click to Expand Solution)
To help you understand these problems, let’s start with a few fundamental concepts. If you are new to linear algebra, think of a matrix as a grid of numbers that represents a transformation. Eigenvalues are special numbers associated with a matrix that tell us how much a vector is stretched or shrunk during that transformation.
Topic 1: Eigenvalues and Determinants of Symmetric Matrices
Terminology for Beginners
- Determinant: A single number calculated from a square matrix. If the determinant is $0$, the matrix has at least one eigenvalue that is $0$.
- Characteristic Equation: The mathematical formula we use to find eigenvalues.
- Positive Definite: A matrix is "positive definite" if all its eigenvalues are greater than zero.
Question
$$G = \begin{bmatrix} 2 & \gamma & 0 \\ \gamma & 2 & \gamma \\ 0 & \gamma & 2 \end{bmatrix}$$ (a) For which values of $\gamma$ will $G$ have only non-zero eigenvalues?(b) If all eigenvalues are positive, what is the range of $\gamma$?
Answer (Problem 1)
Part (a): Finding values for non-zero eigenvalues
A matrix has only non-zero eigenvalues if and only if its determinant is not zero.
$\det(G) = 2(4 - \gamma^2) - \gamma(2\gamma - 0) = 8 - 4\gamma^2$.
Set $\det(G) \neq 0$: $8 - 4\gamma^2 \neq 0 \Rightarrow \gamma^2 \neq 2 \Rightarrow \gamma \neq \pm \sqrt{2}$.
Part (b): Range for positive eigenvalues
For a symmetric matrix like $G$, all eigenvalues are positive if the matrix is Positive Definite. We use Sylvester’s Criterion (all leading principal minors must be positive):
- $M_1 = 2 > 0$ (Satisfied)
- $M_2 = 4 - \gamma^2 > 0 \Rightarrow -2 < \gamma < 2$
- $M_3 = 8 - 4\gamma^2 > 0 \Rightarrow -\sqrt{2} < \gamma < \sqrt{2}$
Conclusion: The strictest range is $-\sqrt{2} < \gamma < \sqrt{2}$.
Topic 2: Properties of Rank-1 Matrices (Outer Products)
Terminology for Beginners
- Outer Product ($rr^T$): A vertical vector multiplied by a horizontal one, creating a square matrix where every row is a multiple of $r$.
- Singular Values: Values that describe the magnitude of the matrix's transformation.
Question
Consider $M = rr^T$ where $r \in \mathbb{R}^k$.(a) How many non-zero eigenvalues? (b) Sum of absolute values of eigenvalues? (c) Non-zero singular values of $M^TM$?
Answer (Problem 2)
Part (a): Non-zero eigenvalues
$M = rr^T$ is a Rank-1 matrix, so it has exactly one non-zero eigenvalue. Testing $v = r$:
$(rr^T)r = r(r^Tr) = (r^Tr)r$.
The non-zero eigenvalue is $\lambda = \|r\|^2$.
Part (b): Sum of absolute values
Since only one eigenvalue is non-zero, the sum is simply $\|r\|^2$.
Part (c): Singular values of $M^TM$
$M^TM = (rr^T)(rr^T) = r(r^Tr)r^T = \|r\|^2(rr^T) = \|r\|^2 M$.
The only non-zero eigenvalue of $M^TM$ is $\|r\|^2 \times \|r\|^2 = \|r\|^4$. For symmetric positive semi-definite matrices, singular values equal eigenvalues.
Conclusion: 1 non-zero singular value equal to $\|r\|^4$.
Topic: Convexity of Functions and Norms (Click to Expand Solution)
Before we dive into the proofs, let's break down the terminology so everything is clear.
💡 Key Terms to Know
- $\mathbb{R}^n$: This represents a space with $n$ dimensions. If $n=2$, it's a flat plane (like a piece of paper). If $n=3$, it's 3D space.
- Convex Function: Imagine a bowl. If you pick any two points on the surface of the bowl and draw a straight line between them, that line will always sit above or on the surface of the bowl.
- Norm ($\|\cdot\|$): A way to measure the "length" or "size" of a vector.
- $L_1$ Norm ($\|x\|_1$): This is the sum of the absolute values of the coordinates. For example, if $x = (3, -4)$, then $\|x\|_1 = |3| + |-4| = 7$.
The Questions
i) $f_1(x) = \|x\|_1$. Prove or disprove that $f_1(x)$ is a convex function by using the properties of norms.ii) Let $g(x)$ be a convex function and $A \in \mathbb{R}^{n \times n}$ and $b \in \mathbb{R}^n$. Prove or disprove that $h(x) = g(Ax + b)$ is a convex function.
Solution for Part (i)
To prove that the $L_1$ norm is convex, we use two fundamental properties of all norms:
- Triangle Inequality: $\|u + v\| \leq \|u\| + \|v\|$
- Absolute Homogeneity: $\|\alpha u\| = |\alpha| \cdot \|u\|$
The Proof:
Let $x$ and $y$ be any two vectors in $\mathbb{R}^n$, and let $\lambda$ be any number such that $0 \leq \lambda \leq 1$. Apply the function to a weighted average of $x$ and $y$:
Apply the Triangle Inequality:
Apply Absolute Homogeneity (note that $\lambda$ and $1-\lambda$ are $\ge 0$):
Conclusion: Since $f_1(\lambda x + (1-\lambda)y) \leq \lambda f_1(x) + (1-\lambda) f_1(y)$, $f_1(x) = \|x\|_1$ is a convex function. In fact, all norms are convex functions.
Solution for Part (ii)
Here, we are plugging a linear transformation ($Ax + b$) into a function $g(x)$ that we already know is convex.
The Proof:
Let $x, y \in \mathbb{R}^n$ and $0 \leq \lambda \leq 1$. We evaluate $h(\lambda x + (1-\lambda)y)$:
- $h(\lambda x + (1-\lambda)y) = g(A(\lambda x + (1-\lambda)y) + b)$
- Distribute $A$: $= g(\lambda Ax + (1-\lambda)Ay + b)$
- Rewrite $b$ as $\lambda b + (1-\lambda)b$: $= g(\lambda Ax + (1-\lambda)Ay + \lambda b + (1-\lambda)b)$
- Group the terms: $= g(\lambda(Ax + b) + (1-\lambda)(Ay + b))$
- Apply $g$'s convexity: $\leq \lambda g(Ax + b) + (1-\lambda) g(Ay + b)$
The inequality holds, so $h(x) = g(Ax + b)$ is a convex function. This is a fundamental rule in optimization: "Convexity is preserved under affine mapping."
Topic: Impact of Feature Scaling on PCA (Click to Expand Solution)
This problem explores how Principal Component Analysis (PCA) behaves when we scale the features of our dataset. Scaling changes the units of our data (e.g., converting meters to centimeters), and as you will see, this has a significant impact on the "directions" the data points in.
Before diving into the questions, let's clarify some core concepts for a beginner:
- Data Point: A single observation (like one person) with several measurements (like height, weight, age).
- Dimensions ($D$): The number of measurements we have for each data point.
- PCA (Principal Component Analysis): A method to find the directions (vectors) in which the data is most "spread out" (has the most variance). These directions are called Principal Components.
- Covariance Matrix: A table that tells us how much two variables change together.
- Eigenvectors: These are the mathematical representation of the Principal Components.
Question 1a
A data scientist works on a problem on $N > 10000$ data points on $D > 100$ dimensions, and obtains the top 10 principal components. Assume the given data is mean-centred. The data scientist’s manager comes to him and tells him that due to a problem with the way in which the data was collected, every data point in $D$ dimensions $(x_1, x_2, \dots, x_D)$ needs to be transformed to $(\alpha_1 x_1, \alpha_2 x_2, \dots, \alpha_D x_D)$ where $(\alpha_1, \alpha_2, \dots, \alpha_D)$ are fixed constants. The data scientist believes that he can simply modify each discovered eigenvector $(b_1, b_2, \dots, b_D)$ to $(\alpha_1 b_1, \alpha_2 b_2, \dots, \alpha_D b_D)$ to get the principal directions for the modified problem. Is the data scientist correct in his belief? Give a mathematical justification for his belief. Otherwise explain why he is wrong.Answer: The Data Scientist is Incorrect.
Conceptual Explanation
PCA is highly sensitive to the scale of the data. If you multiply one feature (like height) by a large number and leave another (like age) alone, the "spread" of the data along the height axis increases dramatically. Consequently, the direction of maximum variance will tilt heavily toward the height axis.
Simply scaling the old eigenvectors doesn't account for how the "shape" of the data cloud has warped. The new principal directions are not just scaled versions of the old ones; they usually point in entirely different directions.
Mathematical Justification
Let $X$ be the original $N \times D$ data matrix. Since it is mean-centered, the original covariance matrix $C$ is:
$$C = \frac{1}{N-1} X^T X$$An eigenvector $v$ of $C$ satisfies the equation $Cv = \lambda v$. Now, let $A$ be a diagonal matrix where the diagonal entries are the scaling constants $\alpha_1, \alpha_2, \dots, \alpha_D$. The transformed data matrix $X'$ can be written as $X' = XA$. The new covariance matrix $C'$ is:
$$C' = \frac{1}{N-1} (XA)^T (XA) = \frac{1}{N-1} A^T X^T X A = A C A$$The scientist believes that if $v$ is an eigenvector of $C$, then $Av$ is an eigenvector of $C'$. Let's check this by multiplying $C'$ by $Av$:
$$C'(Av) = (ACA)(Av) = AC(A^2 v)$$For $Av$ to be an eigenvector of $C'$, the result must be some scalar multiple of $Av$ (i.e., $C'(Av) = \lambda' Av$). However, $AC(A^2 v)$ is generally not proportional to $Av$ unless all $\alpha_i$ are equal (uniform scaling). Therefore, the new principal directions cannot be found by simply scaling the old eigenvectors.
Question 1b
Assume that a data scientist is originally given $N$ points in $D$ dimensions. In order to perform PCA the data scientist computes the covariance matrix. The data scientist is then informed that each given data point of the form $(x_1, x_2, \dots, x_D)$ needs to be transformed to $(\alpha_1 x_1, \alpha_2 x_2, \dots, \alpha_D x_D)$ where $(\alpha_1, \alpha_2, \dots, \alpha_D)$ are fixed constants. The data scientist thinks that he can compute the modified covariance matrix in only $O(D^2)$ time, given that he has the old covariance matrix with him. Is the data scientist justified in his belief? If so, demonstrate the method used by him. Otherwise, show why he is incorrect.Answer: The Data Scientist is Correct.
Conceptual Explanation
The covariance matrix stores the relationship between every pair of features. If you scale every value of feature $i$ by $\alpha_i$ and every value of feature $j$ by $\alpha_j$, the covariance between them is simply multiplied by the product of those two scales ($\alpha_i \times \alpha_j$). Since we already have the old matrix, we just need to visit each of the $D \times D$ cells and perform this simple multiplication.
Demonstration of the Method
Let $C$ be the original covariance matrix. The element in the $i$-th row and $j$-th column is $C_{ij}$. The transformed features are $x'_i = \alpha_i x_i$ and $x'_j = \alpha_j x_j$. The new covariance $C'_{ij}$ is calculated as:
$$C'_{ij} = \text{cov}(x'_i, x'_j) = \text{cov}(\alpha_i x_i, \alpha_j x_j)$$By the properties of covariance:
$$C'_{ij} = \alpha_i \alpha_j \cdot \text{cov}(x_i, x_j) = \alpha_i \alpha_j C_{ij}$$Computational Complexity:
- The covariance matrix $C$ has $D^2$ elements (a $D \times D$ grid).
- For each element $C_{ij}$, we perform a fixed number of multiplications ($\alpha_i \cdot \alpha_j \cdot C_{ij}$).
- Total operations $\approx D^2$.
Therefore, the time complexity is $O(D^2)$. This is much faster than re-calculating the covariance from the raw data, which would take $O(N \cdot D^2)$ time.
Topic: Linear Algebra (SVD and Matrix Properties) (Click to Expand Solution)
This problem explores the relationship between a matrix, its Singular Value Decomposition (SVD), and its Frobenius norm.
The Question
A professor asked students to consider a matrix $A \in \mathbb{R}^{n \times n}$ with SVD: $$A = U\Sigma V^T$$ Given $\|\Sigma\|_F^2 = \gamma$, $B = A^T A$, and $\alpha = \text{trace}(B)$. Student G1 claims $\alpha = \gamma$. Student G2 claims $\alpha = \sqrt{\gamma}$. Also, for $C = AA^T$, let $\beta = \text{trace}(C)$. Prove/disprove if $\beta = \gamma^2$.Understanding the Terminology
- SVD ($A = U\Sigma V^T$): A way of breaking down a matrix. $U$ and $V$ are orthogonal ($U^T U = I, V^T V = I$), and $\Sigma$ is diagonal.
- Frobenius Norm ($\|\cdot\|_F$): The square root of the sum of the squares of all elements. Thus, $\|\Sigma\|_F^2 = \sigma_1^2 + \sigma_2^2 + \dots + \sigma_n^2$.
- Trace ($\alpha$ and $\beta$): The sum of the diagonal elements of a square matrix.
Step-by-Step Solution
(a) Prove or disprove the claim made by G1 ($\alpha = \gamma$)
First, we express $\gamma$: $\|\Sigma\|_F^2 = \sum_{i=1}^{n} \sigma_i^2 = \gamma$.
Next, find $B = A^T A$: $B = (V \Sigma U^T)(U \Sigma V^T) = V \Sigma^2 V^T$.
Using the cyclic property of the trace ($\text{trace}(XYZ) = \text{trace}(ZXY)$):
$$\alpha = \text{trace}(V \Sigma^2 V^T) = \text{trace}(\Sigma^2 V^T V) = \text{trace}(\Sigma^2)$$Since $\Sigma^2$ is diagonal with entries $\sigma_i^2$, $\alpha = \sum \sigma_i^2$.
Conclusion: G1's claim is Correct because $\alpha = \gamma$.
(b) Prove or disprove the claim made by G2 ($\alpha = \sqrt{\gamma}$)
From part (a), we proved $\alpha = \gamma$. The claim $\alpha = \sqrt{\gamma}$ is only true if $\gamma = \sqrt{\gamma}$ (i.e., $\gamma=0$ or $1$).
Conclusion: G2's claim is Disproved.
(c) Prove or disprove that $\beta = \gamma^2$
Here, $\beta$ is the trace of $C = AA^T$. Following the same logic:
$$C = (U \Sigma V^T)(V \Sigma U^T) = U \Sigma^2 U^T$$ $$\beta = \text{trace}(U \Sigma^2 U^T) = \text{trace}(\Sigma^2 U^T U) = \text{trace}(\Sigma^2) = \gamma$$The claim states $\beta = \gamma^2$, but we found $\beta = \gamma$. This is generally false.
Conclusion: The claim $\beta = \gamma^2$ is Disproved.
Topic: Matrix Optimization and the Frobenius Norm (Click to Expand Solution)
To solve this problem, we need to understand a few fundamental concepts in linear algebra: matrices, trace, and the Frobenius norm. Think of this as finding the "closest" possible matrix to a given one while following a specific rule.
Question
The Manager asked the data analyst to find a $2 \times 2$ matrix $M$ with $\text{trace}(M) = 0$ such that $(\|M - A\|_2)^2$ is minimum where $A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$. Help the data analyst in finding $M$.1. Understanding the Terminology
Before we jump into the math, let's break down what the question is asking for:
- $2 \times 2$ Matrix $M$: This is just a grid of four numbers. We can represent it as: $$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
- Trace of $M$ ($\text{trace}(M)$): This is simply the sum of the numbers on the main diagonal (from top-left to bottom-right). $$\text{trace}(M) = a + d$$ The problem tells us $\text{trace}(M) = 0$, so $a + d = 0$. This means $d = -a$.
- Frobenius Norm ($\| \cdot \|_2$): Don't let the symbol scare you! For a matrix, the squared Frobenius norm $(\|M - A\|_2)^2$ is just the sum of the squares of the differences between corresponding elements in two matrices. It’s like the "distance" between two grids of numbers.
2. Setting up the Equation
From our rules above, we can rewrite $M$ using only three variables ($a, b, c$) because $d$ must be $-a$:
$$M = \begin{pmatrix} a & b \\ c & -a \end{pmatrix}$$We are given matrix $A$:
$$A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$$Now, let's find the difference matrix $(M - A)$:
$$M - A = \begin{pmatrix} a - 1 & b - 0 \\ c - 0 & -a - 2 \end{pmatrix} = \begin{pmatrix} a - 1 & b \\ c & -a - 2 \end{pmatrix}$$The value we want to minimize is the sum of the squares of these four elements:
$$f(a, b, c) = (a - 1)^2 + b^2 + c^2 + (-a - 2)^2$$3. Minimizing the Function
To make this total sum as small as possible, we look at each variable:
- For $b$ and $c$: Since $b^2$ and $c^2$ are always positive or zero, the smallest they can ever be is $0$. So, we set $b = 0$ and $c = 0$.
- For $a$: We need to minimize the part of the equation involving $a$: $$g(a) = (a - 1)^2 + (-a - 2)^2$$
Expand the squares:
$$g(a) = (a^2 - 2a + 1) + (a^2 + 4a + 4) = 2a^2 + 2a + 5$$To find the minimum of this quadratic equation, we can take the derivative and set it to zero:
$$g'(a) = 4a + 2 = 0 \implies 4a = -2 \implies a = -0.5$$4. Finding the Final Matrix $M$
Now that we have $a$, we can find $d$:
$$d = -a = -(-0.5) = 0.5$$We already determined that $b = 0$ and $c = 0$. Putting it all together into the matrix $M$:
The matrix the data analyst needs is:
$$M = \begin{pmatrix} -0.5 & 0 \\ 0 & 0.5 \end{pmatrix}$$
Support Vector Machines (SVM) and Lagrangian Duality (Click to Expand Solution)
Question
We are given three points and their associated classifications in the format (x, y, category) as follows: (-1, 3, +), (1, 3, -), (-1, 1, -), and would like to find a separating hyperplane in the form $w^T x + b = 0$ using SVM. Let $\alpha_i, 1 \leq i \leq 3$ be the Lagrangian multipliers for the given points. Set up the Lagrangian dual objective for this problem in terms of only the Lagrangian parameters as a polynomial in the fewest number of variables and the fewest number of terms. If possible, find the optimal separating hyperplane from this expression using the methods of calculus. Give adequate justifications.1. Understanding the Terminology
Before we dive into the math, let's define the terms for someone seeing this for the first time:
- Hyperplane: In a 2D space (like a piece of paper), a hyperplane is simply a straight line that separates two groups of points. In the equation $w^T x + b = 0$, $w$ is a vector that is perpendicular to the line, $x$ represents the coordinates $(x, y)$, and $b$ is the "bias" that shifts the line away from the center.
- SVM (Support Vector Machine): This is a method to find the "best" line. The "best" line is the one that has the largest "margin" (the widest empty corridor) between the two groups of points.
- Lagrangian Multipliers ($\alpha_i$): Instead of solving for the line directly, we assign a "weight" ($\alpha$) to each data point. Points that end up being right on the edge of the corridor are called Support Vectors and will have $\alpha > 0$. Points far away from the boundary will have $\alpha = 0$.
- Lagrangian Dual Objective: This is a mathematical transformation that allows us to find the optimal $\alpha$ values. Once we have the $\alpha$ values, we can easily calculate the line ($w$ and $b$).
2. Setting Up the Data
We have three points, which we will label $x_1, x_2, x_3$ and their classes $y_i$ (where $+$ becomes $+1$ and $-$ becomes $-1$):
- Point 1: $x_1 = \begin{pmatrix} -1 \\ 3 \end{pmatrix}, y_1 = 1$
- Point 2: $x_2 = \begin{pmatrix} 1 \\ 3 \end{pmatrix}, y_2 = -1$
- Point 3: $x_3 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}, y_3 = -1$
3. The Dual Objective Formula
The standard Lagrangian Dual objective function $Q(\alpha)$ that we want to maximize is:
$$Q(\alpha) = \sum_{i=1}^3 \alpha_i - \frac{1}{2} \sum_{i=1}^3 \sum_{j=1}^3 \alpha_i \alpha_j y_i y_j (x_i \cdot x_j)$$Subject to: $\sum_{i=1}^3 \alpha_i y_i = 0$ and $\alpha_i \geq 0$ for all $i$.
Step 3a: Calculate the Dot Products $(x_i \cdot x_j)$
Step 3b: Simplify the Constraint
$\alpha_1(1) + \alpha_2(-1) + \alpha_3(-1) = 0 \implies \mathbf{\alpha_1 = \alpha_2 + \alpha_3}$
4. Expressing the Objective in Fewest Variables
The linear term: $\sum \alpha_i = 2\alpha_2 + 2\alpha_3$.
After substituting $\alpha_1 = \alpha_2 + \alpha_3$ and simplifying the quadratic terms:
$\alpha_2^2$ terms: $10 + 10 - 16 = 4\alpha_2^2$
$\alpha_3^2$ terms: $10 + 2 - 8 = 4\alpha_3^2$
$\alpha_2 \alpha_3$ terms: $20 - 16 - 8 + 4 = 0$
The quadratic term is: $\frac{1}{2} (4\alpha_2^2 + 4\alpha_3^2) = 2\alpha_2^2 + 2\alpha_3^2$.
5. Finding the Optimal Values using Calculus
$\frac{\partial Q}{\partial \alpha_2} = 2 - 4\alpha_2 = 0 \implies \alpha_2 = 0.5$
$\frac{\partial Q}{\partial \alpha_3} = 2 - 4\alpha_3 = 0 \implies \alpha_3 = 0.5$
Then, $\alpha_1 = 0.5 + 0.5 = 1.0$. All points are Support Vectors.
6. Finding the Separating Hyperplane
Step 6a: Calculate $w$
$w = (1)(1)\begin{pmatrix} -1 \\ 3 \end{pmatrix} + (0.5)(-1)\begin{pmatrix} 1 \\ 3 \end{pmatrix} + (0.5)(-1)\begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$
Step 6b: Calculate $b$
Using $x_1$: $(1) [ \begin{pmatrix} -1 & 1 \end{pmatrix} \begin{pmatrix} -1 \\ 3 \end{pmatrix} + b ] = 1 \implies 1 + 3 + b = 1 \implies b = -3$
The optimal separating hyperplane is:
$-1x + 1y - 3 = 0 \implies y = x + 3$
Justification: This line perfectly bisects the distance between the classes, maximizing the margin. All points are support vectors because they lie exactly on the edges of the margin gutters.
Topic: Principal Component Analysis (PCA) and Maximum Variance (Click to Expand Solution)
Question:
A Professor gave a 2 dimensional data matrix $X = \begin{pmatrix} 2 & -1 & -2 & 1 \\ -4 & 2 & 4 & -2 \end{pmatrix}$ for dimension reduction. Student 1 suggested that the direction of maximum variance is $[1, 0]^T$. Student 2 suggested that the direction of maximum variance is $[0, 1]^T$. Student 3 claimed that both student 1 and 2 are wrong. Find the correct $v$ that gives the direction of maximum variance and hence decide on which student is correct.1. Understanding the Goal (For Beginners)
Imagine you have a group of data points plotted on a 2D graph. "Dimension reduction" is the process of trying to represent those 2D points on a 1D line without losing much information.
- To keep the most "information," we look for the direction of maximum variance.
- Variance is simply the "spread" of the data.
- If you draw a line through the data points, the direction of maximum variance is the line where the points are most spread out.
2. Analyzing the Data Matrix
In the matrix $X$, each column represents a single data point, and each row represents a dimension. Let's look at our four data points:
Point 1: $(2, -4)$ | Point 2: $(-1, 2)$ | Point 3: $(-2, 4)$ | Point 4: $(1, -2)$
A Crucial Observation:
Notice the relationship between the first row ($x$) and the second row ($y$). For every point, the $y$-coordinate is exactly $-2$ times the $x$-coordinate. This means all these points lie perfectly on a single straight line defined by the equation $y = -2x$.
3. Finding the Direction of Maximum Variance ($v$)
Method A: The Geometric Intuition
Since all the points lie on the line $y = -2x$, the direction in which they are spread out (the variance) is exactly the direction of that line. A vector $v$ that points along the line $y = -2x$ can be found by picking $x = 1$, which gives $y = -2$.
Therefore, the direction vector is: $v = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$
Method B: The Mathematical Approach (Covariance Matrix)
In technical terms, the direction of maximum variance is the eigenvector associated with the largest eigenvalue of the covariance matrix.
- Check if centered: The mean of both rows is $0$. The data is already centered.
- Compute $XX^T$: $$XX^T = \begin{pmatrix} 2 & -1 & -2 & 1 \\ -4 & 2 & 4 & -2 \end{pmatrix} \begin{pmatrix} 2 & -4 \\ -1 & 2 \\ -2 & 4 \\ 1 & -2 \end{pmatrix} = \begin{pmatrix} 10 & -20 \\ -20 & 40 \end{pmatrix}$$
- Find Eigenvalues ($\lambda$): Solve $\det(XX^T - \lambda I) = 0$: $$(10-\lambda)(40-\lambda) - 400 = 0 \implies \lambda^2 - 50\lambda = 0$$ The eigenvalues are $\lambda_1 = 50$ and $\lambda_2 = 0$.
- Find Eigenvector for $\lambda = 50$: $$\begin{pmatrix} -40 & -20 \\ -20 & -10 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 2x + y = 0 \implies y = -2x$$
4. Conclusion: Who is correct?
- Student 1: Suggested $[1, 0]^T$ (Horizontal). Incorrect.
- Student 2: Suggested $[0, 1]^T$ (Vertical). Incorrect.
- Student 3: Claimed both Student 1 and 2 are wrong. Correct.
The direction of maximum variance is $v = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$. Therefore, Student 3 is correct.
Topic: Dimensionality Reduction and the Relationship Between $XX^T$ and $X^TX$ (Click to Expand Solution)
Question
(2) A data scientist came across a dataset $x_1, x_2, \dots x_N$ where $x_i \in \mathbb{R}^{1024}$. Here $N = 20$. He wants to find the eigenvalues of a new matrix defined as follows: $$S = \sum_{i=1}^N x_ix_i^T$$ Observe that here $S \in \mathbb{R}^{1024 \times 1024}$. The data scientist has access to a piece of octave code containing a function $[P, D] = eig(A)$ which returns eigenvalues and eigenvectors of a square symmetric matrix $A$ as long as it has less than 32 rows. (Note that $eig(A)$ returns matrices $P \in \mathbb{R}^{n \times n}$ of eigenvectors and a diagonal matrix $D \in \mathbb{R}^{n \times n}$ of eigenvalues.)(a) How will you use the $eig(A)$ function to find eigenvalues of $S$ by overcoming the fact that $S$ has 1000 rows? Give a mathematical reasoning of how this can be achieved. [1.5 marks]
(b) How will you use the $eig(A)$ function to find eigenvectors of $S$ by overcoming the fact that $S$ has 1000 rows? Give a mathematical reasoning of how this can be achieved. [1.5 marks]
To understand this problem, let’s first clarify what is happening. We have 20 pieces of data (vectors), and each one is very "long" (1024 dimensions). The matrix $S$ is a Scatter Matrix. It is huge ($1024 \times 1024$), but it is built from only 20 vectors. This means the matrix is "rank-deficient"—it has a lot of information that is redundant or zero.
We are restricted by a computer function that can only handle small matrices (less than 32 rows). Since $20 < 32$, our goal is to turn this "big" problem into a "small" one using the number of samples ($N=20$) instead of the number of dimensions (1024).
Preliminary Setup
Let’s define a matrix $X$ where we stack our data vectors side-by-side:
$$X = [x_1, x_2, \dots, x_{20}]$$The size of $X$ is $1024 \times 20$. The matrix $S$ given in the problem can be written in matrix form as: $S = XX^T$.
(a) Finding the Eigenvalues
The Concept:
There is a fundamental rule in linear algebra: The non-zero eigenvalues of $XX^T$ (a big matrix) are exactly the same as the non-zero eigenvalues of $X^TX$ (a small matrix).
Mathematical Reasoning:
Instead of calculating $S = XX^T$ ($1024 \times 1024$), we calculate a smaller matrix $A = X^TX$. The size of $A$ is $20 \times 20$. Since 20 is less than 32, we can now use the restricted function: [P, D] = eig(A).
If $\lambda$ is an eigenvalue of $X^TX$ with eigenvector $v$, then:
$$X^TXv = \lambda v$$If we multiply both sides by $X$ from the left:
$$X(X^TXv) = X(\lambda v) \implies (XX^T)(Xv) = \lambda (Xv)$$This shows that $\lambda$ is also an eigenvalue for $XX^T$ (which is our matrix $S$).
How to do it:
- Construct $X$ by placing $x_i$ as columns.
- Compute $A = X^TX$.
- Run
[P, D] = eig(A). The diagonal elements of $D$ are the eigenvalues of $S$. (The remaining $1024 - 20 = 1004$ eigenvalues of $S$ will simply be zero).
(b) Finding the Eigenvectors
The Concept:
We can't get the "big" eigenvectors ($1024 \times 1$) directly from the "small" function. However, the math in part (a) gives us a hint: if $v$ is an eigenvector of the small matrix, then $(Xv)$ is an eigenvector of the big matrix.
Mathematical Reasoning:
From the equation $(XX^T)(Xv) = \lambda (Xv)$, we see that the vector $u = Xv$ is an eigenvector of $S$. The vector $v$ is a $20 \times 1$ column from the matrix $P$ we got from eig(A). When we multiply $X$ ($1024 \times 20$) by $v$ ($20 \times 1$), we get a vector $u$ ($1024 \times 1$), which is the correct size for the eigenvectors of $S$.
Note on Normalization: Eigenvectors are usually required to have a length of 1. Even if $v$ is normalized, $Xv$ might not be. We must divide $Xv$ by its magnitude to finish the job.
How to do it:
- Take the matrix $P$ (the eigenvectors of the small $20 \times 20$ matrix).
- Multiply $X$ by $P$: $U_{raw} = XP$.
- Each column in $U_{raw}$ is an eigenvector of $S$. To finalize them, normalize each column so its length equals 1.
Topic: Cholesky Decomposition and Eigenvalues of Triangular Matrices (Click to Expand Solution)
Question
Consider the positive definite matrix $A = \begin{bmatrix} 4 & 2 & 6 \\ 2 & 17 & 7 \\ 6 & 7 & 35 \end{bmatrix}$(i) Calculate a lower triangular matrix $L$ such that $A = LL^T$.
(ii) Calculate all the eigenvalues of $L$ derived in (i).
Answer
To solve this, let's first break down the terminology so the steps make sense.
- Positive Definite Matrix: For our purposes, this is a special kind of symmetric matrix that can always be broken down into the product of a lower triangular matrix and its transpose.
- Lower Triangular Matrix ($L$): This is a square matrix where all the entries above the main diagonal are zero. It looks like this: $$L = \begin{bmatrix} l_{11} & 0 & 0 \\ l_{21} & l_{22} & 0 \\ l_{31} & l_{32} & l_{33} \end{bmatrix}$$
- Transpose ($L^T$): This is the matrix $L$ flipped over its diagonal (rows become columns).
- Cholesky Decomposition ($A = LL^T$): This is the process of finding $L$.
(i) Calculate the lower triangular matrix $L$
We set $A$ equal to the product $LL^T$:
$$\begin{bmatrix} 4 & 2 & 6 \\ 2 & 17 & 7 \\ 6 & 7 & 35 \end{bmatrix} = \begin{bmatrix} l_{11} & 0 & 0 \\ l_{21} & l_{22} & 0 \\ l_{31} & l_{32} & l_{33} \end{bmatrix} \begin{bmatrix} l_{11} & l_{21} & l_{31} \\ 0 & l_{22} & l_{32} \\ 0 & 0 & l_{33} \end{bmatrix}$$By multiplying the matrices on the right, we get equations for each element of $A$. We solve them one column at a time:
Step 1: Solve for the first column of $L$
- $l_{11}^2 = 4 \implies \mathbf{l_{11} = 2}$
- $l_{21} \cdot l_{11} = 2 \implies l_{21} \cdot 2 = 2 \implies \mathbf{l_{21} = 1}$
- $l_{31} \cdot l_{11} = 6 \implies l_{31} \cdot 2 = 6 \implies \mathbf{l_{31} = 3}$
Step 2: Solve for the second column of $L$
- $l_{21}^2 + l_{22}^2 = 17$
- $1^2 + l_{22}^2 = 17 \implies l_{22}^2 = 16 \implies \mathbf{l_{22} = 4}$
- $l_{31}l_{21} + l_{32}l_{22} = 7$
- $(3)(1) + l_{32}(4) = 7 \implies 3 + 4l_{32} = 7 \implies 4l_{32} = 4 \implies \mathbf{l_{32} = 1}$
Step 3: Solve for the third column of $L$
- $l_{31}^2 + l_{32}^2 + l_{33}^2 = 35$
- $3^2 + 1^2 + l_{33}^2 = 35 \implies 9 + 1 + l_{33}^2 = 35 \implies l_{33}^2 = 25 \implies \mathbf{l_{33} = 5}$
Final Matrix $L$:
(ii) Calculate all the eigenvalues of $L$
What are Eigenvalues?
Usually, finding eigenvalues involves a complex calculation (solving $det(L - \lambda I) = 0$). However, there is a very helpful rule for Triangular Matrices (both upper and lower):
The eigenvalues of a triangular matrix are simply the entries on its main diagonal.
Looking at our matrix $L$:
The diagonal entries are $2, 4,$ and $5$.
The eigenvalues of $L$ are $\lambda_1 = 2$, $\lambda_2 = 4$, and $\lambda_3 = 5$.
Topic: Linear Algebra (Properties of Vector Norms) (Click to Expand Solution)
Question
Let $\mathbf{x} \in \mathbb{R}^n$. We define a function $f(\mathbf{x}) : \mathbb{R}^n \to \mathbb{Z}^+ \cup \{0\}$ as $f(\mathbf{x}) =$ total number of non-zero entries in $\mathbf{x}$. Assuming that $f(\mathbf{x})$ satisfies the triangle inequality property of norms, prove or disprove whether $f(\mathbf{x})$ satisfies the other two properties (i.e. absolutely homogeneous, positive definiteness) of norms.Answer
To understand this problem, let's first break down what $f(\mathbf{x})$ actually does and what these "norm properties" mean in plain English.
Understanding the Function
Imagine a vector $\mathbf{x} = (5, 0, -2)$.
The function $f(\mathbf{x})$ simply counts how many numbers in that vector are not zero.
In $(5, 0, -2)$, there are two non-zero numbers ($5$ and $-2$). So, $f(\mathbf{x}) = 2$.
This is often called the $L_0$ norm (though, as we will see, it isn't technically a true "norm").
1. Checking Positive Definiteness
The Concept: For a function to be "positive definite," it must satisfy two things:
- The "size" of the vector is only zero if the vector itself is the zero vector.
- The "size" of any other vector must be greater than zero.
Proof:
- If $\mathbf{x} = \mathbf{0}$: If every entry in the vector is zero, the count of non-zero entries is $0$. Thus, $f(\mathbf{0}) = 0$.
- If $f(\mathbf{x}) = 0$: If the count of non-zero entries is zero, it means there are no non-zero entries. Therefore, every entry must be zero, so $\mathbf{x} = \mathbf{0}$.
- If $\mathbf{x} \neq \mathbf{0}$: If the vector has at least one non-zero entry, the count $f(\mathbf{x})$ will be at least $1$. Since $1 > 0$, the function is always positive for non-zero vectors.
Conclusion: $f(\mathbf{x})$ satisfies Positive Definiteness.
2. Checking Absolute Homogeneity
The Concept: This property says that if you scale a vector by a factor (like doubling it), the "size" should scale by that same factor.
Mathematically: $f(\alpha \mathbf{x}) = |\alpha| \cdot f(\mathbf{x})$ for any number $\alpha$.
Disproof (Counter-example):
Let's test this with a simple example.
Let the vector $\mathbf{x} = (1, 0)$.
The count of non-zero entries is $f(\mathbf{x}) = 1$.
Now, let's scale the vector by $\alpha = 2$.
The new vector is $2\mathbf{x} = (2, 0)$.
The count of non-zero entries for the new vector is $f(2\mathbf{x}) = 1$.
Now, let's check the property:
$$f(2\mathbf{x}) \stackrel{?}{=} |2| \cdot f(\mathbf{x})$$ $$1 \stackrel{?}{=} 2 \cdot 1$$ $$1 \neq 2$$Conclusion: Scaling the vector does not scale the "count" of non-zero entries. Therefore, $f(\mathbf{x})$ does not satisfy Absolute Homogeneity.
Summary Table
| Property | Status | Reasoning |
|---|---|---|
| Positive Definiteness | Satisfied | Only the zero vector has zero non-zero entries. |
| Absolute Homogeneity | Not Satisfied | Scaling a non-zero number doesn't change the fact that it is non-zero. |
Topic: Linear Algebra (Symmetric Matrices and Positive Semi-Definiteness) (Click to Expand Solution)
Question
2) Let $\beta \in \mathbb{R}$ be an unknown constant and let $i = \sqrt{-1}$, a symbol commonly used in definition of complex numbers. Now consider a square matrix $A \in \mathbb{R}^{n \times n}$. It is given to you that $A$ can be factorized as $A = B^T B$ where $B \in \mathbb{R}^{n \times n}$. Let $x \in \mathbb{R}^n$ be an eigenvector of $A$ corresponding to an eigenvalue $\lambda$ i.e. $Ax = \lambda x$. It is also known that this eigenvalue $\lambda$ is of the form $\lambda = 7 + i\beta$.i) Is it possible to find the exact value of $\lambda$ by deriving the constant $\beta$ based on the given information? If yes, derive $\beta$. If no, provide proper reason(s) as to why it is not possible to find $\beta$ with the given information. (1 mark)
ii) Is the matrix $A$ a positive semi-definite matrix? If yes prove it. If no, give proper reason why it cannot be a positive semi-definite matrix. (1 mark)
Answer
To solve this, let's first break down the terminology for a better understanding:
- Eigenvalue ($\lambda$): A special number associated with a matrix that describes how much a vector is "stretched" during a linear transformation.
- Symmetric Matrix: A matrix that is equal to its transpose ($A = A^T$).
- Complex Number ($7 + i\beta$): A number with a real part (7) and an imaginary part ($i\beta$).
- Positive Semi-Definite (PSD): A property of a matrix where it never "flips" a vector in a way that results in a negative dot product with itself ($x^T A x \ge 0$).
i) Deriving the constant $\beta$
Yes, it is possible to find the exact value of $\lambda$ because we can determine that $\beta = 0$.
Reasoning:
- Symmetry of $A$: We are told $A = B^T B$. If we take the transpose of $A$, we get: $$A^T = (B^T B)^T = B^T (B^T)^T = B^T B = A$$ Because $A^T = A$, the matrix $A$ is a symmetric matrix.
- The Real Eigenvalue Theorem: A fundamental property in linear algebra states that all eigenvalues of a real symmetric matrix must be real numbers.
- Solving for $\beta$: The problem states $\lambda = 7 + i\beta$. For $\lambda$ to be a real number (which it must be, since $A$ is symmetric), the imaginary part must be zero. $$i\beta = 0 \implies \beta = 0$$
Result: Therefore, $\lambda = 7 + 0i = 7$.
ii) Is $A$ a Positive Semi-Definite (PSD) matrix?
Yes, the matrix $A$ is a positive semi-definite matrix.
Proof:
A matrix $A$ is positive semi-definite if, for any non-zero vector $x$, the quadratic form $x^T A x$ is greater than or equal to zero. Let's test this using the given factorization $A = B^T B$:
- Substitute $A$ into the quadratic form: $$x^T A x = x^T (B^T B) x$$
- Use the property of transposes $(Bx)^T = x^T B^T$ to regroup the terms: $$x^T B^T B x = (Bx)^T (Bx)$$
- Let the resulting vector $Bx$ be called $y$. The expression becomes: $$y^T y$$
- Magnitude Squared: The term $y^T y$ is the dot product of a vector with itself, which is the square of its magnitude (norm): $$y^T y = \|y\|^2$$
- Non-negativity: Since the square of a magnitude (or any real number) can never be negative: $$\|y\|^2 \ge 0$$
Since $x^T A x \ge 0$ for all $x$, the matrix $A$ is positive semi-definite. Additionally, we already found an eigenvalue $\lambda = 7$. Since $7 > 0$, it is consistent with the property that all eigenvalues of a PSD matrix must be $\ge 0$.
Topic: Singular Value Decomposition (SVD) (Click to Expand Solution)
Singular Value Decomposition (SVD) is a way of breaking down a matrix (a grid of numbers) into three simpler parts. Think of it like "factoring" a number (like $12 = 2 \times 2 \times 3$), but for matrices. It helps us understand the structure of data, reduce noise, and compress information.
The Question
Consider a square matrix $\mathbf{B} \in \mathbb{R}^{2 \times 2}$ defined as follows: $$\mathbf{B} = \begin{bmatrix} 3 & 0 \\ 4 & 0 \end{bmatrix}$$ Assume that $\mathbf{B} = \mathbf{U \Sigma V^T}$ is the singular value decomposition of $\mathbf{B}$.i) Derive the largest singular value $\sigma_1$ of this matrix $\mathbf{B}$. (1 mark)
ii) Derive left singular vector $\mathbf{u}_1$ of this matrix $\mathbf{B}$ corresponding to largest singular value $\sigma_1$. (1 mark)
iii) Derive right singular vector $\mathbf{v}_1$ of this matrix $\mathbf{B}$ corresponding to largest singular value $\sigma_1$. (1 mark)
iv) Derive the rank-1 approximation matrix $\mathbf{B}_1 = \sigma_1(\mathbf{u}_1 \mathbf{v}_1^\mathbf{T})$. (1 mark)
Detailed Answer
Before we start, let's define the parts of SVD ($B = U \Sigma V^T$):
- $\Sigma$ (Singular Values): These numbers tell us the "strength" or importance of specific directions.
- $V$ (Right Singular Vectors): These are the directions in the "input" space.
- $U$ (Left Singular Vectors): These are the directions in the "output" space.
i) Derive the largest singular value $\sigma_1$
To find the singular values of $B$, we first need to look at the product of the matrix with its transpose ($B^T B$). The singular values are the square roots of the eigenvalues of this product.
Step 1: Find $B^T B$
The transpose $B^T$ swaps rows and columns: $B^T = \begin{bmatrix} 3 & 4 \\ 0 & 0 \end{bmatrix}$.
Step 2: Find the eigenvalues ($\lambda$)
Since the matrix $\begin{bmatrix} 25 & 0 \\ 0 & 0 \end{bmatrix}$ is diagonal, the eigenvalues are simply the numbers on the diagonal: $\lambda_1 = 25$ and $\lambda_2 = 0$.
Step 3: Calculate the singular value ($\sigma$)
The singular values are the square roots of the eigenvalues.
$$\sigma_1 = \sqrt{\lambda_1} = \sqrt{25} = 5$$
Result: The largest singular value is $\sigma_1 = 5$.
iii) Derive the right singular vector $\mathbf{v}_1$
(Note: It is easier to solve for $\mathbf{v}_1$ before $\mathbf{u}_1$ because $\mathbf{v}_1$ is the eigenvector of $B^T B$.)
The right singular vector $\mathbf{v}_1$ is the eigenvector associated with the largest eigenvalue ($\lambda_1 = 25$). We solve $(B^T B - \lambda_1 I)\mathbf{v}_1 = 0$:
$$\begin{bmatrix} 25-25 & 0 \\ 0 & 0-25 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies \begin{bmatrix} 0 & 0 \\ 0 & -25 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$This gives us: $0x + 0y = 0$ (always true) and $-25y = 0 \implies y = 0$. Normalized vector (length 1, choosing $x=1$):
Result: $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
ii) Derive the left singular vector $\mathbf{u}_1$
Formula to find the left singular vector: $\mathbf{u}_1 = \frac{1}{\sigma_1} B \mathbf{v}_1$. Substituting our values:
$$\mathbf{u}_1 = \frac{1}{5} \begin{bmatrix} 3 & 0 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} (3\times1 + 0\times0) \\ (4\times1 + 0\times0) \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 0.6 \\ 0.8 \end{bmatrix}$$ Result: $\mathbf{u}_1 = \begin{bmatrix} 0.6 \\ 0.8 \end{bmatrix}$.iv) Derive the rank-1 approximation matrix $\mathbf{B}_1$
A "rank-1 approximation" rebuilds the original matrix using only the most important direction ($\sigma_1, \mathbf{u}_1, \mathbf{v}_1$).
Step 1: Calculate $\mathbf{u}_1 \mathbf{v}_1^T$ (outer product)
$$\mathbf{u}_1 \mathbf{v}_1^T = \begin{bmatrix} 0.6 \\ 0.8 \end{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} = \begin{bmatrix} 0.6\times1 & 0.6\times0 \\ 0.8\times1 & 0.8\times0 \end{bmatrix} = \begin{bmatrix} 0.6 & 0 \\ 0.8 & 0 \end{bmatrix}$$Step 2: Multiply by $\sigma_1 = 5$
$$\mathbf{B}_1 = 5 \times \begin{bmatrix} 0.6 & 0 \\ 0.8 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 4 & 0 \end{bmatrix}$$ Result: $\mathbf{B}_1 = \begin{bmatrix} 3 & 0 \\ 4 & 0 \end{bmatrix}$.Topic: Diagonalization, Matrix Trace, and Cayley-Hamilton Theorem (Click to Expand Solution)
This response breaks down the linear algebra problems provided in the image, explaining the concepts from the ground up for someone new to the subject.
Question 1: Diagonalization and Matrix Trace
Following matrix $\mathbf{Q}$, where $\alpha, \beta, \gamma$ and $\omega$ are real numbers and solutions of the polynomial equation $\sum_{i=0}^{2024} c_ix^{2024-i} = 0$, has 3 linearly independent eigenvectors (as columns) in $\mathbf{P}$. Find the sum of all the elements of $\mathbf{P}^{-1}\mathbf{QP}$. $$\mathbf{Q} = \begin{bmatrix} 3 & \gamma & \alpha + \beta \\ \gamma & 0 & \omega \\ \alpha + \beta & \omega & 3 \end{bmatrix}$$Terminology for Beginners
- Eigenvectors & $\mathbf{P}$: Think of eigenvectors as special directions where the matrix $\mathbf{Q}$ only scales a vector rather than rotating it. If a $3 \times 3$ matrix has 3 "linearly independent" (unique/non-overlapping) eigenvectors, we can pack them into a new matrix called $\mathbf{P}$.
- Diagonalization ($\mathbf{P}^{-1}\mathbf{QP}$): This specific operation $(\mathbf{P}^{-1}\mathbf{QP})$ is called diagonalization. It converts the original matrix $\mathbf{Q}$ into a diagonal matrix (let's call it $\mathbf{D}$). In a diagonal matrix, all numbers are zero except for those on the main diagonal.
- Eigenvalues: The numbers on the diagonal of $\mathbf{D}$ are called the eigenvalues of $\mathbf{Q}$.
- Trace: The "Trace" of a matrix is simply the sum of the numbers on its main diagonal.
Justification & Answer
The Shortcut: A fundamental property in linear algebra is that the sum of the eigenvalues of a matrix is always equal to the Trace of the original matrix.
The Result of $\mathbf{P}^{-1}\mathbf{QP}$: Since $\mathbf{P}$ consists of 3 independent eigenvectors, $\mathbf{P}^{-1}\mathbf{QP} = \mathbf{D}$, where:
$$\mathbf{D} = \begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix}$$The "sum of all elements" of this matrix is just $\lambda_1 + \lambda_2 + \lambda_3$.
Calculation: We find the sum by looking at the diagonal of the original matrix $\mathbf{Q}$:
$$\text{Sum} = \text{Trace}(\mathbf{Q}) = 3 + 0 + 3 = 6$$Note: The information about the polynomial equation and the variables $\alpha, \beta, \gamma, \omega$ are distractors. Regardless of their values, the sum of the elements in the diagonalized matrix depends only on the diagonal entries of $\mathbf{Q}$.
Final Answer: 6
Question 2: Powers of a Matrix (Cayley-Hamilton Theorem)
Compute $\mathbf{A}^9$ without using the conventional matrix multiplication. Show all the computations. $$\mathbf{A} = \begin{bmatrix} 3 & 1 \\ 1 & -3 \end{bmatrix}$$Terminology for Beginners
- Characteristic Equation: A special algebraic equation unique to every square matrix.
- Cayley-Hamilton Theorem: This theorem states that every matrix "satisfies its own characteristic equation." It allows us to turn a matrix power problem (like $\mathbf{A}^9$) into a much simpler algebra problem.
- Identity Matrix ($\mathbf{I}$): The matrix version of the number "1". For a $2 \times 2$ matrix, $\mathbf{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Justification & Answer
Step 1: Find the Characteristic Equation
For a $2 \times 2$ matrix, the equation is $\lambda^2 - \text{Tr}(\mathbf{A})\lambda + \text{Det}(\mathbf{A}) = 0$.
Trace (sum of diagonal): $3 + (-3) = 0$
Determinant ($ad - bc$): $(3 \times -3) - (1 \times 1) = -9 - 1 = -10$
Equation: $\lambda^2 - 0\lambda - 10 = 0 \Rightarrow \lambda^2 - 10 = 0$
Step 2: Apply Cayley-Hamilton Theorem
According to the theorem, we can replace $\lambda$ with the matrix $\mathbf{A}$ and the constant with the Identity matrix $\mathbf{I}$:
This tells us that squaring matrix $\mathbf{A}$ is the same as multiplying the Identity matrix by 10.
Step 3: Compute $\mathbf{A}^9$ using Algebra
We can break $\mathbf{A}^9$ down into smaller parts:
Substitute $\mathbf{A}^2 = 10\mathbf{I}$:
$$\mathbf{A}^9 = \mathbf{A} \cdot (10\mathbf{I})^4 = \mathbf{A} \cdot 10^4 \cdot \mathbf{I}^4$$Since $\mathbf{I}^4 = \mathbf{I}$ and any matrix times $\mathbf{I}$ is itself:
$$\mathbf{A}^9 = 10,000 \cdot \mathbf{A}$$Step 4: Final Calculation
$$\mathbf{A}^9 = 10,000 \begin{bmatrix} 3 & 1 \\ 1 & -3 \end{bmatrix} = \begin{bmatrix} 30,000 & 10,000 \\ 10,000 & -30,000 \end{bmatrix}$$Topic: Linear Algebra — Elementary Matrices and LU Decomposition (Click to Expand Solution)
Question
(5) For the matrix below construct an elementary matrix for every elementary row operation that is performed on A and write the decomposition of A as LU, where L is a lower triangular matrix and U is an upper triangular matrix (2 Marks) $$A = \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix}$$Answer
To solve this problem, we need to break down some terminology first. If you are new to this, think of LU Decomposition as "factoring" a matrix, much like you factor the number $12$ into $3 \times 4$. Here, we are factoring matrix $A$ into two specific types of matrices: $L$ and $U$.
1. Key Terminology
- Upper Triangular Matrix ($U$): A matrix where all entries below the main diagonal are zero.
- Lower Triangular Matrix ($L$): A matrix where all entries above the main diagonal are zero.
- Elementary Row Operation: A basic step used to change a matrix (like subtracting one row from another).
- Elementary Matrix ($E$): A matrix that represents exactly one row operation. If you multiply $E \times A$, it performs that operation on $A$.
Step 1: Perform Row Operations to find $U$
Our goal is to turn $A$ into an upper triangular matrix ($U$). To do this, we need to turn the bottom-left number ($1$) into a $0$.
Matrix $A$:
$$\begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix}$$To turn the $1$ (in Row 2) into $0$ using Row 1, we perform the following operation:
Operation: $R_2 \rightarrow R_2 - \frac{1}{2}R_1$
Calculation:
- New $R_{2,1} = 1 - (\frac{1}{2} \times 2) = 0$
- New $R_{2,2} = 4 - (\frac{1}{2} \times 1) = 3.5 \text{ (or } \frac{7}{2})$
So, our Upper Triangular Matrix ($U$) is:
$$U = \begin{bmatrix} 2 & 1 \\ 0 & 3.5 \end{bmatrix}$$Step 2: Construct the Elementary Matrix ($E$)
An elementary matrix is created by performing the exact same row operation on an Identity Matrix ($I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$).
Applying $R_2 \rightarrow R_2 - \frac{1}{2}R_1$ to the Identity Matrix:
- Row 1 stays: $\begin{bmatrix} 1 & 0 \end{bmatrix}$
- Row 2 becomes: $\begin{bmatrix} 0 - \frac{1}{2}(1) & 1 - \frac{1}{2}(0) \end{bmatrix} = \begin{bmatrix} -1/2 & 1 \end{bmatrix}$
The Elementary Matrix ($E$) is:
$$E = \begin{bmatrix} 1 & 0 \\ -1/2 & 1 \end{bmatrix}$$Step 3: Construct the Lower Triangular Matrix ($L$)
The matrix $L$ is effectively the "inverse" of the operations we performed. If $E$ subtracted $\frac{1}{2}$ of Row 1, $L$ will record that we need to add it back to get the original matrix.
Mathematically, $L = E^{-1}$. For a simple $2 \times 2$ elementary matrix like this, you just change the sign of the number we added/subtracted:
$$L = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix}$$Step 4: Final LU Decomposition
We have now decomposed $A$ into $L \times U$.
$A = LU$
$$\begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 0 & 3.5 \end{bmatrix}$$If you multiply the first row of $L$ by the columns of $U$:
- $(1 \times 2) + (0 \times 0) = 2$
- $(1 \times 1) + (0 \times 3.5) = 1$
And the second row:
- $(1/2 \times 2) + (1 \times 0) = 1$
- $(1/2 \times 1) + (1 \times 3.5) = 0.5 + 3.5 = 4$
The result is $\begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix}$, which is our original matrix A.