Lecture 13: Quiz 1 Review

Author

Chao Ma

Published

October 26, 2025

1. Dimension of Subspaces

Problem: Suppose $u, v, w$ are non-zero vectors in $\mathbb{R}^7$. They span a subspace of $\mathbb{R}^7$. What are the possible dimensions?

Answer: 1, 2, or 3

Explanation:

Minimum dimension: 1 (if all vectors are scalar multiples of each other)
Maximum dimension: 3 (if all three vectors are linearly independent)
Middle case: 2 (if exactly two are independent)
Cannot be 0 (all vectors are non-zero)
Cannot exceed 3 (only three vectors available)

2. Null Space Dimensions

Problem: We have $A$, a $5 \times 3$ matrix with rank $r = 3$. What is the null space?

Answer: $N(A) = \{\mathbf{0}\}$ (just the zero vector)

Calculation:

\[ \dim(N(A)) = n - r = 3 - 3 = 0 \]

Interpretation: Since the rank equals the number of columns, all columns are independent. The only solution to $Ax = \mathbf{0}$ is $x = \mathbf{0}$.

3. Echelon Forms with Block Matrices

Problem 3a: Matrix B

Given:

\[ B = \begin{bmatrix} u \\ 2u \end{bmatrix} \]

where $u$ is a row vector.

Echelon form:

\[ \begin{bmatrix} u \\ \mathbf{0} \end{bmatrix} \]

Explanation: Row 2 is a multiple of row 1, so elimination makes row 2 become zero.

Problem 3b: Matrix C

Given:

\[ C = \begin{bmatrix} u & u \\ u & 0 \end{bmatrix} \]

where $u$ is a $5 \times 3$ matrix with rank 3.

Elimination steps:

\[ \begin{bmatrix} u & u \\ u & 0 \end{bmatrix} \rightarrow \begin{bmatrix} u & u \\ 0 & -u \end{bmatrix} \rightarrow \begin{bmatrix} u & 0 \\ 0 & -u \end{bmatrix} \rightarrow \begin{bmatrix} u & 0 \\ 0 & u \end{bmatrix} \]

Step-by-step:

Subtract row 1 from row 2: eliminates bottom-left block
Subtract column 1 from column 2: eliminates top-right block
Multiply row 2 by -1: normalize

Rank of C:

$u$ is $5 \times 3$ with rank 3
$C$ is $10 \times 6$ (two $5 \times 3$ blocks side by side)
Both $u$ blocks contribute full rank
Rank of C = 6

Dimension of left null space:

\[ \dim(N(C^T)) = m - r = 10 - 6 = 4 \]

4. Complete Solution to Linear Systems

Problem: Given

\[ Ax = \begin{bmatrix} 2 \\ 4 \\ 2 \end{bmatrix}, \quad x = \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} + c\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + d\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \]

Part a: Dimension of Row Space

Answer: 1

Reasoning:

The null space has dimension 2 (two free variables: $c$ and $d$)
$\dim(N(A)) = n - r = 2$
Since $n = 3$ (three columns), $r = 3 - 2 = 1$
Row space dimension = rank = 1

Part b: Find Matrix A

Strategy: Use the particular solution and null space vectors.

Column 1:

\[ A\begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \\ 2 \end{bmatrix} \implies \text{Column 1} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \]

Column 3:

\[ A\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \mathbf{0} \implies \text{Column 3} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

Column 2:

\[ A\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \mathbf{0} \implies \text{Column 1} + \text{Column 2} = \mathbf{0} \]

Therefore, Column 2 = $-$Column 1:

\[ \text{Column 2} = \begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix} \]

Matrix A:

\[ A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & -2 & 0 \\ 1 & -1 & 0 \end{bmatrix} \]

Part c: Which b Can Be Solved?

Answer: $b$ must be a multiple of $\begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix}$

Reasoning:

Column space of $A$ is spanned by $\begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix}$
All columns are multiples of this vector
$Ax = b$ has a solution only if $b \in C(A)$

Therefore: $b = c\begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix}$ for some scalar $c$.

5. Properties of Null Spaces

Problem: If the null space is $\{\mathbf{0}\}$ and $A$ is square, what is the null space of $A^T$?

Answer: $N(A^T) = \{\mathbf{0}\}$

Proof:

$A$ is $n \times n$ (square)
$N(A) = \{\mathbf{0}\}$ means $\dim(N(A)) = 0$
Therefore: $\text{rank}(A) = n - 0 = n$
Since $A$ is square with full rank, $A$ is invertible
For $A^T$: $\dim(N(A^T)) = m - r = n - n = 0$

6. Subspaces of Matrix Spaces

Problem: Consider the space of all $5 \times 5$ matrices (dimension 25). Do the invertible $5 \times 5$ matrices form a subspace?

Answer: No

Reason: Not closed under addition.

Counterexample:

\[ A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \]

Both $A$ and $B$ are invertible, but:

\[ A + B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

is not invertible.

7. Nilpotent Matrices

Problem: If $B^2 = 0$, must $B = \mathbf{0}$?

Answer: False

Counterexample:

\[ B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \]

Verification:

\[ B^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

Explanation: Rows times columns can all be zero even when the matrix itself is non-zero.

8. Solvability of Square Systems

Problem: If an $n \times n$ matrix has rank $n$, does $Ax = b$ always have a solution?

Answer: Yes

Proof:

Rank $n$ for an $n \times n$ matrix means $A$ is invertible
Therefore: $x = A^{-1}b$ always exists
Every $b \in \mathbb{R}^n$ is in the column space

9. Null Space from Matrix Factorization

Problem: Given

\[ B = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & -1 & 2 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

Without multiplication, find the basis for $N(B)$.

Analysis

Dimensions:

Left matrix: $3 \times 3$, full rank (invertible)
Right matrix: $3 \times 4$, rank 2
$B$: $3 \times 4$

Key insight: $\operatorname{rank}(AB) \leq \min(\operatorname{rank}(A), \operatorname{rank}(B))$

Therefore: $\operatorname{rank}(B) = 2$

Null space dimension:

\[ \dim(N(B)) = n - r = 4 - 2 = 2 \]

Finding basis: Since the left matrix is invertible, $N(B) = N(\text{right matrix})$

From the right matrix in RREF:

Pivot columns: 1, 2
Free variables: columns 3, 4

Basis for N(B):

\[ c_1\begin{bmatrix} 1 \\ -1 \\ 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} -2 \\ 1 \\ 0 \\ 1 \end{bmatrix} \]

Verification:

Column 3 = $1 \cdot$Column 1 + $(-1) \cdot$Column 2
Column 4 = $(-2) \cdot$Column 1 + $1 \cdot$Column 2

Solving Bx = b

Problem: Solve $Bx = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$

Analysis:

\[ B_{\text{col1}} = [1, 0, 1]^T \\ B_{\text{col2}} = [1, 1, 0]^T \\ B_{\text{col3}} = [0, 1, -1]^T \\ B_{\text{col4}} = [1, -1, 2]^T \]

Particular solution: $x_p = \begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}$

Complete solution:

\[ x = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + c_1\begin{bmatrix} 1 \\ -1 \\ 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} -2 \\ 1 \\ 0 \\ 1 \end{bmatrix} \]

10. Conceptual Questions

Question 2: If m = n, Are Row Space and Column Space the Same?

Answer: No, not in general

When true: Only when $A$ is symmetric ($A = A^T$)

Counterexample:

\[ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \]

Column space: span of $\begin{bmatrix}1 \\ 0\end{bmatrix}$
Row space: span of $\begin{bmatrix}1 \\ 0\end{bmatrix}$

(In this case they happen to be the same, but that’s coincidental)

Better counterexample:

\[ A = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \]

Column space: $\{c\begin{bmatrix}1 \\ 0\end{bmatrix}\}$ (in $\mathbb{R}^2$)
Row space: $\{c\begin{bmatrix}1 \\ 1\end{bmatrix}\}$ (in $\mathbb{R}^2$)

Different vectors!

Question 3: If A and B Have the Same Four Subspaces, Is A a Multiple of B?

Answer: No

Counterexample: If $A$ and $B$ are both full rank $n \times n$ matrices:

They both have column space = $\mathbb{R}^n$
They both have null space = $\{\mathbf{0}\}$
They both have row space = $\mathbb{R}^n$
They both have left null space = $\{\mathbf{0}\}$

But $A$ and $B$ can be completely different matrices (e.g., different invertible matrices).

Question 4: If I Change Two Rows of A, Which Subspaces Stay the Same?

Answer:

$N(A)$ (null space)
$C(A^T)$ (row space)

Explanation:

Row operations preserve the null space
Row operations preserve the row space (just produce different linear combinations)
Column space and left null space will generally change

Question 5: Can Vector [1, 2, 3] Be in Both Null Space and Row Space?

Answer: No (assuming non-zero matrix)

Reason:

If $v$ is in the null space: $Av = \mathbf{0}$
If $v$ is also a row of $A$: then $Av$ includes the dot product $v \cdot v = \|v\|^2 > 0$
This is a contradiction

Key insight: Null space and row space are orthogonal complements in $\mathbb{R}^n$.

Source: MIT 18.06SC Linear Algebra, Lecture 13

--- title: "Lecture 13: Quiz 1 Review" author: "Chao Ma" date: "2025-10-26" --- {{< video https://www.youtube.com/watch?v=l88D4r74gtM >}} --- ## 1. Dimension of Subspaces **Problem**: Suppose $u, v, w$ are non-zero vectors in $\mathbb{R}^7$. They span a subspace of $\mathbb{R}^7$. What are the possible dimensions? **Answer**: 1, 2, or 3 **Explanation**: - **Minimum dimension**: 1 (if all vectors are scalar multiples of each other) - **Maximum dimension**: 3 (if all three vectors are linearly independent) - **Middle case**: 2 (if exactly two are independent) - Cannot be 0 (all vectors are non-zero) - Cannot exceed 3 (only three vectors available) --- ## 2. Null Space Dimensions **Problem**: We have $A$, a $5 \times 3$ matrix with rank $r = 3$. What is the null space? **Answer**: $N(A) = \{\mathbf{0}\}$ (just the zero vector) **Calculation**: $$ \dim(N(A)) = n - r = 3 - 3 = 0 $$ **Interpretation**: Since the rank equals the number of columns, all columns are independent. The only solution to $Ax = \mathbf{0}$ is $x = \mathbf{0}$. --- ## 3. Echelon Forms with Block Matrices ### Problem 3a: Matrix B **Given**: $$ B = \begin{bmatrix} u \\ 2u \end{bmatrix} $$ where $u$ is a row vector. **Echelon form**: $$ \begin{bmatrix} u \\ \mathbf{0} \end{bmatrix} $$ **Explanation**: Row 2 is a multiple of row 1, so elimination makes row 2 become zero. --- ### Problem 3b: Matrix C **Given**: $$ C = \begin{bmatrix} u & u \\ u & 0 \end{bmatrix} $$ where $u$ is a $5 \times 3$ matrix with rank 3. **Elimination steps**: $$ \begin{bmatrix} u & u \\ u & 0 \end{bmatrix} \rightarrow \begin{bmatrix} u & u \\ 0 & -u \end{bmatrix} \rightarrow \begin{bmatrix} u & 0 \\ 0 & -u \end{bmatrix} \rightarrow \begin{bmatrix} u & 0 \\ 0 & u \end{bmatrix} $$ **Step-by-step**: 1. Subtract row 1 from row 2: eliminates bottom-left block 2. Subtract column 1 from column 2: eliminates top-right block 3. Multiply row 2 by -1: normalize **Rank of C**: - $u$ is $5 \times 3$ with rank 3 - $C$ is $10 \times 6$ (two $5 \times 3$ blocks side by side) - Both $u$ blocks contribute full rank - **Rank of C = 6** **Dimension of left null space**: $$ \dim(N(C^T)) = m - r = 10 - 6 = 4 $$ --- ## 4. Complete Solution to Linear Systems **Problem**: Given $$ Ax = \begin{bmatrix} 2 \\ 4 \\ 2 \end{bmatrix}, \quad x = \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} + c\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + d\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$ ### Part a: Dimension of Row Space **Answer**: 1 **Reasoning**: - The null space has dimension 2 (two free variables: $c$ and $d$) - $\dim(N(A)) = n - r = 2$ - Since $n = 3$ (three columns), $r = 3 - 2 = 1$ - Row space dimension = rank = 1 --- ### Part b: Find Matrix A **Strategy**: Use the particular solution and null space vectors. **Column 1**: $$ A\begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \\ 2 \end{bmatrix} \implies \text{Column 1} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} $$ **Column 3**: $$ A\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \mathbf{0} \implies \text{Column 3} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$ **Column 2**: $$ A\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \mathbf{0} \implies \text{Column 1} + \text{Column 2} = \mathbf{0} $$ Therefore, Column 2 = $-$Column 1: $$ \text{Column 2} = \begin{bmatrix} -1 \\ -2 \\ -1 \end{bmatrix} $$ **Matrix A**: $$ A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & -2 & 0 \\ 1 & -1 & 0 \end{bmatrix} $$ --- ### Part c: Which b Can Be Solved? **Answer**: $b$ must be a multiple of $\begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix}$ **Reasoning**: - Column space of $A$ is spanned by $\begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix}$ - All columns are multiples of this vector - $Ax = b$ has a solution only if $b \in C(A)$ Therefore: $b = c\begin{bmatrix}1 \\ 2 \\ 1\end{bmatrix}$ for some scalar $c$. --- ## 5. Properties of Null Spaces **Problem**: If the null space is $\{\mathbf{0}\}$ and $A$ is square, what is the null space of $A^T$? **Answer**: $N(A^T) = \{\mathbf{0}\}$ **Proof**: - $A$ is $n \times n$ (square) - $N(A) = \{\mathbf{0}\}$ means $\dim(N(A)) = 0$ - Therefore: $\text{rank}(A) = n - 0 = n$ - Since $A$ is square with full rank, $A$ is invertible - For $A^T$: $\dim(N(A^T)) = m - r = n - n = 0$ --- ## 6. Subspaces of Matrix Spaces **Problem**: Consider the space of all $5 \times 5$ matrices (dimension 25). Do the invertible $5 \times 5$ matrices form a subspace? **Answer**: No **Reason**: Not closed under addition. **Counterexample**: $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} $$ Both $A$ and $B$ are invertible, but: $$ A + B = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $$ is not invertible. --- ## 7. Nilpotent Matrices **Problem**: If $B^2 = 0$, must $B = \mathbf{0}$? **Answer**: False **Counterexample**: $$ B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$ **Verification**: $$ B^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $$ **Explanation**: Rows times columns can all be zero even when the matrix itself is non-zero. --- ## 8. Solvability of Square Systems **Problem**: If an $n \times n$ matrix has rank $n$, does $Ax = b$ always have a solution? **Answer**: Yes **Proof**: - Rank $n$ for an $n \times n$ matrix means $A$ is invertible - Therefore: $x = A^{-1}b$ always exists - Every $b \in \mathbb{R}^n$ is in the column space --- ## 9. Null Space from Matrix Factorization **Problem**: Given $$ B = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & -1 & 2 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$ **Without multiplication**, find the basis for $N(B)$. ### Analysis **Dimensions**: - Left matrix: $3 \times 3$, full rank (invertible) - Right matrix: $3 \times 4$, rank 2 - $B$: $3 \times 4$ **Key insight**: $\operatorname{rank}(AB) \leq \min(\operatorname{rank}(A), \operatorname{rank}(B))$ Therefore: $\operatorname{rank}(B) = 2$ **Null space dimension**: $$ \dim(N(B)) = n - r = 4 - 2 = 2 $$ **Finding basis**: Since the left matrix is invertible, $N(B) = N(\text{right matrix})$ From the right matrix in RREF: - Pivot columns: 1, 2 - Free variables: columns 3, 4 **Basis for N(B)**: $$ c_1\begin{bmatrix} 1 \\ -1 \\ 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} -2 \\ 1 \\ 0 \\ 1 \end{bmatrix} $$ **Verification**: - Column 3 = $1 \cdot$Column 1 + $(-1) \cdot$Column 2 - Column 4 = $(-2) \cdot$Column 1 + $1 \cdot$Column 2 --- ### Solving Bx = b **Problem**: Solve $Bx = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$ **Analysis**: $$ B_{\text{col1}} = [1, 0, 1]^T \\ B_{\text{col2}} = [1, 1, 0]^T \\ B_{\text{col3}} = [0, 1, -1]^T \\ B_{\text{col4}} = [1, -1, 2]^T $$ **Particular solution**: $x_p = \begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}$ **Complete solution**: $$ x = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + c_1\begin{bmatrix} 1 \\ -1 \\ 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} -2 \\ 1 \\ 0 \\ 1 \end{bmatrix} $$ --- ## 10. Conceptual Questions ### Question 1: Do A and -A Share the Same Four Fundamental Subspaces? **Answer**: Yes **Proof**: - $C(A) = C(-A)$ (columns are just scaled by -1) - $N(A) = N(-A)$ (if $Ax = \mathbf{0}$, then $(-A)x = \mathbf{0}$) - $C(A^T) = C((-A)^T)$ (row space argument) - $N(A^T) = N((-A)^T)$ (left null space argument) --- ### Question 2: If m = n, Are Row Space and Column Space the Same? **Answer**: No, not in general **When true**: Only when $A$ is symmetric ($A = A^T$) **Counterexample**: $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$ - Column space: span of $\begin{bmatrix}1 \\ 0\end{bmatrix}$ - Row space: span of $\begin{bmatrix}1 \\ 0\end{bmatrix}$ (In this case they happen to be the same, but that's coincidental) Better counterexample: $$ A = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} $$ - Column space: $\{c\begin{bmatrix}1 \\ 0\end{bmatrix}\}$ (in $\mathbb{R}^2$) - Row space: $\{c\begin{bmatrix}1 \\ 1\end{bmatrix}\}$ (in $\mathbb{R}^2$) Different vectors! --- ### Question 3: If A and B Have the Same Four Subspaces, Is A a Multiple of B? **Answer**: No **Counterexample**: If $A$ and $B$ are both full rank $n \times n$ matrices: - They both have column space = $\mathbb{R}^n$ - They both have null space = $\{\mathbf{0}\}$ - They both have row space = $\mathbb{R}^n$ - They both have left null space = $\{\mathbf{0}\}$ But $A$ and $B$ can be completely different matrices (e.g., different invertible matrices). --- ### Question 4: If I Change Two Rows of A, Which Subspaces Stay the Same? **Answer**: - **$N(A)$** (null space) - **$C(A^T)$** (row space) **Explanation**: - Row operations preserve the null space - Row operations preserve the row space (just produce different linear combinations) - Column space and left null space will generally change --- ### Question 5: Can Vector [1, 2, 3] Be in Both Null Space and Row Space? **Answer**: No (assuming non-zero matrix) **Reason**: - If $v$ is in the null space: $Av = \mathbf{0}$ - If $v$ is also a row of $A$: then $Av$ includes the dot product $v \cdot v = \|v\|^2 > 0$ - This is a contradiction **Key insight**: Null space and row space are orthogonal complements in $\mathbb{R}^n$. --- *Source: MIT 18.06SC Linear Algebra, Lecture 13*