Chapter 5: Matrices

Topics: matrix operations, special matrices, rank and column space, null space and rank-nullity

Problems below are collected from past exams; each links back to its full exam. Solutions are in the dropdowns.

Problems

FA25 MT2 · Problem 1
FA25 MT2 · Problem 2
FA25 MT2 · Problem 3
FA25 Final · Problem 5
WN26 MT2 · Problem 2
WN26 MT2 · Problem 3
WN26 MT2 · Problem 5
WN26 Final · Problem 5
SP26 MT2 · Problem 1
SP26 MT2 · Problem 2
SP26 MT2 · Problem 3
SP26 Final · Problem 5

FA25 MT2 · Problem 1: Getting Started 12 pts

3 pts Let $A = \begin{bmatrix} 2 & 4 \\ -3 & -7 \end{bmatrix}$ . Find $\text{det}(A)$ , the determinant of $$A$$ . Give your answer as an integer.

$\text{det}(A) = \_\_\_\_\_\_$

Solution

The determinant of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $$ad - bc$$ . So,

\text{det}(A) = (2)(-7) - (4)(-3) = -14 + 12 = \boxed{-2}

3 pts Using $$A$$ from part a), find $A^{-1}$ , the inverse of $$A$$ . Fully simplify your answer, i.e. don’t leave any constants out front.

$A^{-1} = \_\_\_\_\_\_$

Solution

The inverse of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ . So,

A^{-1} = \frac{1}{(-2)} \begin{bmatrix} -7 & -4 \\\\ 3 & 2 \end{bmatrix} = \boxed{\begin{bmatrix} 7/2 & 2 \\\\ -3/2 & -1 \end{bmatrix}}

2 pts Let $B = \begin{bmatrix} -1 & 2 & -1 \\ 3 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$ . What is the first column of $B^{-1}$ , the inverse of $$B$$ ?

\begin{bmatrix} -1 \\ 0 \\ 0 \end{bmatrix}

\begin{bmatrix} -1 \\ 1/2 \\ -1 \end{bmatrix}

\begin{bmatrix} -1 \\1/3 \\ 1 \end{bmatrix}

\begin{bmatrix} -1/3 \\ 1/3 \\ 0 \end{bmatrix}

\begin{bmatrix} 1/3 \\ -1/3 \\ 0 \end{bmatrix}

\(B\)

is not invertible

Solution

$$B$$ is not invertible

$\boxed{\begin{bmatrix} -1/3 \\ 1/3 \\ 0\end{bmatrix}}$ .

Remember, $B^{-1}$ is the matrix that satisfies $B B^{-1} = I$ . Inverting $$B$$ is not necessary: instead, all one needs to look for is the vector $\vec v$ such that $B \vec v = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ , since $$B$$ multiplied by $B^{-1}$ ’s first column should give the first column of $$I$$ .

And indeed, $B \begin{bmatrix} -1/3 \\ 1/3 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ . So, you could have solved this just by guessing and checking each of the options.

If we asked this as an open-ended question instead, we’d be searching for the vector $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ such that

\begin{bmatrix} -1 & 2 & -1 \\\\ 3 & 3 & 2 \\\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\\\ y \\\\ z \end{bmatrix} = \begin{bmatrix} 1 \\\\ 0 \\\\ 0 \end{bmatrix}

From here, there are two ways to solve for $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ .

You could solve the system of equations directly.
Or, you could notice that $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ must be orthogonal to both the second row and third row of $$B$$ , which means that its in the same direction as the cross product of the second and third rows. If you compute the cross product of the last two rows, you get

\begin{bmatrix} 3(1) - 2(0) \\\\ 2(0) - 3(1) \\\\ 3(0) - 3(0) \end{bmatrix} = \begin{bmatrix} 3 \\\\ -3 \\\\ 0 \end{bmatrix}

which means that $\begin{bmatrix} x \\ y \\ z \end{bmatrix} = c \begin{bmatrix} 3 \\ -3 \\ 0 \end{bmatrix}$ for some constant $$c$$ . To find $$c$$ , solve for the $$c$$ such that the dot product of $c \begin{bmatrix} 3 \\ -3 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 2 \\ -1 \end{bmatrix}$ (the first row of $$B$$ ) is 1. This gives $$c = -1/9$$ , which means that $\begin{bmatrix} x \\ y \\ z \end{bmatrix} = -1/9 \begin{bmatrix} 3 \\ -3 \\ 0 \end{bmatrix} = \begin{bmatrix} -1/3 \\ 1/3 \\ 0 \end{bmatrix}$ .

4 pts This part is independent of the previous parts (i.e. don’t use the specific $$A$$ or $$B$$ from above).

Select all true statements below.

If $$A$$ and $$B$$ are both matrices such that $$AB = I$$ , then $$A$$ and $$B$$ are both invertible.

If $$A$$ and $$B$$ are both invertible matrices, then $(A^TB)^{-1} = \left( (B^{-1})^T A^{-1} \right)^T$ .

If $$A$$ is an invertible matrix, then $\text{rank}(A) = \text{rank}(A^{-1})$ .

If $$A$$ , $$B$$ , and $$C$$ are all symmetric matrices, then $$AB + C$$ is also symmetric.

Solution

If $$A$$ , $$B$$ , and $$C$$ are all symmetric matrices, then $$AB + C$$ is also symmetric.

Only Option 3 is true. Let’s look at each statement one by one.

If $$A$$ and $$B$$ are both matrices such that $$AB = I$$ , then $$A$$ and $$B$$ are both invertible. This is $\boxed{\text{False}}$ , because it’s possible for $$AB = I$$ to be true for two non-square matrices $$A$$ and $$B$$ , meaning they can’t be invertible. For example, suppose $B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}$ and $A = B^T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$ . Then,
$AB = B^TB = \begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\\\ 0 & 1 \end{bmatrix} = I$
If $$A$$ and $$B$$ are both invertible matrices, then $(A^TB)^{-1} = \left( (B^{-1})^T A^{-1} \right)^T$ . This is $\boxed{\text{False}}$ :
- If we expand the right-hand side, we get
$$ ((B^{-1})^T A^{-1})^T = \underbrace{(A^{-1})^T ((B^{-1})^T)^T}_{\text{reverse order of product when taking transpose}} = (A^{-1})^T (B^{-1}) $$
- This is not the same as $(A^TB)^{-1}$ , which is $(A^TB)^{-1} = B^{-1}(A^T)^{-1}$ . Note that $(A^{-1})^T = (A^T)^{-1}$ , but the reason these two expressions aren’t the same is because order matters for matrix multiplication — it’s not commutative.
If $$A$$ is an invertible matrix, then $\text{rank}(A) = \text{rank}(A^{-1})$ . This is $\boxed{\text{True}}$ . If $$A$$ is invertible, then $\text{rank}(A) = n$ . Then, $A^{-1}$ is also invertible (its inverse is $$A$$ ), so it must have a rank of $$n$$ as well.
If $$A$$ , $$B$$ , and $$C$$ are all symmetric matrices, then $$AB + C$$ is also symmetric. This is $\boxed{\text{False}}$ . Recall, what makes a matrix $$A$$ symmetric is that $$A = A^T$$ . Let’s take the transpose of $$AB + C$$ and see if we end up getting back $$AB + C$$ :
$$$ (AB + C)^T = (AB)^T + C^T = B^TA^T + C^T = BA + C $$$
$$AB + C$$ is only symmetric if $$AB + C = BA + C$$ , i.e. if $$AB = BA$$ , which is not true in general, even if $$A$$ and $$B$$ are both symmetric.

FA25 MT2 · Problem 2: Space Jam 20 pts

From FA25 MT2

Let $X = \begin{bmatrix} 1 & -4 & 2 & 2 & 0 \\ 0 & 0 & -3 & 3 & 0 \\ 1 & -4 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ .

(4.5 pts) Determine the values of each of the following. Give your answers as integers.

\begin{array}{lllll} \text{dim}(\text{colsp}(X)) = &\_\_\_\_\_\_ \qquad \qquad & \text{dim}(\text{nullsp}(X)) = &\_\_\_\_\_\_ \\\\ \\\\ \text{dim}(\text{colsp}(X^T)) = &\_\_\_\_\_\_ \qquad \qquad & \text{dim}(\text{nullsp}(X^T)) = &\_\_\_\_\_\_ \\\\ \end{array}

Solution

Recall, the rank-nullity theorem states that for any matrix $$X$$ ,

\text{rank}(X) + \text{dim}(\text{nullsp}(X)) = \text{number of columns of } X

where $\text{rank}(X) = \text{dim}(\text{colsp}(X)) = \text{dim}(\text{colsp}(X^T))$ .

$$X$$ has 3 linearly independent columns: columns 1, 4, and 5. These three columns can be used to create the other two columns:

Column 2 = $\begin{bmatrix} -4 \\ 0 \\ -4 \\ 0 \end{bmatrix} = -4 \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} = (-4) \cdot \text{column 1}$
Column 3 = $\begin{bmatrix} 2 \\ -3 \\ 4 \\ 0 \end{bmatrix} = 4 \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} - \begin{bmatrix} 2 \\ 3\\ 0 \\ 0 \end{bmatrix} = 4 \cdot \text{column 1} - \text{column 4}$

So, $\text{rank}(X) = 3$ , meaning $\text{dim}(\text{colsp}(X)) = \boxed{3}$ and $\text{dim}(\text{colsp}(X^T)) = \boxed{3}$ also.

Since $\text{rank}(X) + \text{dim}(\text{nullsp}(X)) = \text{number of columns of } X$ , we have $\text{dim}(\text{nullsp}(X)) = 5 - 3 = \boxed{2}$ .

And finally, since $\text{rank}(X^T) + \text{dim}(\text{nullsp}(X^T)) = \text{number of columns of } X^T$ , we have $\text{dim}(\text{nullsp}(X^T)) = 5 - 4 = \boxed{1}$ .

(3.5 pts) Suppose $\vec y \in \mathbb{R}^4$ . How many solutions $\vec v \in \mathbb{R}^5$ are there to the system of equations $X \vec v = \vec y$ ? Select all possibilities, since the answer may depend on $\vec y$ .

0 1 2 3 4 5 Infinitely many

Solution

0 1 2 3 4 5 Infinitely many

When solving $X \vec v = \vec y$ for $\vec v$ , there are two possible cases.

$\vec y \notin \text{colsp}(X)$ : This is possible because $\text{dim}(\text{colsp}(X))=3$ , so the columns don’t span all of $\mathbb{R}^4$ . In this case, $\vec v$ has no solutions.
$\vec y \in \text{colsp}(X)$ : The columns of $$X$$ aren’t linearly independent, so there are infinitely many ways to write $\vec y$ as a linear combination of the columns of $$X$$ .

6 pts For some $\vec y \in \mathbb{R}^4$ , the vector $\vec w’ = \begin{bmatrix} 8 \\ 0 \\ 0 \\ 3 \\ 11 \end{bmatrix}$ is such that $X \vec w’$ is the vector in $\text{colsp}(X)$ that is closest to $\vec y$ . State one other vector $\vec \beta$ such that $X \vec \beta = X \vec w’$ . Show your work, and $\boxed{\text{circle}}$ your final answer, which should be a vector with five entries and no variables.

Solution

There’s two ways to approach this problem. The first is adding a vector in $\text{nullsp}(X)$ to $\vec w’$ . Why does this work? Let $\vec \beta = \vec w’ + \vec n$ , where $X\vec n = \vec 0$ :

\begin{align*} X\vec \beta &= X(\vec w' + \vec n) \\\\&=X(\vec w' + \vec n) \\\\&=X\vec w' + X\vec n \\\\&=X\vec w' = \vec y \end{align*}

So, all we have to do is find a vector in the null space of

X = \begin{bmatrix} 1 & -4 & 2 & 2 & 0 \\\\ 0 & 0 & -3 & 3 & 0 \\\\ 1 & -4 & 4 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}

One such vector is $\vec n = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{bmatrix}$ , since $X\vec n = \begin{bmatrix} -4 \\ 0 \\ -4 \\ 0 \end{bmatrix} + \begin{bmatrix} 2 \\ -3 \\ 4 \\ 0 \end{bmatrix} + \begin{bmatrix} 2 \\ 3 \\ 0 \\ 0 \end{bmatrix} = \vec 0$

This leaves us with $\vec \beta=\vec w’ + \vec n = \begin{bmatrix} 8 \\ 0 \\ 0 \\ 3 \\ 11 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{bmatrix} = \boxed{\begin{bmatrix} 8 \\ 1 \\ 1 \\ 4 \\ 11 \end{bmatrix}}$ .

The other way is to “tweak” $\vec w’$ using the relationships we know about in the columns of $$X$$ . Since $\text{column 2} = -4 \cdot \text{column 1}$ , and $\vec w’ = \begin{bmatrix} 8 \\ 0 \\ 0 \\ 3 \\ 11 \end{bmatrix}$ , an easy swap is to change $$w_0$$ from $$8$$ to $$0$$ and $$w_1$$ from 0 to $$-2$$ :

8 \cdot \begin{bmatrix} 1 \\\\ 0 \\\\ 1 \\\\ 0 \end{bmatrix} = -2 \cdot \begin{bmatrix} -4 \\\\ 0 \\\\ -4 \\\\ 0 \end{bmatrix}

Doing this gives $\vec \beta = \boxed{\begin{bmatrix} 0 \\ -2 \\ 0 \\ 3 \\ 11 \end{bmatrix}}$ .

Recall, $X = \begin{bmatrix} 1 & -4 & 2 & 2 & 0 \\ 0 & 0 & -3 & 3 & 0 \\ 1 & -4 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ .

6 pts Find a basis for $\text{nullsp}(X^T)$ (not $\text{nullsp}(X)$ ). Show your work, and $\boxed{\text{circle}}$ your final answer, which should be a list of vectors.

Solution

X^T= \begin{bmatrix} 1 & 0 & 1 & 0 \\\\ -4 & 0 & -4 & 0 \\\\ 2 & -3 & 4 & 0 \\\\ 2 & 3 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \end{bmatrix}

From the rank-nullity theorem, we know that our basis will have exactly one vector, so our goal is to find a non-zero vector where $X^T \vec n = \vec 0$ .

$\text{Column 3}=\text{Column 1} - \frac{2}{3}\cdot \text{Column 2}$ , so one possible basis is $\left\lbrace \begin{bmatrix} 1 \\ -\frac{2}{3} \\ -1 \\ 0\end{bmatrix}\right\rbrace$ .

FA25 MT2 · Problem 3: Nilpotence 12 pts

From FA25 MT2

Suppose $$A$$ is an $n \times n$ matrix such that $A^2 = 0_{n \times n}$ , where $0_{n \times n}$ is an $n \times n$ matrix of all zeros.

6 pts Prove that if $\vec x \in \text{colsp}(A)$ , then $\vec x \in \text{nullsp}(A)$ .

Solution

If $\vec x \in \text{colsp}(A)$ , then $\vec x = A \vec v$ for some $\vec v \in \mathbb{R}^n$ . Then, multiplying both sides of $\vec x = A \vec v$ by $$A$$ on the left gives us:

A \vec x = A (A \vec v) = A^2 \vec v = 0_{n \times n} \vec v = \vec 0

Since $\vec x = A \vec v \implies A \vec x = \vec 0$ , we have $\vec x \in \text{nullsp}(A)$ .

6 pts In part a), you showed that $\text{colsp}(A)$ is a subset of $\text{nullsp}(A)$ . Using this fact, find the maximum possible value of $\text{rank}(A)$ . Show your work and $\boxed{\text{circle}}$ your final answer, which should be an expression involving $$n$$ and/or constants.

Solution

In the previous part, we showed that every element in $\text{colsp}(A)$ is also in $\text{nullsp}(A)$ . (The converse is not true.) Intuitively, this means that the column space is a subset of the null space, so it’s “smaller” than the null space.

This means that

\text{dim}(\text{colsp}(A)) \leq \text{dim}(\text{nullsp}(A))

or in other words

\text{rank}(A) \leq \text{dim}(\text{nullsp}(A))

Let’s add $\text{rank}(A)$ to both sides of the inequality; this will make the right-hand side look like something involved in the rank-nullity theorem.

\text{rank}(A) + \text{rank}(A) \leq \text{rank}(A) + \text{dim}(\text{nullsp}(A)) = n

This tells us that $2\text{rank}(A) \leq n$ , so $\boxed{\text{rank}(A) \leq \frac{n}{2}}$ and so $\frac{n}{2}$ is the maximum possible value of $\text{rank}(A)$ .

FA25 Final · Problem 5 12 pts MT2 Redemption

From FA25 Final

Consider the $n \times 5$ matrix $$A$$ , along with a CR decomposition of it, given below.

A = \begin{bmatrix} 2 & 2 & 2 & 2 & 2 \\\\ 3 & 4 & 5 & 6 & 7 \\\\ 4 & 6 & 8 & 10 & 12 \\\\ 5 & 8 & 11 & 14 & 17 \\\\ 6 & 10 & 14 & 18 & 22 \\\\ \vdots & \vdots & \vdots & \vdots & \vdots \\\\ n+1 & 2n & 3n - 1 & 4n - 2 & 5n - 3 \\\\ \end{bmatrix} = \underbrace{\begin{bmatrix} 2 & ? \\\\ 3 & ? \\\\ 4 & ? \\\\ 5 & ? \\\\ 6 & ? \\\\ \vdots & \vdots \\\\ n + 1 & ? \end{bmatrix}}_{C} \underbrace{\begin{bmatrix} 1 & \boxed{a} & 0 & c & -1 \\\\ 0 & \boxed{b} & 1 & d & 2\end{bmatrix}}_{R}

2 pts Find $\text{rank}(A)$ . Give your answer as an integer with no variables.

$\text{rank}(A) = \_\_\_\_\_\_$

Solution

The CR decomposition writes $$A = CR$$ , where $$C$$ contains linearly independent columns of $$A$$ . Since $$C$$ has 2 columns, $$A$$ has 2 linearly independent columns, so

\text{rank}(A) = \boxed{2}

4 pts Find $$a$$ and $$b$$ . Give your answers as numbers with no variables.

$a = \_\_\_\_\_\_, \qquad b = \_\_\_\_\_\_$

Solution

Because columns 1 and 3 of $$R$$ are the basis of $\text{colsp}(A)$ that we’re using to construct all 5 columns of $$A$$ , column 2 of $$A$$ must be

\text{col}_2(A) = a\,\text{col}_1(A) + b\,\text{col}_3(A)

The “quick” way to spot what $$a$$ and $$b$$ must be is that column 2 is the average of columns 1 and 3: 2 is the average of 2 and 2, 4 is the average of 3 and 5, 6 is the average of 4 and 8, and so on. This alone tells you that $a = b = \frac{1}{2}$ .

Another way to find $$a$$ and $$b$$ more systematically is to set up a system of equations. We have two unknowns — $$a$$ and $$b$$ — so we must need two equations, which we can get from looking at the first two rows of $$A$$ .

\begin{align*} 2 &= 2a + 2b \\\\ 4 &= 3a + 5b \end{align*}

The first equation says $$a+b=1$$ , so $$a=1-b$$ . Substitute into the second:

4 = 3(1-b) + 5b = 3 + 2b \implies b = \frac{1}{2}

Then $a = \frac{1}{2}$ as well. Therefore,

\boxed{a = \frac{1}{2}, \qquad b = \frac{1}{2}}

3 pts State one vector in $\text{nullsp}(A)$ . Give your answer as a vector with no variables. Hint: It is possible to find a vector in $\text{nullsp}(A)$ without using your answer from part b). Try not to rely heavily on your answer from part b) in case it’s incorrect.

$\text{One vector in } \text{nullsp}(A) \text{ is: } \_\_\_\_\_\_$

Solution

To find a vector in $\text{nullsp}(A)$ , we need to find a linear combination of $$A$$ ’s columns that equals $\vec 0$ . One such linear combination can be found from rearranging the linear dependence relationship from the last part:

\begin{align*} \text{col}_2(A) &= \frac{1}{2}\,\text{col}_1(A) + \frac{1}{2}\,\text{col}_3(A) \\\\ \vec 0 &= \frac{1}{2}\,\text{col}_1(A) - \text{col}_2(A) + \frac{1}{2}\,\text{col}_3(A) \end{align*}

The coefficients on columns 1 through 3 are $\frac{1}{2}$ , $$-1$$ , and $\frac{1}{2}$ ; this linear combination doesn’t use columns 4 and 5. So, this tells us that $\begin{bmatrix} 1/2 \\ -1 \\ 1/2 \\ 0 \\ 0 \end{bmatrix}$ is in $\text{nullsp}(A)$ . If we’d like to get rid of the fraction, then we could also say $\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}$ is in $\text{nullsp}(A)$ too.

There are plenty of other answers. For instance, the fact that

\text{col}_3(A) = \frac{1}{2}\,\text{col}_1(A) + \frac{1}{2}\,\text{col}_5(A)

tells us that $\begin{bmatrix} 1/2 \\ 0 \\ -1 \\ 0 \\ 1/2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 0 \\ -2 \\ 0 \\ 1 \end{bmatrix}$ are also in $\text{nullsp}(A)$ .

3 pts Fill in the blanks: $\text{nullsp}(A^T)$ is a __(i)__-dimensional subspace of __(ii)__.

$$i$$	$$2$$	$$3$$	$$4$$	$$5$$	$$n-2$$	$$n-1$$	$$n$$
$$ii$$	$\mathbb{R}^2$	$\mathbb{R}^3$	$\mathbb{R}^4$	$\mathbb{R}^5$	$\mathbb{R}^{n-2}$	$\mathbb{R}^{n-1}$	$\mathbb{R}^n$

Solution

\mathbb{R}^2

\mathbb{R}^3

\mathbb{R}^4

\mathbb{R}^5

\mathbb{R}^{n-2}

\mathbb{R}^{n-1}

\mathbb{R}^n

Since $\text{rank}(A)=2$ and $\text{rank}(A) = \text{rank}(A^T)$ , we also have $\text{rank}(A^T)=2$ . The matrix $$A^T$$ has $$n$$ columns, so rank-nullity gives

\dim(\text{nullsp}(A^T)) = \text{\# columns in }A^T - \text{rank}(A^T) = n - 2

Also, $\text{nullsp}(A^T)$ is a subspace of $\mathbb{R}^n$ , because vectors in $\text{nullsp}(A^T)$ must have one entry for each column of $$A^T$$ (row of $$A$$ ).

WN26 MT2 · Problem 2 10 pts

From WN26 MT2

Suppose $$A$$ is a matrix such that $A \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 6 \\ 0 \\ 1 \end{bmatrix}$ and $\left\lbrace \begin{bmatrix} 1 \\ 4 \\ -2 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0\end{bmatrix} \right\rbrace$ is a basis for $\text{nullsp}(A)$ .

Find one possible matrix $$A$$ . Show your work, and $\boxed{\text{circle}}$ your final answer, which should be a matrix with no variables.

Solution

Let the columns of $$A$$ be $\vec c_1, \vec c_2, \vec c_3, \vec c_4$ . Since

A \begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\\\ 0 \end{bmatrix} = 1\vec c_1 + 0\vec c_2 + 0\vec c_3 + 0\vec c_4 = \begin{bmatrix} 6 \\\\ 0 \\\\ 1 \end{bmatrix},

we know that the first column of $$A$$ must be $\begin{bmatrix} 6 \\ 0 \\ 1 \end{bmatrix}$ .

Now, let’s use the information given about the null space to find the other columns of $$A$$ . Since

\begin{bmatrix} 0 \\\\ 1 \\\\ 1 \\\\ 0 \end{bmatrix} \in \text{nullsp}(A),

we have

\vec c_2 + \vec c_3 = \vec 0 \implies \vec c_3 = -\vec c_2

Also, since

\begin{bmatrix} 1 \\\\ 4 \\\\ -2 \\\\ 0 \end{bmatrix} \in \text{nullsp}(A),

we have

\vec c_1 + 4\vec c_2 - 2\vec c_3 = \vec 0

Substituting $\vec c_3 = -\vec c_2$ gives

\begin{align*} \vec c_1 + 6\vec c_2 &= \vec 0 \\\\ \vec c_2 &= -\frac{1}{6}\vec c_1 = \begin{bmatrix} -1 \\\\ 0 \\\\ -1/6 \end{bmatrix}. \end{align*}

So,

\vec c_3 = -\vec c_2 = \begin{bmatrix} 1 \\\\ 0 \\\\ 1/6 \end{bmatrix}

Finally, $\text{dim}(\text{nullsp}(A)) = 2$ and $$A$$ has 4 columns, so by rank-nullity,

\begin{align*} \text{rank}(A) &= 4 - 2 = 2 \end{align*}

So we should choose $\vec c_4$ to be linearly independent from $\vec c_1$ . One easy choice is

\vec c_4 = \begin{bmatrix} 0 \\\\ 1 \\\\ 0 \end{bmatrix}

This gives one possible matrix.

\boxed{ A = \begin{bmatrix} 6 & -1 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & -1/6 & 1/6 & 0 \end{bmatrix} }

WN26 MT2 · Problem 3 11 pts

From WN26 MT2

Suppose $$A$$ and $$B$$ are both non-zero $6 \times 6$ matrices, such that $\text{rank}(A) = 4$ and that every column of $$B$$ is in $\text{nullsp}(A)$ .

3 pts Fill in the blanks: The third __(i)__ of $$A$$ is __(ii)__ to the fourth __(iii)__ of $$B$$ .

column row
orthogonal parallel
column row

Solution
column row
Every column of $$B$$ is in $\text{nullsp}(A)$ . From Chapter 5.4 in the notes, the row space of $$A$$ and the null space of $$A$$ are orthogonal complements. That means every row of $$A$$ is orthogonal to every vector in $\text{nullsp}(A)$ , and hence orthogonal to every column of $$B$$ .
So, the third row of $$A$$ is orthogonal to the fourth column of $$B$$ .

4 pts Select all possible values of $\text{rank}(AB)$ .

0 1 2 3 4 5 6

Solution

0 1 2 3 4 5 6

Let $\vec b_1, \vec b_2, \ldots, \vec b_6$ be the columns of $$B$$ . Since every column of $$B$$ is in $\text{nullsp}(A)$ , we have

A \vec b_j = \vec 0

for every $$j$$ . But the $$j$$ th column of $$AB$$ is exactly $A \vec b_j$ , so every column of $$AB$$ is $\vec 0$ .

Therefore,

AB = 0_{6 \times 6} \implies \text{rank}(AB) = 0_{6 \times 6}

4 pts Select all possible values of $\text{rank}(B)$ .

0 1 2 3 4 5 6

Solution

0 1 2 3 4 5 6

Since $$A$$ is $6 \times 6$ with rank 4, the rank-nullity theorem gives

\dim(\text{nullsp}(A)) = 6 - 4 = 2

Every column of $$B$$ lies in $\text{nullsp}(A)$ , so

\text{colsp}(B) \subseteq \text{nullsp}(A)

Therefore,

\text{rank}(B) = \dim(\text{colsp}(B)) \leq 2

Also, $$B$$ is non-zero, so $\text{rank}(B) \neq 0$ .

So the only possible values are 1 and 2.

Both are achievable: all columns of $$B$$ could be multiples of one non-zero vector in $\text{nullsp}(A)$ , or they could span all of $\text{nullsp}(A)$ (which is a 2-dimensional subspace of $\mathbb{R}^6$ , since $\text{rank}(A)=4$ ).

WN26 MT2 · Problem 5 13 pts

From WN26 MT2

Let

A = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\\\ 2 & 1 & 0 & 0 & 4 \\\\ 3 & 1 & 0 & -7 & 4 \end{bmatrix}

5 pts Recall, a CR decomposition of an $n \times d$ matrix $$A$$ is a product $$A = CR$$ , where $$C$$ is an $n \times r$ matrix with linearly independent columns and $$R$$ is an $r \times d$ matrix with linearly independent rows, and $r = \text{rank}(A)$ .

Provide a CR decomposition of $$A$$ . Your answers should be matrices with no variables.

$C = \_\_\_\_\_\_, \quad R = \_\_\_\_\_\_$

Solution

The columns of $$A$$ are

\vec c_1 = \begin{bmatrix} 1 \\\\ 2 \\\\ 3 \end{bmatrix}, \quad \vec c_2 = \begin{bmatrix} 0 \\\\ 1 \\\\ 1 \end{bmatrix}, \quad \vec c_3 = \begin{bmatrix} 0 \\\\ 0 \\\\ 0 \end{bmatrix}, \quad \vec c_4 = \begin{bmatrix} 0 \\\\ 0 \\\\ -7 \end{bmatrix}, \quad \vec c_5 = \begin{bmatrix} 0 \\\\ 4 \\\\ 4 \end{bmatrix}

Reading left-to-right, columns 1, 2, and 4 are linearly independent, so we place them in $$C$$ :

C = \begin{bmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 1 & -7 \end{bmatrix}

Now we need to express each column of $$A$$ as a linear combination of the columns of $$C$$ :

\vec c_1 = 1\vec c_1 + 0\vec c_2 + 0\vec c_4, \quad \vec c_2 = 0\vec c_1 + 1\vec c_2 + 0\vec c_4, \quad \vec c_3 = \vec 0,

\vec c_4 = 0\vec c_1 + 0\vec c_2 + 1\vec c_4, \quad \vec c_5 = 0\vec c_1 + 4\vec c_2 + 0\vec c_4

The coefficients in each linear combination are the entries in the corresponding column of $$R$$ . So,

R = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 & 4 \\\\ 0 & 0 & 0 & 1 & 0 \end{bmatrix}

Thus, one CR decomposition is

A = \begin{bmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 1 & -7 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 & 4 \\\\ 0 & 0 & 0 & 1 & 0 \end{bmatrix}

3 pts Find $\text{dim}(\text{nullsp}(A^T))$ . Give your answer as an integer with no variables.

$\text{dim}(\text{nullsp}(A^T)) = \_\_\_\_\_\_$

Solution

The matrix $$A$$ has 3 rows and rank 3. Applying rank-nullity to $$A^T$$ , we get

\begin{align*} \text{rank}(A^T) + \dim(\text{nullsp}(A^T)) &= \text{number of columns of } A^T = 3 \end{align*}

Since $\text{rank}(A^T)=\text{rank}(A)=3$ ,

\begin{align*} 3 + \dim(\text{nullsp}(A^T)) &= 3 \\\\ \dim(\text{nullsp}(A^T)) &= 0 \end{align*}

This means that $$A^T$$ ’s null space is $\left\lbrace \vec 0 \right\rbrace$ .

(5 pts) Suppose we apply the Gram-Schmidt process to the rows of $$A$$ , and place the resulting orthonormal vectors into the rows of a new matrix, $$Q$$ .

Let $$P$$ be the matrix that projects vectors in $\mathbb{R}^5$ onto $\text{colsp}(Q^T)$ (the row space of $$Q$$ ). In other words, if $\vec y \in \mathbb{R}^5$ , then $P\vec y$ is the projection of $\vec y$ onto $\text{colsp}(Q^T)$ .

Find an expression for $$P$$ in terms of $$Q$$ and $$Q^T$$ . Show your work, and $\boxed{\text{circle}}$ your final answer, which should be an expression in terms of $$Q$$ and $$Q^T$$ . Answers that aren’t fully simplified will not be given credit.

P = \_\_\_\_\_\_

Solution

In general, the projection matrix onto $\text{colsp}(X)$ , where $$X$$ is any matrix with linearly independent columns, is

P = X(X^TX)^{-1}X^T

Here, we want to project onto $\text{colsp}(Q^T)$ , so we should use $$X = Q^T$$ :

P = Q^T((Q^T)^TQ^T)^{-1}(Q^T)^T = Q^T(QQ^T)^{-1}Q

But, since $$Q$$ ’s rows are orthonormal, $$QQ^T = I$$ . This is because $$QQ^T$$ is a matrix containing the dot products of the rows of $$Q$$ with each other (the same way $$Q^TQ$$ is a matrix containing the dot products of the columns of $$Q$$ with each other). Since the rows of $$Q$$ are orthonormal, the dot products are all 0 except for the diagonal, which is 1.

So,

$$ P = Q^T I Q = Q^T Q $$

WN26 Final · Problem 5 11 pts MT2 Redemption

From WN26 Final

Suppose $$A$$ is a $6 \times 5$ matrix such that

\text{nullsp}(A) = \text{span}\left( \left\{ \begin{bmatrix}1\\\\0\\\\1\\\\0\\\\0\end{bmatrix}, \begin{bmatrix}0\\\\1\\\\1\\\\0\\\\0\end{bmatrix}, \begin{bmatrix}0\\\\0\\\\0\\\\1\\\\1\end{bmatrix} \right\} \right)

4 pts Find $\text{rank}(A)$ and $\dim(\text{nullsp}(A^T))$ . Give your answers as integers with no variables.

$\text{rank}(A) = \_\_\_\_\_\_ \dim(\text{nullsp}(A^T)) = \_\_\_\_\_\_$

Solution

Recall, the rank-nullity theorem states that for any matrix $$A$$ ,

\text{rank}(A) + \dim(\text{nullsp}(A)) = \text{number of columns of } A

The null space has dimension $$3$$ , since the given basis has $$3$$ vectors. Because $$A$$ has $$5$$ columns, rank-nullity gives

\text{rank}(A) + 3 = 5 \implies \text{rank}(A) = \boxed{2}

Also, $$A^T$$ has $$6$$ columns and $\text{rank}(A^T)=\text{rank}(A)=2$ , so rank-nullity gives

\dim(\text{nullsp}(A^T)) = 6-2 = \boxed{4}

3 pts Which of the following could NOT be the first row of $$A$$ ?

\begin{bmatrix} 2 & 2 & -2 & 3 & -3 \end{bmatrix}

\begin{bmatrix} 1 & 1 & -1 & 4 & -4 \end{bmatrix}

\begin{bmatrix} 2 & 0 & -2 & 5 & -5 \end{bmatrix}

\begin{bmatrix} 3 & 3 & -3 & -2 & 2 \end{bmatrix}

Solution

\begin{bmatrix} 2 & 2 & -2 & 3 & -3 \end{bmatrix}

\begin{bmatrix} 1 & 1 & -1 & 4 & -4 \end{bmatrix}

\begin{bmatrix} 2 & 0 & -2 & 5 & -5 \end{bmatrix}

\begin{bmatrix} 3 & 3 & -3 & -2 & 2 \end{bmatrix}

A key fact is that the row space and null space of a matrix are orthogonal complements, as discussed in Chapter 5.4 (and the linked video). What this means is that every row of $$A$$ is orthogonal to every vector in $\text{nullsp}(A)$ .

So a row

\begin{bmatrix} a & b & c & d & e \end{bmatrix}

must satisfy

a+c = 0, \qquad b+c = 0, \qquad d+e = 0

Equivalently, every row of $$A$$ must have the form

\begin{bmatrix} a & a & -a & d & -d \end{bmatrix}

The first, second, and fourth options all have this form. The third option,

\begin{bmatrix} 2 & 0 & -2 & 5 & -5 \end{bmatrix}

does not. For instance, it is not orthogonal to

\begin{bmatrix}0\\\\1\\\\1\\\\0\\\\0\end{bmatrix} \in \text{nullsp}(A)

since

\begin{bmatrix} 2 & 0 & -2 & 5 & -5 \end{bmatrix} \begin{bmatrix}0\\\\1\\\\1\\\\0\\\\0\end{bmatrix} = -2 \neq 0

So the correct answer is the third option, $\boxed{\begin{bmatrix} 2 & 0 & -2 & 5 & -5 \end{bmatrix}}$ .

4 pts Let $\vec a^{(1)}, \vec a^{(2)}, \vec a^{(3)}, \vec a^{(4)}, \vec a^{(5)} \in \mathbb{R}^6$ be the columns of $$A$$ .

Below, select one possible set of columns of $$A$$ that form a basis for $\text{colsp}(A)$ . You should select the fewest possible number of columns needed to span $\text{colsp}(A)$ .

\begin{array}{c|c} \text{Column} & \text{Include in your basis?} \\\\ \hline \vec a^{(1)} & \square \quad \\\\ \vec a^{(2)} & \square \quad \\\\ \vec a^{(3)} & \square \quad \\\\ \vec a^{(4)} & \square \quad \\\\ \vec a^{(5)} & \square \quad \end{array}

Solution

The vector

\begin{bmatrix}1\\\\0\\\\1\\\\0\\\\0\end{bmatrix} \in \text{nullsp}(A)

tells us

\vec a^{(1)}+\vec a^{(3)}=\vec 0 \implies \vec a^{(3)} = -\vec a^{(1)}

and the vector

\begin{bmatrix}0\\\\1\\\\1\\\\0\\\\0\end{bmatrix} \in \text{nullsp}(A)

tells us

\vec a^{(2)}+\vec a^{(3)}=\vec 0 \implies \vec a^{(3)} = -\vec a^{(2)}

So $\vec a^{(1)}$ , $\vec a^{(2)}$ , and $\vec a^{(3)}$ all lie on the same line and are scalar multiples of each other. Similarly,

\begin{bmatrix}0\\\\0\\\\0\\\\1\\\\1\end{bmatrix} \in \text{nullsp}(A)

tells us

\vec a^{(4)}+\vec a^{(5)}=\vec 0 \implies \vec a^{(5)} = -\vec a^{(4)}

Since $\text{rank}(A)=2$ , the column space is 2-dimensional. A basis for the column space comes from picking one of $\lbrace \vec a^{(1)}, \vec a^{(2)}, \vec a^{(3)} \rbrace$ and one of $\lbrace \vec a^{(4)}, \vec a^{(5)} \rbrace$ . There are therefore 6 possible options; one of them is

\boxed{\{\vec a^{(1)}, \vec a^{(4)}\}}

SP26 MT2 · Problem 1 12 pts

From SP26 MT2

Suppose $$k$$ is a real number. Let

A = \begin{bmatrix} 1 & k+1 \\\\ 1 & 2k+3 \end{bmatrix}

In each part, you are provided with information about $$A$$ . Your job is to find the value of $$k$$ that satisfies the given condition. Show your work in the space provided, and write your final answer in the bottom-right corner of the box. Your answers should be numbers with no variables.

4 pts $\det(A) = 14$ .

k = \_\_\_\_\_\_

Solution

Since $$A$$ is a $2 \times 2$ matrix, its determinant is

\begin{align*} \det(A) &= 1(2k+3) - 1(k+1) \\\\ &= k + 2 \end{align*}

We’re told that $\det(A) = 14$ , so

\begin{align*} k + 2 &= 14 \\\\ k &= 12 \end{align*}

4 pts $$A$$ is not invertible.

k = \_\_\_\_\_\_

Solution

If $$A$$ is not invertible, then $\det(A) = 0$ . From part a),

\det(A) = k + 2

\begin{align*} k + 2 &= 0 \\\\ k &= -2 \end{align*}

4 pts The bottom-right entry of $A^{-1}$ is $$1/4$$ .

k = \_\_\_\_\_\_

Solution

The inverse of a $2 \times 2$ matrix is

\begin{bmatrix} a & b \\\\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\\\ -c & a \end{bmatrix}

Here, $\det(A)=k+2$ , so

A^{-1} = \frac{1}{k+2} \begin{bmatrix} 2k+3 & -(k+1) \\\\ -1 & 1 \end{bmatrix}

The bottom-right entry is $\frac{1}{k+2}$ , and we’re told that this equals $\frac{1}{4}$ . So,

\begin{align*} \frac{1}{k+2} &= \frac{1}{4} \\\\ k+2 &= 4 \\\\ k &= 2 \end{align*}

SP26 MT2 · Problem 2 16 pts

From SP26 MT2

Suppose $$A$$ is a $3 \times 3$ matrix whose null space is the plane

$$ 5x - y + 3z = 0 $$

In other words, $\text{nullsp}(A) = \left\lbrace \begin{bmatrix} x \\ y \\ z \end{bmatrix} \mid 5x - y + 3z = 0 \right\rbrace$ .

3 pts Determine the following values. Give your answers as integers with no variables.

$\text{rank}(A) =$ ______ $\text{dim}(\text{nullsp}(A)) =$ ______

Solution

The null space is a plane in $\mathbb{R}^3$ , which is a $$2$$ -dimensional subspace of $\mathbb{R}^3$ . So,

\dim(\text{nullsp}(A)) = 2

Since $$A$$ has 3 columns, the rank-nullity theorem gives

\begin{align*} \text{rank}(A) + \dim(\text{nullsp}(A)) &= 3 \\\\ \text{rank}(A) + 2 &= 3 \\\\ \text{rank}(A) &= 1 \end{align*}

4 pts State one basis for $\text{nullsp}(A)$ . Your answer should be a list of vectors with no variables.

$\text{one basis for }\text{nullsp}(A) =$ ______

Solution

The null space consists of all vectors satisfying

$$ 5x-y+3z = 0 $$

A basis for the null space, then, consists of any two linearly independent vectors that satisfy this equation. $\begin{bmatrix} 1 \\ 5 \\ 0 \end{bmatrix}$ satisfies it, since $5 \cdot 1 - (5) + 3 \cdot 0 = 0$ , and similarly $\begin{bmatrix} 0 \\ 3 \\ 1 \end{bmatrix}$ satisfies it. Therefore, one basis is

\left\{ \begin{bmatrix} 1 \\\\ 5 \\\\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\\\ 3 \\\\ 1 \end{bmatrix} \right\}

though there are infinitely many possible answers.

3 pts State one basis for $\text{colsp}(A^T)$ , the row space of $$A$$ . Your answer should be a list of vectors with no variables.

$\text{one basis for }\text{colsp}(A^T) =$ ______

Solution

A key fact to remember here is that the row space and null space of a matrix are orthogonal complements, as discussed in Chapter 5.4. This means that every element in the row space must be orthogonal to every element in the null space.

We’re given that the null space consists of all vectors $\vec x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ such that $$5x-y+3z = 0$$ . Equivalently, this means that $\begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix} \cdot \vec x = 0$ . So, this means that every element in the null space is orthogonal to $\begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix}$ , so $\begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix}$ must be in the row space. The row space is $$1$$ -dimensional, since $\text{rank}(A)=1$ . So, one basis is $\left\lbrace \begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix} \right\rbrace$ .

If you didn’t remember this key fact, no problem — you could have arrived at this conclusion from scratch on the exam. We know $$A$$ is $3 \times 3$ , so suppose it looks like

A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \end{bmatrix}

If $\vec x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is in the null space, it must mean that $A \vec x = \vec 0$ :

A \vec x = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \end{bmatrix} \begin{bmatrix} x \\\\ y \\\\ z \end{bmatrix} = \begin{bmatrix} a_{11}x + a_{12}y + a_{13}z \\\\ a_{21}x + a_{22}y + a_{23}z \\\\ a_{31}x + a_{32}y + a_{33}z \end{bmatrix} = \begin{bmatrix} 0 \\\\ 0 \\\\ 0 \end{bmatrix}

From here, you see that the dot product of each row of $$A$$ with $\vec x$ must be 0, so each row of $$A$$ must be orthogonal to $\vec x$ . The plane form of the null space, $$5x-y+3z=0$$ , tells you that $\begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix}$ is orthogonal to every vector in the null space, so putting these facts together gives us that $\begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix}$ is in the row space. Together with the fact that the row space is $$1$$ -dimensional, since $\text{rank}(A)=1$ , we have that a basis is $\left\lbrace \begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix} \right\rbrace$ .

All other possible answers involve (non-zero) scalar multiples of $\begin{bmatrix} 5 \\ -1 \\ 3 \end{bmatrix}$ .

Recall, $$A$$ is a $3 \times 3$ matrix whose null space is the plane $$5x-y+3z=0$$ .

6 pts Suppose that

A \begin{bmatrix} 3 \\\\ 0 \\\\ 0 \end{bmatrix} = \begin{bmatrix} 15 \\\\ 30 \\\\ 0 \end{bmatrix}

Find $$A$$ . Show your work, and $\boxed{\text{circle}}$ your final answer, which should be a matrix with no variables.

Solution

Since $\text{rank}(A)=1$ and the row space is

\text{span}\left( \left\{ \begin{bmatrix} 5 \\\\ -1 \\\\ 3 \end{bmatrix} \right\} \right),

each row of $$A$$ must be a scalar multiple of $\begin{bmatrix} 5 & -1 & 3 \end{bmatrix}$ . So, for some constants $$a$$ , $$b$$ , and $$c$$ ,

A = \begin{bmatrix} 5a & -a & 3a \\\\ 5b & -b & 3b \\\\ 5c & -c & 3c \end{bmatrix}

We’re told that

A \begin{bmatrix} 3 \\\\ 0 \\\\ 0 \end{bmatrix} = \begin{bmatrix} 15 \\\\ 30 \\\\ 0 \end{bmatrix}

The left-hand side is 3 times the first column of $$A$$ , so the first column of $$A$$ is

\begin{bmatrix} 5 \\\\ 10 \\\\ 0 \end{bmatrix}

This gives

5a = 5, \qquad 5b = 10, \qquad 5c = 0

so $$a=1$$ , $$b=2$$ , and $$c=0$$ . Therefore,

A = \begin{bmatrix} 5 & -1 & 3 \\\\ 10 & -2 & 6 \\\\ 0 & 0 & 0 \end{bmatrix}

SP26 MT2 · Problem 3 12 pts

From SP26 MT2

Suppose $$A$$ is an $n \times n$ matrix.

For each statement below, determine whether it is true or false. If true, prove that it is true. If false, give a counterexample or a short explanation.

4 pts If $$A$$ is symmetric, then $$A^2$$ must be symmetric.

True False

Solution

True False

This is true. Since $$A$$ is symmetric, $$A^T = A$$ . So,

$$ (A^2)^T = (AA)^T = A^T A^T = AA = A^2 $$

Therefore, $$A^2$$ is symmetric.

4 pts If $$A^2$$ is symmetric, then $$A$$ must be symmetric.

True False

Solution

True False

This is false. For example, let

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

This matrix is not symmetric, but

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

which is symmetric.

4 pts If $\vec x \in \text{nullsp}(A^T)$ and $\vec y \in \text{colsp}(A)$ , then $\vec x$ and $\vec y$ are orthogonal.

True False

Solution

True False

This is true. If $\vec y \in \text{colsp}(A)$ , then $\vec y = A\vec v$ for some vector $\vec v$ . Since $\vec x \in \text{nullsp}(A^T)$ , we know $A^T \vec x = \vec 0$ . So,

\vec x \cdot \vec y = \vec x^T A \vec v = (A^T \vec x)^T \vec v = \vec 0^T \vec v = 0

Therefore, $\vec x$ and $\vec y$ are orthogonal.

SP26 Final · Problem 5 4 pts MT2 Redemption

From SP26 Final

Suppose $$A$$ is a $7 \times 12$ matrix. Fill in each blank with an integer with no variables.

(2 pts) What is the minimum possible value of $\text{dim}(\text{nullsp}(A))$ ? ______
(2 pts) What is the maximum possible value of $\text{dim}(\text{nullsp}(A))$ ? ______

Solution
By the rank-nullity theorem from Chapter 5.4,
$\text{rank}(A)+\text{dim}(\text{nullsp}(A))=12$
The rank of a $7\times 12$ matrix is at least $$0$$ and at most $$7$$ . So the dimension of the null space is
$\text{dim}(\text{nullsp}(A))=12-\text{rank}(A)$
This is as small as possible when $\text{rank}(A)=7$ , giving minimum $\text{dim}(\text{nullsp}(A)) = 5$ , and as large as possible when $\text{rank}(A)=0$ , giving maximum $\text{dim}(\text{nullsp}(A)) = 12$ .