Homework 6

1. Consider the linear transformation considered in the previous homework. Determine if they are surjective or injective.

(i) $T(x_1, x_2, x_3) = (3x_1 - x_2, x_2 + x_3, x_1 - x_2 - x_3)$

$$\operatorname{rref} \begin{pmatrix} 3 & -1 & 0 \\ 0 & 1 & 1 \\ 1 & -1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

Full rank. Therefore, bijective.

(ii) $T$ maps $\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}$, and$\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}$ and$\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$ respectively to$\begin{bmatrix} 0 \\ 1\end{bmatrix}$ ,$\begin{bmatrix} 1 \\ 1\end{bmatrix}$ ,$\begin{bmatrix} 1 \\ -1\end{bmatrix}$

$$\operatorname{rref} \begin{pmatrix} 0 & 1 & 1 \\ 1 & 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \end{pmatrix}$$

No free variables. Nullity is not zero. Therefore, not injective.

Rank two, which is equal to the dimension of the codomain. Therefore, surjective.

(iii) $T(x_1, x_2) = x_1 \begin{bmatrix} 1 \\ 1 \\ 2\end{bmatrix} + x_2 \begin{bmatrix} -1 \\ 1 \\ 5\end{bmatrix}$

$$\operatorname{rref} \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$$

Nullity is zero. Therefore, injective.

Rank two but codomain is in $\R^3$. Therefore, not surjective.

2. Determine if the following statements are true or false. Give explanation.

(a) Suppose that there are 6 vectors in $\R^4$, it must be linearly decpendent.

True. Only four linearly independent vectors are needed to span $\R^4$.

(b) Suppose that there are 6 vectors in $\R^4$, it must span $\R^4$.

False. At least four linearly independent vectors are needed to span $\R^4$. If at least four of the six are the same or are multiples of each other, then they cannot $\R^4$.

(c) Suppose that there are 4 vectors in $\R^6$, it must be linearly independent.

False. If they are not distinct vectors or are multiples of each other, then they would be linearly dependent.

(d) Suppose that there are 4 vectors in $\R^6$, it cannot span $\R^6$.

True. At least six linearly independent vectors are needed to span $\R^6$.

3. Find a basis for the kernel and image of the following matrices and compute its dimensions. $$A =\begin{bmatrix} 1 & 2 & 4 & −2 & 2 \\ 2 & 4 & 6 & 1 & 1 \\ 2 & 3 & 4 & 1 & 1\end{bmatrix},B =\begin{bmatrix} 1 & −2 & 3 \\ −3 & 6 & −9 \\ −2 & 4 & −6 \\ 3 & 0 & −1\end{bmatrix}$$

Using our brainpower, we find that the RREF of $A$ and $B$ are given by:

$$\operatorname{rref}(A) = \begin{bmatrix} 1 & 0 & 0 & -2 & 2 \\ 0 & 1 & 0 & 5 & -3 \\ 0 & 0 & 1 & -\frac{5}{2} & \frac{3}{2} \end{bmatrix} \\ \therefore x_3 = \frac{5}{2} x_4 - \frac{3}{2} x_5 \\ x_2= -5x_4 + 3x_5 \\ x_1 = 2x_4 - 2x_5 \\ x_4,x_5 \in \R \\ \therefore\ker(A) = \Set{ \begin{pmatrix} 2x_4 - 2x_5 \\ -5x_4 + 3x_5 \\ \frac{5}{2} x_4 - \frac{3}{2} x_5 \\ x_4 \\ x_5 \end{pmatrix}: x_4, x_5 \in \R } = \Set{ x\begin{pmatrix} 2 \\ -5 \\ \frac{5}{2} \\ 1 \\ 0 \end{pmatrix} + y\begin{pmatrix} -2 \\ 3 \\ -\frac{3}{2} \\ 0 \\ 1 \end{pmatrix} : x, y \in \R }$$

A basis of $\ker(A)$ is $\Set{ \begin{pmatrix} 2 \\ -5 \\ \frac{5}{2} \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -2 \\ 3 \\ -\frac{3}{2} \\ 0 \\ 1 \end{pmatrix} }$ and its dimension is $2$.

Note that the pivots in $\operatorname{rref}(A)$ are located in columns one, two, and three. As such, the basis of $\operatorname{Im}(A)$ are the corresponding column vectors in $A$.

A basis of $\operatorname{Im}(A)$ is $\Set{\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix},\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix},\begin{pmatrix} 4 \\ 6 \\ 4 \end{pmatrix}}$ and its dimension is $3$.

$$\operatorname{rref}(B) = \begin{bmatrix} 1 & 0 & -\frac{1}{3} \\ 0 & 1 & -\frac{5}{3} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ \therefore x_2 = \frac{5}{3}x_3 \\ x_1 = \frac{1}{3}x_3 \\ x_3\in\R \\ \therefore\ker(B) = \Set{ \begin{pmatrix} \frac{1}{3}x_3 \\ \frac{5}{3}x_3 \\ x_3 \end{pmatrix}: x_3 \in \R } = \Set{ x \begin{pmatrix} \frac{1}{3} \\ \frac{5}{3} \\ 1 \end{pmatrix}: x\in\R }$$

A basis of $\ker(B)$ is $\Set{ \begin{pmatrix} \frac{1}{3} \\ \frac{5}{3} \\ 1 \end{pmatrix} }$ and its dimension is $1$.

Note that the pivots in $\operatorname{rref}(B)$ are located in columns one and two. As such, the basis of $\operatorname{Im}(B)$ are the corresponding column vectors in $B$.

A basis of $\operatorname{Im}(B)$ is $\Set{\begin{pmatrix} 1 \\ -3 \\ -2 \\ 3 \end{pmatrix},\begin{pmatrix} -2 \\ 6 \\ 4 \\ 0 \end{pmatrix}}$ and its dimension is $2$.

4. Let $S = \Set{ \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}, \begin{bmatrix} 7 \\ 8 \\ 9 \end{bmatrix}, \begin{bmatrix} 10 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 13 \\ 4 \\ 5 \end{bmatrix}}$. Is it possible to extract a basis for $\R^3$ from the set $S$? Explain.

Let $P$ be a matrix composed of the vectors in $S$ such that

$$P = \begin{pmatrix} 1 & 4 & 7 & 10 & 13 \\ 2 & 5 & 8 & 1 & 4 \\ 3 & 6 & 9 & 2 & 5 \end{pmatrix}.$$

Then,

$$\operatorname{rref}(P) = \begin{pmatrix} 1 & 0 & -1 & 0 & -1 \\ 0 & 1 & 2 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 \end{pmatrix}.$$

Notice $\operatorname{rank}(P) = 3$. As such, there exists three linearly independent vectors in $S$ (which are the basis of $\operatorname{Im}(P)$), namely:

$$\Set{ \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}, \begin{pmatrix} 10 \\ 1 \\ 2 \end{pmatrix} }$$

which are sufficient to span $\R^3$.

5. Let A be the $6\times4$ matrix with $A = \begin{bmatrix} | & | & | & | \\ \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \mathbf{v}_4 \\ | & | & | & |\end{bmatrix}$. Suppose that after Gaussian elimination, the row echelon form of $A$ is given by $$\begin{bmatrix} 1 & 2 & 0 & 3 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$$

(i) Find the rank of $A$.

The three pivots in the row echelon form tells us that $\operatorname{rank}(A)=3$.

(ii) Find a basis for the $\operatorname{ker}(A)$. What is its dimension?

$$\operatorname{rref}\begin{bmatrix} 1 & 2 & 0 & 3 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \\ \therefore\ker(A) = \Set{ \begin{pmatrix} -2x_2 \\ x_2 \\ 0 \\ 0 \end{pmatrix}: x_2\in\R } = \Set{ x\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix}: x\in\R }$$

A basis for $\ker(A)$ is $\Set{\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix}}$ and its dimension is $1$.

(iii) Find the subset of the columns of $A$ so that it forms a basis for the $\operatorname{Im}(A)$. What is the dimension of $\operatorname{Im}(A)$?

The pivots in the row echelon form of $A$ are at columns one, three, and four. Correspondingly, the basis for $\operatorname{Im}(A)$ is $\set{\mathbf{v}_1, \mathbf{v}_3, \mathbf{v}_4}$ and its dimension is $3$.

6. Book Question 26, 27, 53, 55, 56.

In Exercises 25 through 30, find the matrix $B$ of the linear transformation $T(\vec{x})=A\vec{x}$ with respect to the basis $\mathfrak{B}=(\vec{v}_1,\ldots,\vec{v}_m)$.

26. $A = \begin{bmatrix} 0 & 1 \\ 2 & 3\end{bmatrix};\vec{v}_1 =\begin{bmatrix} 1 \\ 2\end{bmatrix},\vec{v}_2 =\begin{bmatrix} 1 \\ 1\end{bmatrix}$

Let $P$ be a matrix composed of column vectors $\set{\vec{v}_1, \vec{v}_2}$ such that $P = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}$.

Then:

$$P^{-1}AP = B \\ \begin{align*} \\ \therefore B &= \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}^{-1} \begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \\ &= \begin{pmatrix} -1 & 1 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 6 & 4 \\ -4 & 3 \end{pmatrix} \end{align*}$$

27. $A = \begin{bmatrix} 4 & 2 & −4 \\ 2 & 1 & −2 \\ −4 & −2 & 4\end{bmatrix};\vec{v}_1 = \begin{bmatrix} 2 \\ 1 \\ −2\end{bmatrix},\vec{v}_2 =\begin{bmatrix} 0 \\ 2 \\ 1\end{bmatrix},\vec{v}_3 =\begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}$

Let $P$ be a matrix composed of column vectors $\set{\vec{v}_1,\vec{v}_2,\vec{v}_3}$ such that $P=\begin{pmatrix} 2 & 0 & 1 \\ 1 & 2 & 0 \\ -2 & 1 & 1 \end{pmatrix}$.

Then:

$$P^{-1}AP = B \\ \begin{align*} \therefore B &= \begin{pmatrix} 2 & 0 & 1 \\ 1 & 2 & 0 \\ -2 & 1 & 1 \end{pmatrix}^{-1} \begin{pmatrix} 4 & 2 & -4 \\ 2 & 1 & -2 \\ -4 & -2 & 4 \end{pmatrix} \begin{pmatrix} 2 & 0 & 1 \\ 1 & 2 & 0 \\ -2 & 1 & 1 \end{pmatrix} \\ &= \frac{1}{9}\begin{pmatrix} 2 & 1 & -2 \\ -1 & 4 & 1 \\ 5 & -2 & 4 \end{pmatrix} \begin{pmatrix} 4 & 2 & -4 \\ 2 & 1 & -2 \\ -4 & -2 & 4 \end{pmatrix} \begin{pmatrix} 2 & 0 & 1 \\ 1 & 2 & 0 \\ -2 & 1 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 9 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{align*}$$

53. Consider the basis $\mathfrak{B}$ of $\R^2$ consisting of the vectors $\begin{bmatrix} 1 \\ 2\end{bmatrix}$ and $\begin{bmatrix} 3 \\ 4\end{bmatrix}$. We are told that $\begin{bmatrix} \vec{x}\end{bmatrix}_\mathfrak{B} = \begin{bmatrix} 7 \\ 11\end{bmatrix}$ for a certain vector $\vec{x}$ in $\R^2$. Find $\vec{x}$.

Let $P = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$. Then:

$$\begin{align*} \vec{x} &= P[\vec{x}]_\mathfrak{B} \\ &= \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} 7 \\ 11 \end{pmatrix} \\ &= \begin{pmatrix} 40 \\ 58 \end{pmatrix} \end{align*}$$

55. Consider the basis $\mathfrak{B}$ of $\R^2$ consisting of the vectors $\begin{bmatrix} 1 \\ 1\end{bmatrix}$ and $\begin{bmatrix} 1 \\ 2\end{bmatrix}$ and let $\mathfrak{R}$ be the basis consisting of $\begin{bmatrix} 1 \\ 2\end{bmatrix}$ and $\begin{bmatrix} 3 \\ 4\end{bmatrix}$. Find a matrix $P$ such that $\begin{bmatrix} \vec{x}\end{bmatrix}_\mathfrak{R} = P \begin{bmatrix} \vec{x}\end{bmatrix}_\mathfrak{B}$, for all $\vec{x}$ in $\R^2$.

Let $U$ and $V$ be a matrix composed of the basis vectors of $\mathfrak{B}$ and $\mathfrak{R}$ where

$$U = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix},\quad V = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}.$$

By definition:

$$\vec{x} = U[\vec{x}]_\mathfrak{B} \\ \vec{x} = V[\vec{x}]_\mathfrak{R}$$

Applying the inverse for one of them yields:

$$V^{-1}\vec{x} = [\vec{x}]_\mathfrak{R}$$

Then, writing $\vec{x}$ as a $U$ transformation:

$$\begin{align*} [\vec{x}]_\mathfrak{R} &= V^{-1}\vec{x} \\ &= \underbrace{V^{-1}U}_P [\vec{x}]_\mathfrak{B} \end{align*}$$

By comparison, $P=V^{-1}U$.

$$\begin{align*} \therefore P &= V^{-1}U \\ &= \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}^{-1} \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \\ &= \begin{pmatrix} -2 & \frac{3}{2} \\ 1 & -\frac{1}{2} \end{pmatrix}^{-1} \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \\ &= \begin{pmatrix} -\frac{1}{2} & 1 \\ \frac{1}{2} & 0 \end{pmatrix} \\ &= \frac{1}{2} \begin{pmatrix} -1 & 2 \\ 1 & 0 \end{pmatrix} \end{align*}$$

56. Find a basis $\mathfrak{B}$ of $\R^2$ such that $\begin{bmatrix} 1 \\ 2\end{bmatrix}_\mathfrak{B} = \begin{bmatrix} 3 \\ 5\end{bmatrix}$ and $\begin{bmatrix} 3\\ 4\end{bmatrix}_\mathfrak{B} = \begin{bmatrix} 2 \\ 3\end{bmatrix}$.

Let $P = \begin{pmatrix} | & | \\ \vec{v}_1 & \vec{v}_2 \\ | & | \end{pmatrix}$ for some vectors $\vec{v}_1,\vec{v}_2\in\R^2$.

Then, by $\vec{x}\overset{\Delta}{=}P[\vec{x}]_\mathfrak{B}$, we have:

$$\begin{cases} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = P\begin{pmatrix} 3 \\ 5 \end{pmatrix} \\ \begin{pmatrix} 3 \\ 4 \end{pmatrix} = P\begin{pmatrix} 2 \\ 3 \end{pmatrix} \end{cases} \implies P\begin{pmatrix} 3 & 2 \\ 5 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} \\ \begin{align*} \therefore P &= \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} 3 & 2 \\ 5 & 3 \end{pmatrix}^{-1} \\ &= \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix} \\ &= \begin{pmatrix} 12 & -7 \\ 14 & -8 \end{pmatrix} \end{align*}$$

As such, $\mathfrak{B} = \Set{\begin{pmatrix} 12 \\ 14 \end{pmatrix}, \begin{pmatrix} -7 \\ -8 \end{pmatrix}}$.