# 5.E: Exercises

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## Exercise $$\PageIndex{1}$$

Show the map $$T$$: $$\mathbb{R}^n → \mathbb{R}^m$$ defined by $$T (\vec{x}) = A\vec{x}$$ where $$A$$ is an $$m\times n$$ matrix and $$\vec{x}$$ is an $$m\times 1$$ column vector is a linear transformation.

This result follows from the properties of matrix multiplication.

## Exercise $$\PageIndex{2}$$

Show that the function $$T_{\vec{u}}$$ defined by $$T_{\vec{u}} (\vec{v}) =\vec{v}−proj_{\vec{u}} (\vec{v})$$ is also a linear transformation.

\begin{aligned}T_{\vec{u}}(a\vec{v}+b\vec{w})&=a\vec{v}+b\vec{w}-\frac{(a\vec{v}+b\vec{w}\bullet\vec{u})}{||\vec{u}||^2}\vec{u} \\ &=a\vec{v}-a\frac{(\vec{v}\bullet\vec{u})}{||\vec{u}||^2}\vec{u}+b\vec{w}-b\frac{(\vec{w}\bullet\vec{u})}{||\vec{u}||^2}\vec{u} \\ &=aT_{\vec{u}}(\vec{v})+bT_{\vec{u}}(\vec{w})\end{aligned}

## Exercise $$\PageIndex{3}$$

Let $$\vec{u}$$ be a fixed vector. The function $$T_{\vec{u}}$$ defined by $$T_{\vec{u}}\vec{v} =\vec{u}+\vec{v}$$ has the effect of translating all vectors by adding $$\vec{u}\neq\vec{0}$$. Show this is not a linear transformation. Explain why it is not possible to represent $$T_{\vec{u}}$$ in $$\mathbb{R}^3$$ by multiplying by a $$3×3$$ matrix.

Linear transformations take $$\vec{0}$$ to $$\vec{0}$$ which $$T$$ does not. Also $$T_{\vec{a}} (\vec{u}+\vec{v})\neq T_{\vec{a}}\vec{u}+T_{\vec{a}}\vec{v}$$.

## Exercise $$\PageIndex{4}$$

Consider the following functions which map $$\mathbb{R}^n$$ to $$\mathbb{R}^n$$.

1. $$T$$ multiplies the $$j$$th component of $$\vec{x}$$ by a nonzero number $$b$$.
2. $$T$$ replaces the $$i$$th component of $$\vec{x}$$ with $$b$$ times the $$j$$th component added to the $$i$$th component.
3. $$T$$ switches the $$i$$th and $$j$$th components.

Show these functions are linear transformations and describe their matrices $$A$$ such that $$T (\vec{x}) = A\vec{x}$$.

1. The matrix of $$T$$ is the elementary matrix which multiplies the $$j$$th diagonal entry of the identity matrix by $$b$$.
2. The matrix of $$T$$ is the elementary matrix which takes $$b$$ times the $$j$$th row and adds to the $$i$$th row.
3. The matrix of $$T$$ is the elementary matrix which switches the $$i$$th and the $$j$$th rows where the two components are in the $$i$$th and $$j$$th positions.

## Exercise $$\PageIndex{5}$$

You are given a linear transformation $$T$$ : $$\mathbb{R}^n → \mathbb{R}^m$$ and you know that $T(A_i)=B_i\nonumber$ where $$[A_1\cdots a_n]^{-1}$$ exists. Show that the matrix of $$T$$ is of the form $[B_1\cdots B_n][A_1\cdots A_n]^{-1}\nonumber$

Suppose $\left[\begin{array}{c}\vec{c}_1^{T} \\ \vdots \\ \vec{c}_n^T \end{array}\right]=\left[\begin{array}{ccc}\vec{a}_1&\cdots &\vec{a}_n\end{array}\right]^{-1}\nonumber$ Thus $$\vec{c}_i^T\vec{a}_j=\delta_{ij}$$. Therefore \begin{aligned} \left[\begin{array}{ccc}\vec{b}_1&\cdots&\vec{b}_n\end{array}\right]\left[\begin{array}{ccc}\vec{a}_1&\cdots&\vec{a}_n\end{array}\right]^{-1}\vec{a}_i&=\left[\begin{array}{ccc}\vec{b}_1&\cdots&\vec{b}_n\end{array}\right]\left[\begin{array}{c}\vec{c}_1^T \\ \vdots \\ \vec{c}_n^T\end{array}\right]\vec{a}_i\\ &=\left[\begin{array}{ccc}\vec{b}_1&\cdots&\vec{b}_n\end{array}\right]\vec{e}_i \\ &=\vec{b}_i\end{aligned} Thus $$T\vec{a}_i=\left[\begin{array}{ccc}\vec{b}_1&\cdots&\vec{b}_n\end{array}\right]\left[\begin{array}{ccc}\vec{a}_1&\cdots&\vec{a}_n\end{array}\right]^{-1}\vec{a}_i=A\vec{a}_i$$. If $$\vec{x}$$ is arbitrary, then since the matrix $$\left[\begin{array}{ccc}\vec{a}_1&\cdots&\vec{a}_n\end{array}\right]$$ is invertible, there exists a unique $$\vec{y}$$ such that $$\left[\begin{array}{ccc}\vec{a}_1&\cdots&\vec{a}_n\end{array}\right]\vec{y}=\vec{x}$$ Hence $T\vec{x}=T\left(\sum\limits_{i=1}^n y_i\vec{a}_i\right)=\sum\limits_{i=1}^n y_iT\vec{a}_i=\sum\limits_{i=1}^n y_iA\vec{a}_i=A\left(\sum\limits_{i=1}^ny_i\vec{a}_i\right)=A\vec{x}\nonumber$

## Exercise $$\PageIndex{6}$$

Suppose $$T$$ is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\2\\-6\end{array}\right]&=\left[\begin{array}{r}5\\1\\3\end{array}\right] \\ T\left[\begin{array}{r}-1\\-1\\5\end{array}\right]&=\left[\begin{array}{r}1\\1\\5\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\2\end{array}\right]&=\left[\begin{array}{r}5\\3\\-2\end{array}\right]\end{aligned} Find the matrix of $$T$$. That is find $$A$$ such that $$T(\vec{x}) = A\vec{x}$$.

$\left[\begin{array}{ccc}5&1&5\\1&1&3\\3&5&-2\end{array}\right]\left[\begin{array}{ccc}3&2&1\\2&2&1\\4&1&1\end{array}\right]=\left[\begin{array}{ccc}37&17&11 \\ 17&7&5\\11&14&6\end{array}\right]\nonumber$

## Exercise $$\PageIndex{7}$$

Suppose $$T$$ is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\1\\-8\end{array}\right]&=\left[\begin{array}{r}1\\3\\1\end{array}\right] \\ T\left[\begin{array}{r}-1\\0\\6\end{array}\right]&=\left[\begin{array}{r}2\\4\\1\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\3\end{array}\right]&=\left[\begin{array}{r}6\\1\\-1\end{array}\right]\end{aligned} Find the matrix of $$T$$. That is find $$A$$ such that $$T(\vec{x})=A\vec{x}$$.

$\left[\begin{array}{ccc}1&2&6\\3&4&1\\1&1&-1\end{array}\right]\left[\begin{array}{ccc}6&3&1\\5&3&1\\6&2&1\end{array}\right]=\left[\begin{array}{ccc}52&21&9\\44&23&8\\5&4&1\end{array}\right]\nonumber$

## Exercise $$\PageIndex{8}$$

Suppose $$T$$ is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\3\\-7\end{array}\right]&=\left[\begin{array}{r}-3\\1\\3\end{array}\right] \\ T\left[\begin{array}{r}-1\\-2\\6\end{array}\right]&=\left[\begin{array}{r}1\\3\\-3\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\2\end{array}\right]&=\left[\begin{array}{r}5\\3\\-3\end{array}\right]\end{aligned} Find the matrix of $$T$$. That is find $$A$$ such that $$T(\vec{x})=A\vec{x}$$.

$\left[\begin{array}{ccc}-3&1&5\\1&3&3\\3&-3&-3\end{array}\right]\left[\begin{array}{ccc}2&2&1\\1&2&1\\4&1&1\end{array}\right]=\left[\begin{array}{ccc}15&1&3\\17&11&7\\-9&-3&-3\end{array}\right]\nonumber$

## Exercise $$\PageIndex{9}$$

Suppose $$T$$ is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\1\\-7\end{array}\right]&=\left[\begin{array}{r}3\\3\\3\end{array}\right] \\ T\left[\begin{array}{r}-1\\0\\6\end{array}\right]&=\left[\begin{array}{r}1\\2\\3\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\2\end{array}\right]&=\left[\begin{array}{r}1\\3\\-1\end{array}\right]\end{aligned} Find the matrix of $$T$$. That is find $$A$$ such that $$T(\vec{x})=A\vec{x}$$.

$\left[\begin{array}{ccc}3&1&1\\3&2&3\\3&3&-1\end{array}\right]\left[\begin{array}{ccc}6&2&1\\5&2&1\\6&1&1\end{array}\right]=\left[\begin{array}{ccc}29&9&5\\46&13&8\\27&11&5\end{array}\right]\nonumber$

## Exercise $$\PageIndex{10}$$

Suppose $$T$$ is a linear transformation such that \begin{aligned}T\left[\begin{array}{r}1\\2\\-18\end{array}\right]&=\left[\begin{array}{r}5\\2\\5\end{array}\right] \\ T\left[\begin{array}{r}-1\\-1\\15\end{array}\right]&=\left[\begin{array}{r}3\\3\\5\end{array}\right] \\ T\left[\begin{array}{r}0\\-1\\4\end{array}\right]&=\left[\begin{array}{r}2\\5\\-2\end{array}\right]\end{aligned} Find the matrix of $$T$$. That is find $$A$$ such that $$T(\vec{x})=A\vec{x}$$.

$\left[\begin{array}{ccc}5&3&2\\2&3&5\\5&5&-2\end{array}\right]\left[\begin{array}{ccc}11&4&1\\10&4&1\\12&3&1\end{array}\right]=\left[\begin{array}{ccc}109&38&10\\112&35&10\\81&34&8\end{array}\right]\nonumber$

## Exercise $$\PageIndex{11}$$

Consider the following functions $$T : \mathbb{R}^3 → \mathbb{R}^2$$. Show that each is a linear transformation and determine for each the matrix $$A$$ such that $$T(\vec{x}) = A\vec{x}$$.

1. $$T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}x+2y+3z \\ 2y-3x+z\end{array}\right]$$
2. $$T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}7x+2y+z \\ 3x-11y+2z\end{array}\right]$$
3. $$T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}3x+2y+z \\ x+2y+6z\end{array}\right]$$
4. $$T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}2y-5x+z \\ x+y+z\end{array}\right]$$

## Exercise $$\PageIndex{12}$$

Consider the following functions $$T : \mathbb{R}^3 → \mathbb{R}^2$$. Explain why each of these functions $$T$$ is not linear.

1. $$T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}x+2y+3z+1 \\ 2y-3x+z\end{array}\right]$$
2. $$T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}x+2y^2+3z \\ 2y+3x+z\end{array}\right]$$
3. $$T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}\sin x+2y+3z \\ 2y+3x+z\end{array}\right]$$
4. $$T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}x+2y+3z \\ 2y+3x-\ln z\end{array}\right]$$

## Exercise $$\PageIndex{13}$$

Suppose $[A_1\cdots A_n]^{-1}\nonumber$ exists where each $$A_j ∈ \mathbb{R}^n$$ and let vectors $$\{B_1,\cdots ,B_n\}$$ in $$\mathbb{R}^m$$ be given. Show that there always exists a linear transformation $$T$$ such that $$T(A_i) = B_i$$.

## Exercise $$\PageIndex{14}$$

Find the matrix for $$T (\vec{w}) = proj_{\vec{v}}(\vec{w})$$ where $$\vec{v}=\left[\begin{array}{ccc}1&-2&3\end{array}\right]^T$$.

Recall that $$proj_{\vec{u}}(\vec{v})=\frac{\vec{v}\bullet\vec{u}}{||\vec{u}||^2}\vec{u}$$ and so the desired matrix has $$i$$th column equal to $$proj_{\vec{u}}(\vec{e}_i)$$. Therefore, the matrix desired is $\frac{1}{14}\left[\begin{array}{ccc}1&-2&3\\-2&4&-6\\3&-6&9\end{array}\right]\nonumber$

## Exercise $$\PageIndex{15}$$

Find the matrix for $$T (\vec{w}) = proj_{\vec{v}}(\vec{w})$$ where $$\vec{v}=\left[\begin{array}{ccc}1&5&3\end{array}\right]^T$$.

$\frac{1}{35}\left[\begin{array}{ccc}1&5&3\\5&25&15\\3&15&9\end{array}\right]\nonumber$

## Exercise $$\PageIndex{16}$$

Find the matrix for $$T (\vec{w}) = proj_{\vec{v}}(\vec{w})$$ where $$\vec{v}=\left[\begin{array}{ccc}1&0&3\end{array}\right]^T$$.

$\frac{1}{10}\left[\begin{array}{ccc}1&0&3\\0&0&0\\3&0&9\end{array}\right]\nonumber$

## Exercise $$\PageIndex{17}$$

Show that if a function $$T :\mathbb{R}^n → \mathbb{R}^m$$ is linear, then it is always the case that $$T(\vec{0})=\vec{0}$$.

## Exercise $$\PageIndex{18}$$

Let $$T$$ be a linear transformation induced by the matrix $$A=\left[\begin{array}{cc}3&1\\-1&2\end{array}\right]$$ and $$S$$ a linear transformation induced by $$B=\left[\begin{array}{cc}0&-2\\4&2\end{array}\right]$$. Find matrix of $$S\circ T$$ and find $$(S\circ T)(\vec{x})$$ for $$\vec{x}=\left[\begin{array}{r}2\\-1\end{array}\right]$$.

The matrix of $$S\circ T$$ is given by $$BA$$. $\left[\begin{array}{cc}0&-2\\4&2\end{array}\right]\left[\begin{array}{cc}3&1\\-1&2\end{array}\right]=\left[\begin{array}{cc}2&-4\\10&8\end{array}\right]\nonumber$ Now, $$(S\circ T)(\vec{x})=(BA)\vec{x}$$. $\left[\begin{array}{cc}2&-4\\10&8\end{array}\right]\left[\begin{array}{c}2\\-1\end{array}\right]=\left[\begin{array}{c}8\\12\end{array}\right]\nonumber$

## Exercise $$\PageIndex{19}$$

Let $$T$$ be a linear transformation and suppose $$T\left(\left[\begin{array}{r}1\\-4\end{array}\right]\right)=\left[\begin{array}{r}2\\-3\end{array}\right]$$. Suppose $$S$$ is a linear transformation induced by the matrix $$B=\left[\begin{array}{cc}1&2\\-1&3\end{array}\right]$$. Find $$(S\circ T)(\vec{x})$$ for $$\vec{x}=\left[\begin{array}{r}1\\-4\end{array}\right]$$.

To find $$(S\circ T)(\vec{x})$$ we compute $$S(T(\vec{x}))$$. $\left[\begin{array}{cc}1&2\\-1&3\end{array}\right]\left[\begin{array}{c}2\\-3\end{array}\right]=\left[\begin{array}{c}-4\\-11\end{array}\right]\nonumber$

## Exercise $$\PageIndex{20}$$

Let $$T$$ be a linear transformation induced by the matrix $$A=\left[\begin{array}{cc}2&3\\1&1\end{array}\right]$$ and $$S$$ a lienar transformation induced by $$B=\left[\begin{array}{cc}-1&3\\1&-2\end{array}\right]$$. Find matrix of $$S\circ T$$ and find $$(S\circ T)(\vec{x})$$ for $$\vec{x}=\left[\begin{array}{c}5\\6\end{array}\right]$$.

## Exercise $$\PageIndex{21}$$

Let $$T$$ be a linear transformation induced by the matrix $$A=\left[\begin{array}{cc}2&1\\5&2\end{array}\right]$$. Find the matrix of $$T^{-1}$$.

The matrix of $$T^{-1}$$ is $$A^{-1}$$. $\left[\begin{array}{cc}2&1\\5&2\end{array}\right]^{-1}=\left[\begin{array}{cc}-2&1\\5&-2\end{array}\right]\nonumber$

## Exercise $$\PageIndex{22}$$

Let $$T$$ be a linear transformation induced by the matrix $$A=\left[\begin{array}{cc}4&-3\\2&-2\end{array}\right]$$. Find the matrix of $$T^{-1}$$.

## Exercise $$\PageIndex{23}$$

Let $$T$$ be a linear transformation and suppose $$T\left(\left[\begin{array}{c}1\\2\end{array}\right]\right)=\left[\begin{array}{c}9\\8\end{array}\right],\:T\left(\left[\begin{array}{r}0\\-1\end{array}\right]\right)=\left[\begin{array}{c}-4\\-3\end{array}\right]$$. Find the matrix of $$T^{-1}$$.

## Exercise $$\PageIndex{24}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$π/3$$.

$$\left[\begin{array}{cc}\cos\left(\frac{\pi}{3}\right)&-\sin\left(\frac{\pi}{3}\right) \\ \sin\left(\frac{\pi}{3}\right)&\cos\left(\frac{\pi}{3}\right)\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&-\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3}&\frac{1}{2}\end{array}\right]$$

## Exercise $$\PageIndex{25}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$π/4$$.

$$\left[\begin{array}{cc}\cos\left(\frac{\pi}{4}\right)&-\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right)&\cos\left(\frac{\pi}{4}\right)\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{2}\end{array}\right]$$

## Exercise $$\PageIndex{26}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$-π/3$$.

$$\left[\begin{array}{cc}\cos\left(-\frac{\pi}{3}\right)&-\sin\left(-\frac{\pi}{3}\right) \\ \sin\left(-\frac{\pi}{3}\right)&\cos\left(-\frac{\pi}{3}\right)\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&\frac{1}{2}\sqrt{3} \\ -\frac{1}{2}\sqrt{3}&\frac{1}{2}\end{array}\right]$$

## Exercise $$\PageIndex{27}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$2π/3$$.

$$\left[\begin{array}{cc}\cos\left(\frac{2\pi}{3}\right)&-\sin\left(\frac{2\pi}{3}\right) \\ \sin\left(\frac{2\pi}{3}\right)&\cos\left(\frac{2\pi}{3}\right)\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}&-\frac{1}{2}\sqrt{3} \\ \frac{1}{2}\sqrt{3}&-\frac{1}{2}\end{array}\right]$$

## Exercise $$\PageIndex{28}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$π/12$$. Hint: Note that $$\pi /12=\pi /3-\pi /4$$.

\begin{aligned}&\left[\begin{array}{cc}\cos\left(\frac{\pi}{3}\right)&-\sin\left(\frac{\pi}{3}\right) \\ \sin\left(\frac{\pi}{3}\right)&\cos\left(\frac{\pi}{3}\right)\end{array}\right]\left[\begin{array}{cc}\cos\left(-\frac{\pi}{4}\right)&-\sin\left(-\frac{\pi}{4}\right) \\ \sin\left(-\frac{\pi}{4}\right)&\cos\left(-\frac{\pi}{4}\right)\end{array}\right] \\ =&\left[\begin{array}{cc}\frac{1}{4}\sqrt{2}\sqrt{3}+\frac{1}{4}\sqrt{2}&\frac{1}{4}\sqrt{2}-\frac{1}{4}\sqrt{2}\sqrt{3} \\ \frac{1}{4}\sqrt{2}\sqrt{3}-\frac{1}{4}\sqrt{2}&\frac{1}{4}\sqrt{2}\sqrt{3}+\frac{1}{4}\sqrt{2}\end{array}\right]\end{aligned}

## Exercise $$\PageIndex{29}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$2π/3$$ and then reflects across the $$x$$ axis.

$\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]\left[\begin{array}{cc}\cos\left(\frac{2\pi}{3}\right)&-\sin\left(\frac{2\pi}{3}\right) \\ \sin\left(\frac{2\pi}{3}\right)&\cos\left(\frac{2\pi}{3}\right)\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}&-\frac{1}{2}\sqrt{3}\\-\frac{1}{2}\sqrt{3}&\frac{1}{2}\end{array}\right]\nonumber$

## Exercise $$\PageIndex{30}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$π/3$$ and then reflects across the $$x$$ axis.

$\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]\left[\begin{array}{cc}\cos\left(\frac{\pi}{3}\right)&-\sin\left(\frac{\pi}{3}\right) \\ \sin\left(\frac{\pi}{3}\right)&\cos\left(\frac{\pi}{3}\right)\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}&-\frac{1}{2}\sqrt{3}\\-\frac{1}{2}\sqrt{3}&-\frac{1}{2}\end{array}\right]\nonumber$

## Exercise $$\PageIndex{31}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$π/4$$ and then reflects across the $$x$$ axis.

$\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]\left[\begin{array}{cc}\cos\left(\frac{\pi}{4}\right)&-\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right)&\cos\left(\frac{\pi}{4}\right)\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2}\\-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2}\end{array}\right]\nonumber$

## Exercise $$\PageIndex{32}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$π/6$$ and then reflects across the $$x$$ axis followed by a reflection across the $$y$$ axis.

$\left[\begin{array}{cc}-1&0\\0&1\end{array}\right]\left[\begin{array}{cc}\cos\left(\frac{\pi}{6}\right)&-\sin\left(\frac{\pi}{6}\right) \\ \sin\left(\frac{\pi}{6}\right)&\cos\left(\frac{\pi}{6}\right)\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}\sqrt{3}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\sqrt{3}\end{array}\right]\nonumber$

## Exercise $$\PageIndex{33}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ across the $$x$$ axis and then rotates every vector through an angle of $$\pi /4$$.

$\left[\begin{array}{cc}\cos\left(\frac{\pi}{4}\right)&-\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right)&\cos\left(\frac{\pi}{4}\right)\end{array}\right]\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2}\end{array}\right]\nonumber$

## Exercise $$\PageIndex{34}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ across the $$y$$ axis and then rotates every vector through an angle of $$\pi /4$$.

$\left[\begin{array}{cc}\cos\left(\frac{\pi}{4}\right)&-\sin\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right)&\cos\left(\frac{\pi}{4}\right)\end{array}\right]\left[\begin{array}{cc}-1&0\\0&1\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}\sqrt{2}&-\frac{1}{2}\sqrt{2} \\ -\frac{1}{2}\sqrt{2}&\frac{1}{2}\sqrt{2}\end{array}\right]\nonumber$

## Exercise $$\PageIndex{35}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ across the $$x$$ axis and then rotates every vector through an angle of $$\pi /6$$.

$\left[\begin{array}{cc}\cos\left(\frac{\pi}{6}\right)&-\sin\left(\frac{\pi}{6}\right) \\ \sin\left(\frac{\pi}{6}\right)&\cos\left(\frac{\pi}{6}\right)\end{array}\right]\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}\sqrt{3}&\frac{1}{2} \\ \frac{1}{2}&-\frac{1}{2}\sqrt{3}\end{array}\right]\nonumber$

## Exercise $$\PageIndex{36}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ across the $$y$$ axis and then rotates every vector through an angle of $$\pi /6$$.

$\left[\begin{array}{cc}\cos\left(\frac{\pi}{6}\right)&-\sin\left(\frac{\pi}{6}\right) \\ \sin\left(\frac{\pi}{6}\right)&\cos\left(\frac{\pi}{6}\right)\end{array}\right]\left[\begin{array}{cc}-1&0\\0&1\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{2}\sqrt{3}&-\frac{1}{2} \\ -\frac{1}{2}&\frac{1}{2}\sqrt{3}\end{array}\right]\nonumber$

## Exercise $$\PageIndex{37}$$

Find the matrix for the linear transformation which rotates every vector in $$\mathbb{R}^2$$ through an angle of $$5\pi /12$$. Hint: Note that $$5\pi /12=2\pi /3-\pi /4$$.

$\left[\begin{array}{cc}\cos\left(\frac{2\pi}{3}\right)&-\sin\left(\frac{2\pi}{3}\right) \\ \sin\left(\frac{2\pi}{3}\right)&\cos\left(\frac{2\pi}{3}\right)\end{array}\right]\left[\begin{array}{cc}\cos\left(-\frac{\pi}{4}\right)&-\sin\left(-\frac{\pi}{4}\right) \\ \sin\left(-\frac{\pi}{4}\right)&\cos\left(-\frac{\pi}{4}\right)\end{array}\right]=\nonumber$ $\left[\begin{array}{cc}\frac{1}{4}\sqrt{2}\sqrt{3}-\frac{1}{4}\sqrt{2}&-\frac{1}{4}\sqrt{2}\sqrt{3}-\frac{1}{4}\sqrt{2} \\ \frac{1}{4}\sqrt{2}\sqrt{3}+\frac{1}{4}\sqrt{2}&\frac{1}{4}\sqrt{2}\sqrt{3}-\frac{1}{4}\sqrt{2}\end{array}\right]\nonumber$ Note that it doesn't matter about the order in this case.

## Exercise $$\PageIndex{38}$$

Find the matrix of the linear transformation which rotates every vector in $$\mathbb{R}^3$$ counter clockwise about the $$z$$ axis when viewed from the positive $$z$$ axis through an angle of $$30^◦$$ and then reflects through the $$xy$$ plane.

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&-1\end{array}\right]\left[\begin{array}{ccc}\cos\left(\frac{\pi}{6}\right)&-\sin\left(\frac{\pi}{6}\right)&0 \\ \sin\left(\frac{\pi}{6}\right)&\cos\left(\frac{\pi}{6}\right)&0 \\ 0&0&1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}\sqrt{3}&-\frac{1}{2}&0\\ \frac{1}{2}&\frac{1}{2}\sqrt{3}&0\\0&0&-1\end{array}\right]\nonumber$

## Exercise $$\PageIndex{39}$$

Let $$\vec{u}=\left[\begin{array}{c}a\\b\end{array}\right]$$ be a unit vector in $$\mathbb{R}^2$$. Find the matrix which reflects all vectors across this vector, as shown in the following picture.

Hint: Notice that $$\left[\begin{array}{c}a\\b\end{array}\right]=\left[\begin{array}{c}\cos\theta \\ \sin\theta\end{array}\right]$$ for some $$\theta$$. First rotate through $$-\theta$$. Next reflect through the $$x$$ axis. Finally rotate through $$\theta$$.

\begin{aligned} &\left[\begin{array}{cc}\cos (\theta )&-\sin(\theta) \\ \sin(\theta)&\cos(\theta)\end{array}\right]\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]\left[\begin{array}{cc}\cos(-\theta)&-\sin(\theta) \\ \sin(-\theta)&\cos(-\theta)\end{array}\right] \\ =&\left[\begin{array}{cc}\cos^2\theta-\sin^2\theta &2\cos\theta\sin\theta \\ 2\cos\theta\sin\theta&\sin^2\theta-\cos^2\theta\end{array}\right]\end{aligned} Now to write in terms of $$(a,b)$$, note that $$a/\sqrt{a^2+b^2}=\cos\theta$$, $$b/\sqrt{a^2+b^2}=\sin\theta$$. Now plug this in to the above. The result is $\left[\begin{array}{cc}\frac{a^2-b^2}{a^2+b^2}&2\frac{ab}{a^2+b^2} \\ 2\frac{ab}{a^2+b^2}&\frac{b^2-a^2}{a^2+b^2}\end{array}\right]=\frac{1}{a^2+b^2}\left[\begin{array}{cc}a^2-b^2&2ab \\ 2ab&b^2-a^2\end{array}\right]\nonumber$ Since this is a unit vector, $$a^2+b^2=1$$ and so you get $\left[\begin{array}{cc}a^2-b^2&2ab \\ 2ab&b^2-a^2\end{array}\right]\nonumber$

## Exercise $$\PageIndex{40}$$

Let $$T$$ be a linear transformation given by $T\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}2&1\\0&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\nonumber$ Is $$T$$ one to one? Is $$T$$ onto?

## Exercise $$\PageIndex{41}$$

Let $$T$$ be a linear transformation given by $T\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}-1&2\\2&1\\1&4\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\nonumber$ Is $$T$$ one to one? Is $$T$$ onto?

## Exercise $$\PageIndex{42}$$

Let $$T$$ be a linear transformation given by $T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}2&0&1\\1&2&-1\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]\nonumber$ Is $$T$$ one to one? Is $$T$$ onto?

## Exercise $$\PageIndex{43}$$

Let $$T$$ be a linear transformation given by $T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&3&-5\\2&0&2\\2&4&-6\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]\nonumber$ Is $$T$$ one to one? Is $$T$$ onto?

## Exercise $$\PageIndex{44}$$

Give an example of a $$3\times 2$$ matrix with the property that the linear transformation determined by this matrix is one to one but not onto.

$\left[\begin{array}{cc}1&0\\0&1\\0&0\end{array}\right]\nonumber$

## Exercise $$\PageIndex{45}$$

Suppose $$A$$ is an $$m\times n$$ matrix in which $$m ≤ n$$. Suppose also that the rank of $$A$$ equals $$m$$. Show that the transformation $$T$$ determined by $$A$$ maps $$\mathbb{R}^n$$ onto $$\mathbb{R}^m$$. Hint: The vectors $$\vec{e}_1,\cdots ,\vec{e}_m$$ occur as columns in the reduced row-echelon form for $$A$$.

This says that the columns of $$A$$ have a subset of $$m$$ vectors which are linearly independent. Therefore, this set of vectors is a basis for $$\mathbb{R}^m$$. It follows that the span of the columns is all of $$\mathbb{R}^m$$. Thus $$A$$ is onto.

## Exercise $$\PageIndex{46}$$

Suppose $$A$$ is an $$m\times n$$ matrix in which $$m ≥ n$$. Suppose also that the rank of $$A$$ equals $$n$$. Show that $$A$$ is one to one. Hint: If not, there exists a vector, $$\vec{x}$$ such that $$A\vec{x} = 0$$, and this implies at least one column of $$A$$ is a linear combination of the others. Show this would require the rank to be less than $$n$$.

The columns are independent. Therefore, $$A$$ is one to one.

## Exercise $$\PageIndex{47}$$

Explain why an $$n\times n$$ matrix $$A$$ is both one to one and onto if and only if its rank is $$n$$.

The rank is $$n$$ is the same as saying the columns are independent which is the same as saying $$A$$ is one to one which is the same as saying the columns are a basis. Thus the span of the columns of $$A$$ is all of $$\mathbb{R}^n$$ and so $$A$$ is onto. If $$A$$ is onto, then the columns must be linearly independent since otherwise the span of these columns would have dimension less than $$n$$ and so the dimension of $$\mathbb{R}^n$$ would be less than $$n$$.

## Exercise $$\PageIndex{48}$$

Let $$V$$ and $$W$$ be subspaces of $$\mathbb{R}^n$$ and $$\mathbb{R}^m$$ respectively and let $$T : V → W$$ be a linear transformation. Suppose that $$\{T\vec{v}_1,\cdots ,T\vec{v}_r\}$$ is linearly independent. Show that it must be the case that $$\{T\vec{v}_1,\cdots ,T\vec{v}_r\}$$ is also linearly independent.

If $$\sum_i^r a_i\vec{v}_r=0$$, then using linearity properties of $$T$$ we get $0=T(0)=T\left(\sum\limits_i^ra_i\vec{v}_r\right)=\sum\limits_i^ra_iT(\vec{v}_r).\nonumber$ Since we assume that $$\{T\vec{v}_a,\cdots ,T\vec{v}_r\}$$ is linearly independent, we must have all $$a_i = 0$$, and therefore we conclude that $$\{\vec{v}_1,\cdots ,\vec{v}_r\}$$ is also linearly independent.

## Exercise $$\PageIndex{49}$$

Let $V=span\left\{\left[\begin{array}{c}1\\1\\2\\0\end{array}\right],\:\left[\begin{array}{c}0\\1\\1\\1\end{array}\right],\:\left[\begin{array}{c}1\\1\\0\\1\end{array}\right]\right\}\nonumber$ Let $$T\vec{x}=A\vec{x}$$ where $$A$$ is the matrix $\left[\begin{array}{cccc}1&1&1&1\\0&1&1&0\\0&1&2&1\\1&1&1&2\end{array}\right]\nonumber$ Give a basis for $$im(T)$$.

## Exercise $$\PageIndex{50}$$

Let $V=span\left\{\left[\begin{array}{c}1\\0\\0\\1\end{array}\right],\:\left[\begin{array}{c}1\\1\\1\\1\end{array}\right],\:\left[\begin{array}{c}1\\4\\4\\1\end{array}\right]\right\}\nonumber$ Let $$T\vec{x}=A\vec{x}$$ where $$A$$ is the matrix $\left[\begin{array}{cccc}1&1&1&1\\0&1&1&0\\0&1&2&1\\1&1&1&2\end{array}\right]\nonumber$ Find a basis for $$im(T)$$. In this case, the original vectors do not form an independent set.

Since the third vector is a linear combinations of the first two, then the image of the third vector will also be a linear combinations of the image of the first two. However the image of the first two vectors are linearly independent (check!), and hence form a basis of the image.

Thus a basis for $$im(T)$$ is: $V=span\left\{\left[\begin{array}{c}2\\0\\1\\3\end{array}\right],\:\left[\begin{array}{c}4\\2\\4\\5\end{array}\right]\right\}\nonumber$

## Exercise $$\PageIndex{51}$$

If $$\{\vec{v}_1,\cdots ,\vec{v}_r\}$$ is linearly independent and $$T$$ is a one to one linear transformation, show that $$\{T\vec{v}_1,\cdots ,T\vec{v}_r\}$$ is also linearly independent. Give an example which shows that if $$T$$ is only linear, it can happen that, although $$\{\vec{v}_1,\cdots ,\vec{v}_r\}$$ is linearly independent, $$\{T\vec{v}_1,\cdots ,T\vec{v}_r\}$$ is not. In fact, show that it can happen that each of the $$T\vec{v}_j$$ equals $$0$$.

## Exercise $$\PageIndex{52}$$

Let $$V$$ and $$W$$ be subspaces of $$\mathbb{R}^n$$ and $$\mathbb{R}^m$$ respectively and let $$T : V → W$$ be a linear transformation. Show that if $$T$$ is onto $$W$$ and if $$\{\vec{v}_1,\cdots ,\vec{v}_r\}$$ is a basis for $$V$$, then $$span\{T\vec{v}_1,\cdots ,T\vec{v}_r\} = W$$.

## Exercise $$\PageIndex{53}$$

Define $$T$$: $$\mathbb{R}^4\to\mathbb{R}^3$$ as follows. $T\vec{x}=\left[\begin{array}{cccc}3&2&1&8\\2&2&-2&6\\1&1&-1&3\end{array}\right]\vec{x}\nonumber$ Find a basis for $$im(T)$$. Also find a basis for $$\text{ker}(T)$$.

## Exercise $$\PageIndex{54}$$

Define $$T$$: $$\mathbb{R}^3\to\mathbb{R}^3$$ as follows. $T\vec{x}=\left[\begin{array}{ccc}1&2&0\\1&1&1\\0&1&1\end{array}\right]\vec{x}\nonumber$ where on the right, it is just matrix multiplication of the vector $$\vec{x}$$ which is meant. Explain why $$T$$ is an isomorphism of $$\mathbb{R}^3$$ to $$\mathbb{R}^3$$.

## Exercise $$\PageIndex{55}$$

Suppose $$T$$: $$\mathbb{R}^3\to\mathbb{R}^3$$ is a linear transformation given by $T\vec{x}=A\vec{x}\nonumber$ where $$A$$ is a $$3\times 3$$ matrix. Show that $$T$$ is an isomorphism if and only if $$A$$ is invertible.

## Exercise $$\PageIndex{56}$$

Suppose $$T$$: $$\mathbb{R}^n\to\mathbb{R}^m$$ is a linear transformation given by $T\vec{x}=A\vec{x}\nonumber$ where $$A$$ is an $$m\times n$$ matrix. Show that $$T$$ is never an ismorphism if $$m\neq n$$. In particular, show that if $$m>n$$, $$T$$ cannot be onto and if $$m<n$$, then $$T$$ cannot be one to one.

## Exercise $$\PageIndex{57}$$

Define $$T$$: $$\mathbb{R}^2\to\mathbb{R}^3$$ as follows. $T\vec{x}=\left[\begin{array}{cc}1&0\\1&1\\0&1\end{array}\right]\vec{x}\nonumber$ where on the right, it is just matrix multiplication of the vector $$\vec{x}$$ which is meant. Show that $$T$$ is one to one. Next let $$W = im(T)$$. Show that $$T$$ is an isomorphism of $$\mathbb{R}^2$$ and $$im (T)$$.

## Exercise $$\PageIndex{58}$$

In the above problem, find a $$2\times 3$$ matrix $$A$$ such that the restriction of $$A$$ to $$im(T)$$ gives the same result as $$T^{−1}$$ on $$im(T)$$. Hint: You might let $$A$$ be such that $A\left[\begin{array}{c}1\\1\\0\end{array}\right]=\left[\begin{array}{c}1\\0\end{array}\right],\:A\left[\begin{array}{c}0\\1\\1\end{array}\right]=\left[\begin{array}{c}0\\1\end{array}\right]\nonumber$ now find another vector $$\vec{v}\in\mathbb{R}^3$$ such that $\left\{\left[\begin{array}{c}1\\1\\0\end{array}\right],\:\left[\begin{array}{c}0\\1\\1\end{array}\right],\:\vec{v}\right\}\nonumber$ is a basis. You could pick $\vec{v}=\left[\begin{array}{c}0\\0\\1\end{array}\right]\nonumber$ for example. Explain why this one works or one of your choice works. Then you could define $$A\vec{v}$$ to equal some vector in $$\mathbb{R}^2$$. Explain why there will be more than one such matrix $$A$$ which will deliver the inverse isomorphism $$T^{−1}$$ on $$im(T)$$.

## Exercise $$\PageIndex{59}$$

Now let $$V$$ equan $$span\left\{\left[\begin{array}{c}1\\0\\1\end{array}\right],\:\left[\begin{array}{c}0\\1\\1\end{array}\right]\right\}$$ and let $$T$$: $$V\to W$$ be a linear transformation where $W=span\left\{\left[\begin{array}{c}1\\0\\1\\0\end{array}\right],\:\left[\begin{array}{c}0\\1\\1\\1\end{array}\right]\right\}\nonumber$ and $T\left[\begin{array}{c}1\\0\\1\end{array}\right]=\left[\begin{array}{c}1\\0\\1\\0\end{array}\right],\:T\left[\begin{array}{c}0\\1\\1\end{array}\right]=\left[\begin{array}{c}0\\1\\1\\1\end{array}\right]\nonumber$ Explain why $$T$$ is an isomorphism. Determine a matrix $$A$$ which, when multiplied on the left gives the same result as $$T$$ on $$V$$ and a matrix $$B$$ which delivers $$T^{−1}$$ on $$W$$. Hint: You need to have $A\left[\begin{array}{cc}1&0\\0&1\\1&1\end{array}\right]=\left[\begin{array}{cc}1&0\\0&1\\1&1\\0&1\end{array}\right]\nonumber$ Now enlarge $$\left[\begin{array}{c}1\\0\\1\end{array}\right],\:\left[\begin{array}{c}0\\1\\1\end{array}\right]$$ to obtain a basis for $$\mathbb{R}^3$$. You could add in $$\left[\begin{array}{c}0\\0\\1\end{array}\right]$$ for example, and then pick another vector in $$\mathbb{R}^4$$ and let $$A\left[\begin{array}{c}0\\0\\1\end{array}\right]$$ equal this other vector. Then you would have $A\left[\begin{array}{ccc}1&0&0\\0&1&0\\1&1&1\end{array}\right]=\left[\begin{array}{ccc}1&0&0\\0&1&0\\1&1&0\\0&1&1\end{array}\right]\nonumber$ This would involve picking for the new vector in $$\mathbb{R}^4$$ the vector $$\left[\begin{array}{cccc}0&0&0&1\end{array}\right]^T$$. Then you could find $$A$$. You can do something similar to find a matrix for $$T^{−1}$$ denoted as $$B$$.

## Exercise $$\PageIndex{60}$$

Let $$V=\mathbb{R}^3$$ and let $W=span(S),\text{ where }S=\left\{\left[\begin{array}{r}1\\-1\\1\end{array}\right],\:\left[\begin{array}{r}-2\\2\\-2\end{array}\right],\:\left[\begin{array}{r}-1\\1\\1\end{array}\right],\:\left[\begin{array}{r}1\\-1\\3\end{array}\right]\right\}\nonumber$ Find a basis of $$W$$ consisting of vectors in $$S$$.

In this case $$\text{dim}(W) = 1$$ and a basis for $$W$$ consisting of vectors in $$S$$ can be obtained by taking any (nonzero) vector from $$S$$.

## Exercise $$\PageIndex{61}$$

Let $$T$$ be a linear transformation given by $T\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}1&1\\1&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\nonumber$ Find a basis for $$\text{ker}(T)$$ and $$im(T)$$.

A basis for $$\text{ker}(T)$$ is $$\left\{\left[\begin{array}{r}1\\-1\end{array}\right]\right\}$$ and a basis for $$im(T)$$ is $$\left\{\left[\begin{array}{c}1\\1\end{array}\right]\right\}$$. There are many other possibilities for the specific bases, but in this case $$\text{dim}(\text{ker}(T))=1$$ and $$\text{dim}(im(T))=1$$.

## Exercise $$\PageIndex{62}$$

Let $$T$$ be a linear transformation given by $T\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{cc}1&0\\1&1\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\nonumber$ Find a basis for $$\text{ker}(T)$$ and $$im(T)$$.

In this case $$\text{ker}(T)=\{0\}$$ and $$im(T)=\mathbb{R}^2$$ (pick any basis of $$\mathbb{R}^2$$).

## Exercise $$\PageIndex{63}$$

Let $$V=\mathbb{R}^3$$ and let $W=span\left\{\left[\begin{array}{c}1\\1\\1\end{array}\right],\:\left[\begin{array}{r}-1\\2\\-1\end{array}\right]\right\}\nonumber$ Extend this basis of $$W$$ to a basis of $$V$$.

There are many possible such extensions, one is (how do we know?): $\left\{\left[\begin{array}{c}1\\1\\1\end{array}\right],\:\left[\begin{array}{r}-1\\2\\-1\end{array}\right],\:\left[\begin{array}{c}0\\0\\1\end{array}\right]\right\}\nonumber$

## Exercise $$\PageIndex{64}$$

Let $$T$$ be a linear transformation given by $T\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&1&1\\1&1&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]\nonumber$ What is $$\text{dim}(\text{ker}(T))$$?

We can easily see that $$\text{dim}(im(T))=1$$, and thus $$\text{dim}(\text{ker}(T))=3-\text{dim}(im(T))=3-1=2$$.

## Exercise $$\PageIndex{65}$$

Let $$B=\left\{\left[\begin{array}{r}2\\-1\end{array}\right],\:\left[\begin{array}{r}3\\2\end{array}\right]\right\}$$ be a basis of $$\mathbb{R}^2$$ and let $$\vec{x}=\left[\begin{array}{r}5\\-7\end{array}\right]$$ be a vector in $$\mathbb{R}^2$$. Find $$C_B(\vec{x})$$.

## Exercise $$\PageIndex{66}$$

Let $$B=\left\{\left[\begin{array}{r}1\\-1\\2\end{array}\right],\:\left[\begin{array}{r}2\\1\\2\end{array}\right],\:\left[\begin{array}{r}-1\\0\\2\end{array}\right]\right\}$$ be a basis of $$\mathbb{R}^3$$ and let $$\vec{x}=\left[\begin{array}{r}5\\-1\\4\end{array}\right]$$ be a vector in $$\mathbb{R}^2$$. Find $$C_B(\vec{x})$$.

$$C_B(\vec{x})=\left[\begin{array}{r}2\\1\\-1\end{array}\right]$$

## Exercise $$\PageIndex{67}$$

Let $$T$$: $$\mathbb{R}^2→\mathbb{R}^2$$ be a linear transformation defined by $$T\left(\left[\begin{array}{c}a\\b\end{array}\right)\right]=\left[\begin{array}{c}a+b\\a-b\end{array}\right]$$.

Consider the two bases $B_1=\{\vec{v}_1,\vec{v}_2\}=\left\{\left[\begin{array}{c}1\\0\end{array}\right],\:\left[\begin{array}{r}-1\\1\end{array}\right]\right\}\nonumber$ and $B_2=\left\{\left[\begin{array}{c}1\\1\end{array}\right],\:\left[\begin{array}{r}1\\-1\end{array}\right]\right\}\nonumber$

Find the matrix $$M_{B_2,B_1}$$ of $$T$$ with respect to the bases $$B_1$$ and $$B_2$$.

$$M_{B_2B_1}=\left[\begin{array}{rr}1&0\\-1&1\end{array}\right]$$

## Exercise $$\PageIndex{68}$$

Write the solution set of the following system as a linear combination of vectors $\left[\begin{array}{rrr}1&-1&2\\1&-2&1\\3&-4&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{r}-3\hat{t} \\ -\hat{t} \\ \hat{t}\end{array}\right],\:\hat{t}_3\in\mathbb{R}$$. A basis for the solution space is $$\left[\begin{array}{r}-3\\-1\\1\end{array}\right]\nonumber$$

## Exercise $$\PageIndex{69}$$

Using Exercise $$\PageIndex{68}$$ find the general solution to the following linear system. $\left[\begin{array}{rrr}1&-1&2\\1&-2&1\\3&-4&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\2\\4\end{array}\right]\nonumber$

Note that this has the same matrix as the above problem. Solution is: $$\left[\begin{array}{r}-3\hat{t}_3 \\ -\hat{t}_3 \\ \hat{t}_3\end{array}\right]+\left[\begin{array}{r}0\\-1\\0\end{array}\right],\:\hat{t}_3\in\mathbb{R}$$

## Exercise $$\PageIndex{70}$$

Write the solution set of the following system as a linear combination of vectors $\left[\begin{array}{rrr}0&-1&2\\1&-2&1\\1&-4&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{c}3\hat{t} \\ 2\hat{t} \\ \hat{t}\end{array}\right]$$, A basis is $$\left[\begin{array}{c}3\\2\\1\end{array}\right]$$

## Exercise $$\PageIndex{71}$$

Using Exercise $$\PageIndex{70}$$ find the general solution to the following linear system. $\left[\begin{array}{rrr}0&-1&2\\1&-2&1\\1&-4&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\-1\\1\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{c}3\hat{t} \\ 2\hat{t} \\ \hat{t}\end{array}\right] +\left[\begin{array}{c}-3\\-1\\0\end{array}\right],\:\hat{t}\in\mathbb{R}$$

## Exercise $$\PageIndex{72}$$

Write the solution set of the following system as a linear combination of vectors. $\left[\begin{array}{ccc}1&-1&2\\1&-2&0\\3&-4&4\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{r}-4\hat{t} \\ -2\hat{t} \\ \hat{t}\end{array}\right]$$. A basis is $$\left[\begin{array}{r}-4\\-2\\1\end{array}\right]$$

## Exercise $$\PageIndex{73}$$

Using Exercise $$\PageIndex{72}$$ find the general solution to the following linear system. $\left[\begin{array}{ccc}1&-1&2\\1&-2&0\\3&-4&4\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\2\\4\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{r}-4\hat{t} \\ -2\hat{t} \\ \hat{t}\end{array}\right]+\left[\begin{array}{r}0\\-1\\0\end{array}\right],\:\hat{t}\in\mathbb{R}$$

## Exercise $$\PageIndex{74}$$

Write the solution set of the following system as a linear combination of vectors $\left[\begin{array}{ccc}0&-1&2\\1&0&1\\1&-2&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}0\\0\\0\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{r}-\hat{t} \\ 2\hat{t} \\ \hat{t}\end{array}\right],\:\hat{t}\in\mathbb{R}$$

## Exercise $$\PageIndex{75}$$

Using Exercise $$\PageIndex{74}$$ find the general solution to the following linear system. $\left[\begin{array}{ccc}0&-1&2\\1&0&1\\1&-2&5\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\-1\\1\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{r}-\hat{t} \\ 2\hat{t} \\ \hat{t}\end{array}\right]+\left[\begin{array}{r}-1\\-1\\0\end{array}\right]$$

## Exercise $$\PageIndex{76}$$

Write the solution set of the following system as a linear combination of vectors $\left[\begin{array}{cccc}1&0&1&1\\1&-1&1&0\\3&-1&3&2\\3&3&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{r}0\\ -\hat{t} \\ -\hat{t} \\ \hat{t}\end{array}\right],\:\hat{t}\in\mathbb{R}$$

## Exercise $$\PageIndex{77}$$

Using Exercise $$\PageIndex{76}$$ find the general solution to the following linear system. $\left[\begin{array}{cccc}1&0&1&1\\1&-1&1&0\\3&-1&3&2\\3&3&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}1\\2\\4\\3\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{r}0\\ -\hat{t} \\ -\hat{t} \\ \hat{t}\end{array}\right]+\left[\begin{array}{r}2\\-1\\-1\\0\end{array}\right]$$

## Exercise $$\PageIndex{78}$$

Write the solution set of the following system as a linear combination of vectors $\left[\begin{array}{cccc}1&1&0&1\\2&1&1&2\\1&0&1&1\\0&0&0&0\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{c}-s-t \\ s\\s\\t\end{array}\right],\:s,t\in\mathbb{R}$$. A basis is $\left\{\left[\begin{array}{r}-1\\1\\1\\0\end{array}\right],\:\left[\begin{array}{r}-1\\0\\0\\1\end{array}\right]\right\}\nonumber$

## Exercise $$\PageIndex{79}$$

Using Exercise $$\PageIndex{78}$$ find the general solution to the following linear system. $\left[\begin{array}{rrrr}1&1&0&1\\2&1&1&2\\1&0&1&1\\0&-1&1&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{r}2\\-1\\-3\\0\end{array}\right]\nonumber$

Solution is: $\left[\begin{array}{r}-\hat{t}\\ \hat{t} \\ \hat{t}\\0\end{array}\right]+\left[\begin{array}{c}-8\\5\\0\\5\end{array}\right]\nonumber$

## Exercise $$\PageIndex{80}$$

Write the solution set of the following system as a linear combination of vectors $\left[\begin{array}{rrrr}1&1&0&1\\1&-1&1&0\\3&1&1&2\\3&3&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]\nonumber$

Solution is: $\left[\begin{array}{c}-\frac{1}{2}s-\frac{1}{2}t \\ \frac{1}{2}s-\frac{1}{2}t \\ s\\t\end{array}\right]\nonumber$ for $$s,t\in\mathbb{R}$$. A basis is $\left\{\left[\begin{array}{r}-1\\1\\2\\0\end{array}\right],\:\left[\begin{array}{r}-1\\1\\0\\1\end{array}\right]\right\}\nonumber$

## Exercise $$\PageIndex{81}$$

Using Exercise $$\PageIndex{80}$$ find the general solution to the following linear system. $\left[\begin{array}{rrrr}1&1&0&1\\1&-1&1&0\\3&1&1&2\\3&3&0&3\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}1\\2\\4\\3\end{array}\right]\nonumber$

Solution is: $\left[\begin{array}{r}\frac{3}{2} \\ -\frac{1}{2}\\0\\0\end{array}\right]+\left[\begin{array}{c}-\frac{1}{2}s-\frac{1}{2}t \\ \frac{1}{2}s-\frac{1}{2}t \\ s\\t\end{array}\right]\nonumber$

## Exercise $$\PageIndex{82}$$

Write the solution set of the following system as a linear combination of vectors $\left[\begin{array}{rrrr}1&1&0&1\\2&1&1&2\\1&0&1&1\\0&-1&1&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{r}-\hat{t} \\ \hat{t} \\ \hat{t}\\0\end{array}\right]$$, a basis is $$\left[\begin{array}{c}1\\1\\1\\0\end{array}\right]$$.

## Exercise $$\PageIndex{83}$$

Using Exercise $$\PageIndex{82}$$ find the general solution to the following linear system. $\left[\begin{array}{rrrr}1&1&0&1\\2&1&1&2\\1&0&1&1\\0&-1&1&1\end{array}\right]\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]=\left[\begin{array}{r}2\\-1\\-3\\1\end{array}\right]\nonumber$

Solution is: $$\left[\begin{array}{r}-\hat{t} \\ \hat{t} \\ \hat{t}\\0\end{array}\right]+\left[\begin{array}{r}-9\\5\\0\\6\end{array}\right],t\in\mathbb{R}$$.

## Exercise $$\PageIndex{84}$$

Suppose $$A\vec{x}=\vec{b}$$ has a solution. Explain why the solution is unique precisely when $$A\vec{x}=\vec{0}$$ has only the trivial solution.

If not, then there would be a infintely many solutions to $$A\vec{x}=\vec{0}$$ and each of these added to a solution to $$A\vec{x}=\vec{b}$$ would be a solution to $$A\vec{x}=\vec{b}$$.