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Section 6.3 #1: Let $\vec{x}=\begin{bmatrix}2 \\3 \end{bmatrix} \in \mathbb{R}^{2 \times 1}$. Let $\mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right\}$. Let $\mathcal{C} = \left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix} \right\}$. Find $[\vec{x}]_{\mathcal{B}}$ and $[\vec{x}]_{\mathcal{C}}$. Find the change of basis matrix $P_{\mathcal{C}\leftarrow \mathcal{B}}$. Compute $[\vec{x}]_{\mathcal{C}}$ using $P_{\mathcal{C}\leftarrow\mathcal{B}}$. Find $P_{\mathcal{B}\leftarrow\mathcal{C}}$. Compute $[\vec{x}]_{\mathcal{B}}$ using $P_{\mathcal{B}\leftarrow\mathcal{C}}$.
Solution: To find $[\vec{x}]_{\mathcal{B}}$, we must find the coordinate vector $\begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$ where the coordinates are the solutions to the vector equation formed when trying to write $\vec{x}$ as a linear combination from $\mathcal{B}$:
$$c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}.$$
Note that the augmented matrix is already in reduced echelon form: $\begin{bmatrix} 1&0&2 \\ 0&1&3 \end{bmatrix}$. Therefore $c_1=2$ and $c_2=3$. Hence we have shown
$$[\vec{x}]_{\mathcal{B}} = \begin{bmatrix} 2 \\ 3 \end{bmatrix},$$
(note here that $\vec{x}$ equals its $\mathcal{B}$-coordinates -- this is why we call the basis $\mathcal{B}$ the "standard basis" in $\mathbb{R}^{2 \times 1}$). Similarly, to find $[\vec{x}]_{\mathcal{C}}$ we must find the solution to
$$c_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}.$$
Solve this by putting the appropriate augmented matrix into reduced echelon form: compute
$$\begin{bmatrix} 1&1&2 \\ 1&-1&3 \end{bmatrix} \sim \begin{bmatrix} 1&0&\frac{5}{2} \\ 0&1&-\frac{1}{2} \end{bmatrix}.$$
Therefore we have shown
$$[\vec{x}]_{\mathcal{C}} = \begin{bmatrix} \frac{5}{2} \\ -\frac{1}{2} \end{bmatrix}.$$
Now we will find $P_{\mathcal{C}\leftarrow \mathcal{B}}$ -- this is done by finding the $\mathcal{C}$-coordinates of the basis vectors in $\mathcal{B}$. By definition (pg. 465 in the text):
$$P_{\mathcal{C}\leftarrow \mathcal{B}} = \begin{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}_{\mathcal{C}} & \begin{bmatrix} 0 \\ 1 \end{bmatrix}_{\mathcal{C}} \end{bmatrix} \stackrel{\mathrm{compute}}{=} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}.$$
Now we find $P_{\mathcal{B} \leftarrow \mathcal{C}}$:
$$P_{\mathcal{B} \leftarrow \mathcal{C}} = \begin{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix}_{\mathcal{B}} & \begin{bmatrix} 1 \\ -1 \end{bmatrix}_{\mathcal{B}} \end{bmatrix} \stackrel{\mathrm{compute}}{=} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}.$$
Using property a.) in Theorem 6.12 (pg. 465 in the text): we may compute $[\vec{x}]_{\mathcal{B}}$ by using the matrix $P_{\mathcal{B}\leftarrow \mathcal{C}}$ and the vector $[\vec{x}]_{\mathcal{C}}$ using the formula
$$P_{\mathcal{B} \leftarrow \mathcal{C}} [\vec{x}]_{\mathcal{C}} = [\vec{x}]_{\mathcal{B}}.$$
Therefore we compute
$$\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} \frac{5}{2} \\ -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix},$$
as was to be shown.
Section 6.3 #7: Let $p(x)=2-x \in \mathcal{P}_1$. Let $\mathcal{B}=\{1,x\}$. Let $\mathcal{C}=\{x,1+x\}$. Find $[p]_{\mathcal{B}}$ and $[p]_{\mathcal{C}}$. Find $P_{\mathcal{C}\leftarrow \mathcal{B}}$. Use that to compute $[p]_{\mathcal{C}}$. Find $P_{\mathcal{B}\leftarrow \mathcal{C}}$. Use that to compute $[p]_{\mathcal{B}}$.
Solution: To find $[p]_{\mathcal{B}}$ we must solve
$$c_1 (1) + c_2(x) = 2-x.$$
The solution is $c_1=2$ and $c_2=-1$. Therefore
$$[p]_{\mathcal{B}} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}.$$
To find $[p]_{\mathcal{C}}$ we must solve
$$c_1 (x) + c_2(1+x) = 2-x.$$
The solution is $c_1=-3$ and $c_2=2$. Therefore
$$[p]_{\mathcal{C}} = \begin{bmatrix} -3 \\ 2 \end{bmatrix}.$$
To find $P_{\mathcal{C} \leftarrow \mathcal{B}}$, compute
$$P_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} [1]_{\mathcal{C}} & [x]_{\mathcal{C}} \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix}.$$
To find $P_{\mathcal{B} \leftarrow \mathcal{C}}$, compute
$$P_{\mathcal{B} \leftarrow \mathcal{C}} = \begin{bmatrix} [x]_{\mathcal{B}} & [1+x]_{\mathcal{B}} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}.$$
Now we may compute
$$[p]_{\mathcal{B}} = P_{\mathcal{B}\leftarrow\mathcal{C}} [p]_{\mathcal{C}}= \begin{bmatrix} 0 & 1 \\ 1 &1 \end{bmatrix} \begin{bmatrix} -3 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \end{bmatrix},$$
as expected.
Section 6.4 #1: $T \colon \mathbb{R}^{2 \times 2} \rightarrow \mathbb{R}^{2 \times 2}$ is defined by
$$T \left( \begin{bmatrix} a & b \\ c&d \end{bmatrix} \right) = \begin{bmatrix} a+b & 0 \\ 0& c+d \end{bmatrix}.$$
Is $T$ a linear transformation?
Solution: Yes -- let $\vec{u}=\begin{bmatrix} u_{11} & u_{12} \\ u_{21} & u_{22} \end{bmatrix} \in \mathbb{R}^{2 \times 2}$ and $\vec{v}=\begin{bmatrix} v_{11} & v_{12} \\ v_{21} & v_{22} \end{bmatrix} \in \mathbb{R}^{2 \times 2}$ and let $\alpha, \beta \in \mathbb{R}$ be scalars. Now compute
$$\begin{array}{ll}
T \left( \alpha \vec{u} + \beta \vec{v} \right) &= T \left( \alpha \begin{bmatrix} u_{11} & u_{12} \\ u_{21} & u_{22} \end{bmatrix} + \beta \begin{bmatrix} v_{11} & v_{12} \\ v_{21} & v_{22} \end{bmatrix} \right) \\
&=T \left( \begin{bmatrix} \alpha u_{11} + \beta v_{11} & \alpha u_{12} + \beta v_{12} \\ \alpha u_{21} + \beta v_{21} & \alpha u_{22} + \beta v_{22} \end{bmatrix} \right) \\
&= \begin{bmatrix} (\alpha u_{11} + \beta v_{11}) + (\alpha u_{12}+\beta v_{12}) & 0 \\ 0 & (\alpha u_{21} + \beta v_{21}) + (\alpha u_{22} + \beta v_{22}) \end{bmatrix} \\
&= \alpha \begin{bmatrix} u_{11} + u_{12} & 0 \\ 0 & u_{21}+u_{22} \end{bmatrix} + \beta \begin{bmatrix} v_{11}+v_{12} & 0 \\ 0 & v_{21} + v_{22} \end{bmatrix} \\
&= \alpha T(\vec{u}) + \beta T(\vec{v}),
\end{array}$$
therefore $T$ is a linear transformation.
Section 6.4 #3: $T \colon \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n}$ is defined by $T(A)=AB$, where $B$ is a fixed element of $\mathbb{R}^{n \times n}$. Is $T$ a linear transformation?
Solution: Let $X, Y \in \mathbb{R}^{n \times n}$ and let $\alpha, \beta$ be scalars. Then using the rules of matrix multiplication,
$$\begin{array}{ll}
T(\alpha X + \beta Y) &= (\alpha X + \beta Y)B \\
&= (\alpha X)B + (\beta Y)B \\
&= \alpha (XB) + \beta (YB) \\
&= \alpha T(X) + \beta T(Y),
\end{array}$$
hence $T$ is a linear transformation.
Section 6.4 #6: $T \colon \mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ is defined by $T(A)=a_{11}a_{22}\ldots a_{nn}$. Is $T$ a linear transformation?
Solution: No. Consider $n=2$ and $A=\begin{bmatrix} 3&3 \\ 3&3 \end{bmatrix}$. Then $T(A+A)=T \left( \begin{bmatrix} 6&6 \\ 6&6 \end{bmatrix} \right) = 6^2 = 36$ while $T(A)+T(A) = 9+9=18$.
Section 6.4 #10: $T \colon \mathscr{F} \rightarrow \mathscr{F}$ is defined by $T(f)=f(x^2)$. Is $T$ a linear transformation?
Solution: Let $f,g \in \mathscr{F}$ and $\alpha, \beta$ be scalars. Calculate
$$\begin{array}{ll}
T(\alpha f + \beta g) &= (\alpha f + \beta g)(x^2) \\
&= \alpha f(x^2) + \beta g(x^2) \\
&= \alpha T(f) + \beta T(g),
\end{array}$$
hence $T$ is a linear transformation.
Section 6.4 #14: Let $T \colon \mathbb{R}^{2 \times 1} \rightarrow \mathbb{R}^{3 \times 1}$ be a linear transformation for which
$$T \left( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right) = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$$
and
$$T \left( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) = \begin{bmatrix} 3 \\ 0 \\ 4 \end{bmatrix}.$$
Find $T \left( \begin{bmatrix} 5 \\ 2 \end{bmatrix} \right)$ and $T \left( \begin{bmatrix} a \\ b \end{bmatrix} \right)$.
Solution: We begin by expressing $\begin{bmatrix} a \\ b \end{bmatrix}$ as a linear combination of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Luckily this is the standard basis of $\mathbb{R}^{2 \times 1}$ so we get
$$\begin{bmatrix} a \\ b \end{bmatrix} = a \begin{bmatrix} 1 \\ 0 \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \end{bmatrix}.$$
Therefore we may compute
$$\begin{array}{ll}
T \left( \begin{bmatrix} a \\ b \end{bmatrix} \right) &= T \left( a \begin{bmatrix} 1 \\ 0 \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) \\
&\stackrel{T \mathrm{\hspace{2pt} linear}}{=} a T \left( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right) + b T \left( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) \\
&\stackrel{\mathrm{given \hspace{2pt} value}}{=} a \begin{bmatrix} 1 \\ 2 \\ - 1 \end{bmatrix} + b \begin{bmatrix} 3 \\ 0 \\ 4 \end{bmatrix} \\
&= \begin{bmatrix} a+3b \\ 2a \\ -a+4b \end{bmatrix}.
\end{array}$$
Finding the image of $\begin{bmatrix} 5 \\ 2 \end{bmatrix}$ under $T$ is just a matter of substituting $a=5$ and $b=2$.
Section 6.4 #27: Define linear transformations $S \colon \mathscr{P}_n \rightarrow \mathscr{P}_n$ and $T \colon \mathscr{P}_n \rightarrow \mathscr{P}_n$ by
$$S(p(x)) = p(x+1)$$
and
$$T(p(x))=p'(x).$$
Find $(S \circ T)(p(x))$ and $(T \circ S)(p(x))$.
Solution: Calculate
$$\begin{array}{ll}
(S \circ T)(p(x)) &= S(T(p(x))) \\
&= S(p'(x)) \\
&= p'(x+1).
\end{array}$$
Now calculate
$$\begin{array}{ll}
(T\circ S)(p(x)) &= T(S(p(x))) \\
&= T(p(x+1)) \\
&= \dfrac{\mathrm{d}}{\mathrm{d}x} p(x+1) \\
&= p'(x+1) \dfrac{\mathrm{d}}{\mathrm{d}x} (x+1) \\
&= p'(x+1).
\end{array}$$
