Answer
True, True, False, False, False
Work Step by Step
(a) True. This is one of the main ideas from this section. Here works. We know that every vector $\mathbf{v}$ in $\mathbb{R}^{n}$ can be written in terms of the columns of the $n \times n$ identity matrix since those columns are just the vectors $(1,0, \ldots, 0),(0,1, \ldots, 0),$ . Thus if $\mathbf{v}=\left(v_{1}, v_{2}, \ldots, v_{n}\right),$ then
\[
\mathbf{v}=v_{1} \mathbf{e}_{1}+v_{2} \mathbf{e}_{2}+\cdots+v_{n} \mathbf{e}_{n}
\]
where $\mathbf{e}_{i}$ is the $i$ th column of the identity matrix for each $i$
Now we know the 'effect [of $T]$ on the columns of the $n \times n$ identity matrix'. This means that we know how to calculate $T\left(\mathbf{e}_{i}\right)$ for each
i. Then we can use that information to calculate $T(\mathbf{v})$ for each $\mathbf{v}$ just by decomposing v and using the linearity of $T:$
\[
\begin{aligned}
T(\mathbf{v}) &=T\left(v_{1} \mathbf{e}_{1}+v_{2} \mathbf{e}_{2}+\cdots+v_{n} \mathbf{e}_{n}\right) \\
&=v_{1} T\left(\mathbf{e}_{1}\right)+v_{2} T\left(\mathbf{e}_{2}\right)+\cdots+v_{n} T\left(\mathbf{e}_{n}\right)
\end{aligned}
\]
But we already know how to calculate each $T\left(\mathbf{e}_{i}\right),$ so we just plug in those vectors and the numbers $v_{1}, v_{2}, \ldots, v_{n},$ add up, and we're done! Easy.
(b) True. In Example $3,$ the book finds the standard matrix for this transformation. A transformation is linear if and only if it has a standard matrix, so we conclude that rotations (by a fixed angle) about the origin are linear.
(c) False. We can show that if $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ and $g: \mathbb{R}^{p} \rightarrow \mathbb{R}^{n}$ are both linear transformations, then $f \circ g: \mathbb{R}^{p} \rightarrow \mathbb{R}^{n}$ is also a linear transformation.* $^{*}$
Here's how: Let $\mathbf{u}$ and $\mathbf{v}$ be any two vectors in the domain of $g$ and let $c_{1}$ and $c_{2}$ be some arbitrary numbers. Then, by linearity of $g,$ we know that
\[
(f \circ g)\left(c_{1} \mathbf{u}+c_{2} \mathbf{v}\right)=f\left(g\left(c_{1} \mathbf{u}+c_{2} \mathbf{v}\right)\right)=f\left(c_{1} g(\mathbf{u})+c_{2} g(\mathbf{v})\right)
\]
But $g(\mathbf{u})$ and $g(\mathbf{v})$ are two vectors in the domain of $f .$ Temporarily, rename those vectors $\mathbf{x}=g(\mathbf{u})$ and $\mathbf{y}=g(\mathbf{v})$. Then we conclude from the linearity of $f$ that
\[
f\left(c_{1} \mathbf{x}+c_{2} \mathbf{y}\right)=c_{1} f(\mathbf{x})+c_{2} f(\mathbf{y})=c_{1} f(g(\mathbf{u}))+c_{2} f(g(\mathbf{v}))=c_{1}(f \circ g)(\mathbf{u})+c_{2}(f \circ g)(\mathbf{v})
\]
We have thus just shown that
\[
(f \circ g)\left(c_{1} \mathbf{u}+c_{2} \mathbf{v}\right)=c_{1}(f \circ g)(\mathbf{u})+c_{2}(f \circ g)(\mathbf{v})
\]
and hence that indeed $f \circ g$ is linear.
$*$ The o symbol is an operation called 'composition' and it is used to construct the function that you get if you first transformation a number by the function $g$ and then $f($ so it kind of works right-to-left). In particular, we define $(f \circ g)(x)=f(g(x))$ for all $x$ in the domain of $g$
(e) False. Consider the matrix $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1 \\ 0 & 0\end{array}\right] .$ I claim that the linear transformation $T(\mathbf{x})=A \mathbf{x}$ is one-to-one. Let's prove it:
From Theorem $12,$ we know that $T$ is one-to-one if and only if the columns of $A$ are linearly independent. From the text in the purple box on page
$59,$ we know that the columns of $A$ are linearly independent if and only if neither of the two vectors are a scalar multiple of the other. By inspection it is clear that the two columns are not scalar multiples of each other. Hence $T$ is one-to-one as claimed. $\square$
Note that the proof I give above gives you a way to construct infinitely many counterexamples: just construct any $3 \times 2$ matrix where the columns are not scalar multiples of each other and you've got yourself a counterexample.
