Problem 3-35
(a) Let $g:\mathbb{R}^n\to\mathbb{R}^n$ be a linear transformation of one of the following types:
$$\begin{cases} g(e_i)=e_i & i\neq j \\ g(e_j)=ae_j & \end{cases}$$ $$\begin{cases} g(e_i)=e_i & i\neq j \\ g(e_j)=e_j+e_k & \end{cases}$$ $$\begin{cases} g(e_k)=e_k & k\neq i,j\\ g(e_i)=e_j &\\ g(e_j)=e_i & \end{cases}$$ If $U$ is a rectangle, show that the volume of $g(U)$ is $|\det g|\cdot v(U)$.
(b) Prove that $|\det g|\cdot v(U)$ is the volume of $g(U)$ for any linear transformation $g:\mathbb{R}^n\to\mathbb{R}^n$. Hint: If $\det g\neq 0$, then $g$ is the composition of linear transformations of the type considered in (a).
Let $g:\mathbb{R}^n\to\mathbb{R}^n$ be a linear transformation.
Let $g=g_m\circ g_{m-1}\circ\cdots\circ g_1$, where each $g_i$ is a linear transformation of one of the three types in (a).
Let $i=\min\{j:g_j\text{ is a linear transformation of the second type in (a).}\}$.
$v(g_1(U))=|\det g_1|\cdot v(U)$.
$g_1(U)$ is also a rectangle.
So, $v(g_2(g_1(U)))=|\det g_2|\cdot v(g_1(U))=|\det g_2|\cdot |\det g_1|\cdot v(U)$.
$\cdots$
$v((g_{i-1}\circ\cdots\circ g_1)(U))=|\det g_{i-1}|\cdot\cdots\cdot |\det g_1|\cdot v(U)=|\det (g_{i-1}\circ\cdots\circ g_1)|\cdot v(U)$.
But $g_i((g_{i-1}\circ\cdots\circ g_1)(U))$ is not a rectangle.
So, we cannot continue this calculation using the result (a) anymore.
Please tell me how to complete (b) using the author's hint.
Let $U$ be an arbitrary Jordan-measurable set.
I guess if $g$ is the first type in (a), then we can show $v(g(U))=|a|\cdot v(U)$.
I guess if $g$ is the third type in (a), then we can show very easily $v(g(U))=v(U)$.
So, if $g_{i+1}$ is the first type in (a) or the third type in (a), then $v((g_{i+1}\circ\cdots\circ g_1)(U))=|\det (g_{i+1}\circ\cdots\circ g_1)|\cdot v(U)$ holds.
But if $g_{i+1}$ is also the second type in (a), $\cdots$.