这是一份warwick华威大学MA140-10/MA152-15的成功案例

Proof As is shown in linear algebra, the matrix $A$ that represents $T$ is a product of elementary matrices
$$
A=E_{1} \cdots E_{k} .
$$
Each elementary $2 \times 2$ matrix is one of the following types:
$$
\left[\begin{array}{ll}
\lambda & 0 \
0 & 1
\end{array}\right] \quad\left[\begin{array}{ll}
1 & 0 \
0 & \lambda
\end{array}\right] \quad\left[\begin{array}{ll}
0 & 1 \
1 & 0
\end{array}\right] \quad\left[\begin{array}{ll}
1 & \sigma \
0 & 1
\end{array}\right]
$$
where $\lambda>0$. The first three matrices represent isomorphisms whose effect on $I^{2}$ is obvious: $I^{2}$ is converted to the rectangles $\lambda I \times I, I \times \lambda I, I^{2}$. In each case, the area agrees with the magnitude of the determinant. The fourth isomorphism converts $I^{2}$ to the parallelogram$\Pi$ is Riemann measurable since its boundary is a zero set. By Fubini’s Theorem, we get
$$
|\Pi|=\int \chi_{\Pi}=\int_{0}^{1}\left[\int_{x=\sigma x}^{x=1+\sigma y} 1 d x\right] d y=1=\operatorname{det} E .
$$

MA140-10/MA152-15 COURSE NOTES ：

$$
d x_{I}: \varphi \mapsto \int_{l^{k}} \frac{\partial \varphi_{I}}{\partial u} d u
$$
where this integral notation is shorthand for
$$
\int_{0}^{1} \ldots \int_{0}^{1} \frac{\partial\left(\varphi_{i_{1}}, \ldots, \varphi_{i_{k}}\right)}{\partial\left(u_{1}, \ldots, u_{k}\right)} d u_{1 \ldots} . d v_{k}
$$
If $f$ is a smooth function on $\mathbb{R}^{n}$ then $f d x_{l}$ is the functional
$$
f d x_{I}: \varphi \mapsto \int_{I^{k}} f(\varphi(u)) \frac{\partial \varphi_{I}}{\partial u} d u
$$