我正在学习线性代数，但目前很难直观理解为什么坐标变换矩阵可以是对合矩阵（即自身的逆）。

设$\mathbb{R}^2$中的两个有序基：
$$
A = \left{\begin{bmatrix}1 \0\end{bmatrix}, \begin{bmatrix}1 \1\end{bmatrix}\right}, \quad B = \left{\begin{bmatrix}1 \1\end{bmatrix}, \begin{bmatrix}1 \0\end{bmatrix}\right}
$$

为了求$P_{A\to B}$，我将A作为增广矩阵的左半部分，B作为右半部分：
$$
\begin{pmatrix}1 & 1 & \bigm | & 1 & 1\1 & 0 & \bigm | & 0 & 1\\end{pmatrix}\xrightarrow{\substack{r_2-r_1 \ -1r_2 \ r_1-r_2}}\begin{pmatrix}1 & 0 & \bigm | & 0 & 1\0 & 1 & \bigm | & 1 & 0\\end{pmatrix}
$$
因此得到：
$$
P_{A\to B} = \begin{pmatrix}0 & 1 \1 & 0\end{pmatrix}
$$

类似地，求$P_{B\to A}$时：
$$
\begin{pmatrix}1 & 1 & \bigm | & 1 & 1\0 & 1 & \bigm | & 1 & 0\\end{pmatrix}\xrightarrow{\substack{r_1-r_2}}\begin{pmatrix}1 & 0 & \bigm | & 0 & 1\0 & 1 & \bigm | & 1 & 0\\end{pmatrix}
$$
同样得到：
$$
P_{B\to A} = \begin{pmatrix}0 & 1 \1 & 0\end{pmatrix}
$$

显然$P_{A\to B}$和$P_{B\to A}$互为逆矩阵，满足$(P_{A\to B})(P_{B\to A}) = (P_{B\to A})(P_{A\to B}) = I$。但我很难直观理解：为什么从基B回到A的逆变换，和从A到B的变换是同一个矩阵？是不是仅仅因为基A和B包含的向量集合是相同的？

我不确定我的问题是否合理，但希望有人能帮我理清思路。谢谢！

Hey there! Great question—this is totally a common intuitive hangup when first diving into coordinate transformations, so rest assured, it makes perfect sense to ask.

Your guess is exactly right: the key reason this transformation matrix is its own inverse (we call this an involutory matrix) is that bases A and B are the exact same set of vectors—they just have their order reversed. Let’s break this down intuitively:

先回忆坐标变换矩阵的本质 #

$P_{A\to B}$的作用是：把一个向量在基A下的坐标，转换成它在基B下的坐标。

• 对于基A = ${v_1, v_2}$，基B = ${v_2, v_1}$（也就是A的向量顺序反过来）：
  • 假设某个向量$\mathbf{x}$在基A下的坐标是$\begin{bmatrix}a \ b\end{bmatrix}$，这意味着$\mathbf{x} = a v_1 + b v_2$。
  • 换到基B下，这个向量可以写成$\mathbf{x} = b v_2 + a v_1$，对应的坐标就是$\begin{bmatrix}b \ a\end{bmatrix}$——而这正是用矩阵$\begin{pmatrix}0 & 1 \1 & 0\end{pmatrix}$乘$\begin{bmatrix}a \ b\end{bmatrix}$的结果。

为什么逆变换是同一个矩阵？ #

Now think about going back from B to A:

• If $\mathbf{x}$ has coordinates $\begin{bmatrix}c \ d\end{bmatrix}$ in basis B, that means $\mathbf{x} = c v_2 + d v_1$.
• To write this in basis A, we rearrange it to $\mathbf{x} = d v_1 + c v_2$, so the coordinates become $\begin{bmatrix}d \ c\end{bmatrix}$—again, this is exactly what the same swap matrix does!

In short: swapping the order of the basis vectors means "swapping the coordinate positions" both ways. Do that swap twice, and you’re back to the original coordinates (which is why $P \times P = I$—applying the transformation twice gives the identity).

更一般的对合变换小补充 #

This isn’t the only case where a coordinate transformation matrix is involutory. For example, if you have basis $C = {v, w}$ and basis $D = {v, -w}$, the transformation matrix would be $\begin{pmatrix}1 & 0 \0 & -1\end{pmatrix}$, which is also its own inverse (since squaring -1 gives 1). But in your specific example, the core cause is definitely that the two bases are identical sets—only their ordering is flipped.

So your intuition was spot-on here! The transformation works the same both ways because reversing the basis order means the "forward" and "backward" coordinate swaps are the exact same operation.

备注：内容来源于stack exchange，提问作者DPhysics_Student