vault backup: 2026-01-21 00:46:21

编写小组/讲义/矩阵相似变换（解析版）.md

@@ -198,7 +198,7 @@ $$
 由 $\left[\boldsymbol{\beta}_1\;\boldsymbol{\beta}_2\;\dots\;\boldsymbol{\beta}_n\right]=\left[\boldsymbol{\alpha}_1\;\boldsymbol{\alpha}_2\;\dots\;\boldsymbol{\alpha}_n\right]C$，且  $$
 \left[T(\boldsymbol{\alpha}_1)\;T(\boldsymbol{\alpha}_2)\;\dots\;T(\boldsymbol{\alpha}_n)\right]=\left[\boldsymbol{\alpha}_1\;\boldsymbol{\alpha}_2\;\dots\;\boldsymbol{\alpha}_n\right]A,$$  $$
 \left[T(\boldsymbol{\beta}_1)\;T(\boldsymbol{\beta}_2)\;\dots\;T(\boldsymbol{\beta}_n)\right]=\left[\boldsymbol{\beta}_1\;\boldsymbol{\beta}_2\;\dots\;\boldsymbol{\beta}_n\right]B.$$
-由 $\boldsymbol \beta_i=c_{1i}\boldsymbol\alpha_1+\cdots+c_{ni}\boldsymbol\alpha_n$ 及线性变换的定义可知$$T(\boldsymbol\beta_i)=c_{1i}T(\boldsymbol\alpha_1)+\cdots+c_{ni}T(\boldsymbol\alpha_n),$$
+由 $\boldsymbol \beta_i=c_{1i}\boldsymbol\alpha_1+\cdots+c_{ni}\boldsymbol\alpha_n$ 及线性变换的定义可知$$T(\boldsymbol\beta_i)=c_{1i}T(\boldsymbol\alpha_1)+\cdots+c_{ni}T(\boldsymbol\alpha_n)=[T(\boldsymbol\alpha_1)\ \cdots T(\boldsymbol\alpha_n)]\begin{bmatrix}c_{1i}\\\vdots\\c_{ni}\end{bmatrix},$$
 从而：$$
 \begin{aligned}
 \left[T(\boldsymbol{\beta}_1)\;T(\boldsymbol{\beta}_2)\;\dots\;T(\boldsymbol{\beta}_n)\right]&=\left[T(\boldsymbol{\alpha}_1)\;T(\boldsymbol{\alpha}_2)\;\dots\;T(\boldsymbol{\alpha}_n)\right]C \\