From 239e961389b7cf7f79f4e422cd50e0d0cef214e1 Mon Sep 17 00:00:00 2001
From: Cym10x <Cym_em10x@outlook.com>
Date: Sun, 18 Jan 2026 17:59:34 +0800
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---
 素材/特征值与相似对角化.md | 11 +++++++++++
 1 file changed, 11 insertions(+)

diff --git a/素材/特征值与相似对角化.md b/素材/特征值与相似对角化.md
index 5dfc85e..dec5f20 100644
--- a/素材/特征值与相似对角化.md
+++ b/素材/特征值与相似对角化.md
@@ -116,3 +116,14 @@ D. $\boldsymbol{A}^\text{T}+\boldsymbol{A}$与$\boldsymbol{B}^\text{T}+\boldsymb
 > 而 $(5E-A)\boldsymbol x=\boldsymbol 0$，即$\begin{bmatrix}-2&-1&-2\\0&4&0\\-2&b&-2\end{bmatrix}\boldsymbol x=\boldsymbol 0$，对应的特征向量是 $(1,0,1)^\mathrm{T}$；
 > 因此，$P=\begin{bmatrix}1&1&1\\-2&0&0\\0&-1&1\end{bmatrix}$. 
 
+>[!example] 例题6
+>设 $n$ 阶方阵 $A$ 满足  $A^2 - 3A + 2E = O$ ，证明 $A$ 可相似对角化。
+
+>[!note] 解析
+>$(A - 2E)(A - E) = 0$，
+>$\therefore \text{rank}(A - 2E) + \text{rank}(A - E) \leq n$
+>$\text{rank}((A - E) - (A - 2E)) = n \leq \text{rank}(A - E)$
+>$\therefore \text{rank}(A - 2E) + \text{rank}(A - E) = n$
+>
+>则 $A$ 有两个特征值 $1,2$，几何重数和为 $n$，故有 $n$ 个线性无关的特征向量
+>证毕
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