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一些有趣的线代题目.md

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+设 $n$ 阶方阵 $A$ 满足  
+$$ A^2 - 3A + 2E = O $$  
+证明 $A$ 可相似对角化。
+$$
+\lambda^2 - 3\lambda + 2 = 0
+$$
+
+$$
+\lambda_1 = 2 \quad \lambda_2 = 1
+$$
+
+$$
+(A - 2E)(A - E) = 0
+$$
+
+$$
+\therefore \text{rank}(A - 2E) + \text{rank}(A - E) \leq n
+$$
+
+$$
+\text{rank}((A - E) - (A - 2E)) = n \leq \text{rank}(A - E)
+$$
+
+
+$$
+\therefore \text{rank}(A - 2E) + \text{rank}(A - E) = n
+$$
+设 A - 2E对应几何重数为 $k$，A - E对应几何重数为 $n-k$
+
+则：
+
+$$
+n - \text{rank}(A - 2E) \leq k
+$$
+
+$$
+n - \text{rank}(A - E) \leq n - k
+$$
+
+$$
+\therefore \text{rank}(A - 2E) \geq n - k
+$$
+
+$$
+\text{rank}(A - E) \geq k
+$$
+
+$$
+\therefore \text{rank}(A - 2E) = n - k
+$$
+$$\text{rank}(A - E) =  k$$
+
+证毕