例题 1.3

设对角矩阵

A = a_{11} 0 ⋮ 0 0 a_{22} ⋮ 0 \dots \dots \dots 00 ⋮ a_{nn}

其中 $a_{ii} \neq = a_{jj} (i \neq = j; i, j = 1, 2, \dots, n)$ , 证明和 $A$ 可交换的矩阵只能是对角矩阵. 证明设矩阵 $B = b_{11} b_{21} ⋮ b_{n 1} b_{12} b_{22} ⋮ b_{n 2} \dots \dots \dots b_{1 n} b_{2 n} ⋮ b_{nn}$ 和 $A$ 可交换, 那么有

b_{11} b_{21} ⋮ b_{n 1} b_{12} b_{22} ⋮ b_{n 2} \dots \dots \dots b_{1 n} b_{2 n} ⋮ b_{nn} a_{11} 0 ⋮ 0 0 a_{22} ⋮ 0 \dots \dots \dots 00 ⋮ a_{nn} = a_{11} 0 ⋮ 0 0 a_{22} ⋮ 0 \dots \dots \dots 00 ⋮ a_{nn} b_{11} b_{21} ⋮ b_{n 1} b_{12} b_{22} ⋮ b_{n 2} \dots \dots \dots b_{1 n} b_{2 n} ⋮ b_{nn}

即有

a_{11} b_{11} a_{11} b_{21} ⋮ a_{11} b_{n 1} a_{22} b_{12} a_{22} b_{22} ⋮ a_{22} b_{n 2} \dots \dots \dots a_{nn} b_{1 n} a_{nn} b_{2 n} ⋮ a_{nn} b_{nn} = a_{11} b_{11} a_{22} b_{21} ⋮ a_{nn} b_{n 1} a_{11} b_{12} a_{22} b_{22} ⋮ a_{nn} b_{n 2} \dots \dots \dots a_{11} b_{1 n} a_{22} b_{2 n} ⋮ a_{nn} b_{nn}

依次比较等式两边第 1 行，第 2 行， $\dots$ ，第 $n$ 行相应位置上的元素，可以得到

b_{12} = b_{13} = \dots = b_{1 n} = 0 b_{21} = b_{23} = \dots = b_{2 n} = 0 \dots b_{n 1} = b_{n 2} = \dots = b_{n, n - 1} = 0

故结论成立。

Youliang Zhong

例题 1.3

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