设A=（1 1 0）可交换矩阵 0 1 1 0 0 1

记 A=
1 0 0 0 1 0
0 1 0 + 0 0 1
0 0 1 0 0 0
= E + B
则 AX=XA
<=> EX+BX = XE+XB
<=> X+BX=X+XB
<=> BX=XB
所以求出与B交换的矩阵即可
令 X=
x11 x12 x12
x21 x22 x23
x31 x32 x33
则 由 BX=XB 得
0 x11 x12 x21 x22 x23
0 x21 x22 = x31 x32 x33
0 x31 x32 0 0 0
得
x11=x22=x33
x12=x23
x21=x31=x32=0
所以与A可交换的矩阵为
a b c
0 a b
0 0 a

设可交换的矩阵 X =
[x11 x12 x13]
[x21 x22 x23]
[x31 x32 x33]
AX =
[x11+x21+x31 x12+x22+x32 x13+x23+x33]
[x21+x31 x22+x32 x23+x33]
[x31 x32 x33]
XA =
[x11 x11+x12 x11+x12+x13]
[x21 x21+x22 x21+x22+x23]
[x31 x31+x32 x31+x32+x33]
AX= XA，
比较第 1 列得: x21 = x31 = 0
比较第 2 列得: x32 = x21 = 0, x11= x22
比较第 3 列得: x22 = x33 , x11 + x12 = x23 + x33
则 x11 = x22 = x33, x12 = x13
得可交换的矩阵 X =
[a b c]
[0 a b]
[0 0 a]
其中 a, b, c 为任意常数。