Question

设A=1 1 0 1 求所有与A可交换的矩阵

Answer 1

设B
=
b1
b2
b3
b4
若
AB=BA,
则有
b1+b3
b2+b4
b3
b4
=
b1
b2+b1
b3
b4+b3
所以有
b1+b3
=
b1
b2+b4
=
b2+b1
b4
=
b4+b3
解得:
b3=0,
b1=b4
所以,所有与A可交换的矩阵为
a
b
0
a
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Answer 2

设任意与a可交换的矩阵b是
a
b
c
d
e
f
g
h
i
ab=
0
1
0
0
0
1
0
0
0
×
a
b
c
d
e
f
g
h
i
=
d
e
f
g
h
i
0
0
0
ba=
a
b
c
d
e
f
g
h
i
×
0
1
0
0
0
1
0
0
0
=
0
a
b
0
d
e
0
g
h
则
d=g=h=0
a=e=i
f=b
即b=
a
b
c
0
a
b
0
0
a