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证明 不妨设
φ
(
A
)
\varphi(\boldsymbol{A})
φ(A) 是矩阵
A
\boldsymbol{A}
A 的
m
1
m_1
m1 次多项式，
f
(
A
)
f(\boldsymbol{A})
f(A) 是矩阵
A
\boldsymbol{A}
A 的
m
2
m_2
m2 次多项式，令
m
=
max
(
m
1
,
m
2
)
m = \max(m_1,m_2)
m=max(m1,m2)，于是
φ
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A
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\varphi(\boldsymbol{A})
φ(A) 和
f
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A
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f(\boldsymbol{A})
f(A) 可以表示为
φ
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=
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0
E
+
a
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+
⋯
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m
A
m
f
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=
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⋯
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m
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m
\begin{align*} \varphi(\boldsymbol{A}) = a_0 \boldsymbol{E} + a_1 \boldsymbol{A} + \cdots a_m \boldsymbol{A}^m \\ f(\boldsymbol{A}) = b_0 \boldsymbol{E} + b_1 \boldsymbol{A} + \cdots b_m \boldsymbol{A}^m \\ \end{align*}
φ(A)=a0E+a1A+⋯amAmf(A)=b0E+b1A+⋯bmAm 于是根据矩阵乘法的分配律，有
φ
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A
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f
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=
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=
(
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⋅
b
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+
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⋅
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A
+
⋯
+
a
0
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⋅
b
m
A
m
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+
(
a
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⋅
b
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E
+
a
1
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⋅
b
1
A
+
⋯
+
a
1
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⋅
b
m
A
m
)
+
⋯
+
(
a
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A
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⋅
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+
a
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⋅
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+
a
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⋅
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m
)
(1)
\begin{align*} \varphi(\boldsymbol{A}) f(\boldsymbol{A}) = & (a_0 \boldsymbol{E} + a_1 \boldsymbol{A} + \cdots a_m \boldsymbol{A}^m) (b_0 \boldsymbol{E} + b_1 \boldsymbol{A} + \cdots b_m \boldsymbol{A}^m) \\ = & (a_0 \boldsymbol{E} \cdot b_0 \boldsymbol{E} + a_0 \boldsymbol{E} \cdot b_1 \boldsymbol{A} + \cdots + a_0 \boldsymbol{E} \cdot b_m \boldsymbol{A}^m) \\ & + (a_1 \boldsymbol{A} \cdot b_0 \boldsymbol{E} + a_1 \boldsymbol{A} \cdot b_1 \boldsymbol{A} + \cdots + a_1 \boldsymbol{A} \cdot b_m \boldsymbol{A}^m) \\ & + \cdots \\ & + (a_m \boldsymbol{A}^m \cdot b_0 \boldsymbol{E} + a_m \boldsymbol{A}^m \cdot b_1 \boldsymbol{A} + \cdots + a_m \boldsymbol{A}^m \cdot b_m \boldsymbol{A}^m) \end{align*} \tag{1}
φ(A)f(A)==(a0E+a1A+⋯amAm)(b0E+b1A+⋯bmAm)(a0E⋅b0E+a0E⋅b1A+⋯+a0E⋅bmAm)+(a1A⋅b0E+a1A⋅b1A+⋯+a1A⋅bmAm)+⋯+(amAm⋅b0E+amAm⋅b1A+⋯+amAm⋅bmAm)(1) 因为定理 2，所以上式
(
1
)
(1)
(1) 中所有包含
E
\boldsymbol{E}
E 的项，满足
a
i
E
⋅
b
j
A
k
=
b
j
A
k
⋅
a
i
E
a
i
A
k
⋅
b
j
E
=
b
j
E
⋅
a
i
A
k
(2)
\begin{align*} a_i \boldsymbol{E} \cdot b_j \boldsymbol{A}^k = b_j \boldsymbol{A}^k \cdot a_i \boldsymbol{E} \\ a_i \boldsymbol{A}^k \cdot b_j \boldsymbol{E} = b_j \boldsymbol{E} \cdot a_i \boldsymbol{A}^k \\ \end{align*} \tag{2}
aiE⋅bjAk=bjAk⋅aiEaiAk⋅bjE=bjE⋅aiAk(2) 因为引理 1 和矩阵乘法的结合律，所以上式
(
1
)
(1)
(1) 中所有只包含
A
\boldsymbol{A}
A 的幂的项，满足
a
i
A
k
⋅
b
j
A
l
=
b
j
A
l
⋅
a
i
A
k
(3)
a_i \boldsymbol{A}^k \cdot b_j \boldsymbol{A}^l = b_j \boldsymbol{A}^l \cdot a_i \boldsymbol{A}^k \tag{3}
aiAk⋅bjAl=bjAl⋅aiAk(3) 将式
(
2
)
(2)
(2) 和
(
3
)
(3)
(3) 代入式
(
1
)
(1)
(1)，则有
φ
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A
)
f
(
A
)
=
f
(
A
)
φ
(
A
)
\varphi(\boldsymbol{A}) f(\boldsymbol{A}) = f(\boldsymbol{A}) \varphi(\boldsymbol{A})
φ(A)f(A)=f(A)φ(A) 得证。
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