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$\bf命题1：$设$A,B \in {M_n}\left( F\right)$且矩阵$A$各特征值互异，若$AB=BA$，则
$(1)$$A,B$可同时相似对角化
$(2)$$A,B$有公共的特征向量
$(3)$存在唯一的次数不超过$n-1$的多项式$f\left( x \right) \in F\left[ x \right]$，使得$B=f(A)$
证明：$(1)$由矩阵$A$各特征值互异知，存在可逆阵$R$，使得
\[{R^{ - 1}}AR = diag\left( {{\lambda _1}, \cdots ,{\lambda _n}} \right)\]
其中${{\lambda _1}, \cdots ,{\lambda _n}}$为$A$互异的特征值
由$AB=BA$知，${R^{ - 1}}AR \cdot {R^{ - 1}}BR = {R^{ - 1}}BR \cdot {R^{ - 1}}AR$，从而可知
\[{R^{ - 1}}BR = diag\left( {{\mu _1}, \cdots ,{\mu _n}} \right)\]
即$A,B$可同时相似对角化
$(2)$由$(1)$知，存在可逆阵$R = \left( {{\alpha _1}, \cdots ,{\alpha _n}} \right)$，使得
\[AR = Rdiag\left( {{\lambda _1}, \cdots ,{\lambda _n}} \right),BR = Rdiag\left( {{\mu _1}, \cdots ,{\mu _n}} \right)\]
即\[A{\alpha _i} = {\lambda _i}{\alpha _i},B{\alpha _i} = {\mu _i}{\alpha _i},i = 1,2, \cdots ,n\]
$(3)$由于
\[\begin{array}{l}
&B = {a_0}E + {a_1}A + {a_2}{A^2} + \cdots + {a_{n - 1}}{A^{n - 1}}\\
\Leftrightarrow& {R^{ - 1}}BR = {a_0}E + {a_1}\left( {{R^{ - 1}}AR} \right) + {a_2}{\left( {{R^{ - 1}}AR} \right)^2} + \cdots + {a_{n - 1}}{\left( {{R^{ - 1}}AR} \right)^{n - 1}}\\
\Leftrightarrow &\left\{ {\begin{array}{*{20}{c}}
{{a_0}{\lambda _1} + {a_1}{\lambda _1} + {a_2}{\lambda _1}^2 + \cdots + {a_{n - 1}}{\lambda _1}^{n - 1} = {\mu _1}}\\
\cdots \\
{{a_0}{\lambda _n} + {a_1}{\lambda _1} + {a_2}{\lambda _n}^2 + \cdots + {a_{n - 1}}{\lambda _n}^{n - 1} = {\mu _n}}
\end{array}} \right.\end{array}\]
而上述线性方程组的系数矩阵的$\bf{Vandermonde行列式}$$D=$$\prod\limits_{1 \le j < i \le n} {\left( {{\lambda _i} - {\lambda _j}} \right)} \ne 0$，故存在唯一解${{a_0},{a_1},{a_2}, \cdots ,{a_{n - 1}}}$，即存在唯一的多项式$f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + \cdots + {a_{n - 1}}{x^{n - 1}}$满足$B=f(A)$
$\bf注1：$$A,B$有公共的特征向量$\Leftrightarrow $$AB=BA$
$\bf注2：$
