$f命题1：$设$A,B in {M_n}left( F ight)$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$fleft( x ight) in Fleft[ x ight]$，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 = diagleft( {{lambda _1}, cdots ,{lambda _n}}
ight)]
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 = diagleft( {{mu _1}, cdots ,{mu _n}}
ight)]
即$A,B$可同时相似对角化
$(2)$由$(1)$知，存在可逆阵$R = left( {{alpha _1}, cdots ,{alpha _n}}
ight)$，使得
[AR = Rdiagleft( {{lambda _1}, cdots ,{lambda _n}}
ight),BR = Rdiagleft( {{mu _1}, cdots ,{mu _n}}
ight)]
即[A{alpha _i} = {lambda _i}{alpha _i},B{alpha _i} = {mu _i}{alpha _i},i = 1,2, cdots ,n]
$(3)$由于
[egin{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}
ight) + {a_2}{left( {{R^{ - 1}}AR}
ight)^2} + cdots + {a_{n - 1}}{left( {{R^{ - 1}}AR}
ight)^{n - 1}}\
Leftrightarrow &left{ {egin{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}}
ight.end{array}]
而上述线性方程组的系数矩阵的$f{Vandermonde行列式}$$D=$$prodlimits_{1 le j < i le n} {left( {{lambda _i} - {lambda _j}} ight)} e 0$，故存在唯一解${{a_0},{a_1},{a_2}, cdots ,{a_{n - 1}}}$，即存在唯一的多项式$fleft( x ight) = {a_0} + {a_1}x + {a_2}{x^2} + cdots + {a_{n - 1}}{x^{n - 1}}$满足$B=f(A)$
$f注1：$$A,B$有公共的特征向量$Leftrightarrow $$AB=BA$
$f注2：$
