设\(\phi\)的极小多项式为
\begin{equation*}
m_{\phi}(\lambda) = (\lambda-\lambda_1)^{e_1}\cdots(\lambda-\lambda_t)^{e_t},
\end{equation*}
则
\begin{equation*}
V= R(\lambda_1)\oplus\cdots\oplus R(\lambda_t),
\end{equation*}
其中\(R(\lambda_i)= {\rm Ker}(\phi-\lambda_i {\rm id}_V)^{e_i}\ (i=1,\dots t)\)。
由于
\((\lambda-\lambda_1)^{e_1},\dots,(\lambda-\lambda_t)^{e_t}\)两两互素,根据孙子定理(
定理 1.3.18 ),存在多项式
\(g(\lambda)\)和
\(q_i(\lambda), (i=1,\dots t)\),使得对任意的
\(i\),
\begin{equation*}
g(\lambda)-\lambda_i=q_i(\lambda)(\lambda-\lambda_i)^{e_i}.
\end{equation*}
记\(h(\lambda) = \lambda -g(\lambda)\)。令\(\psi = g(\phi)\),\(\delta = h(\phi)\),则有
\begin{equation*}
\phi = \psi+\delta,\quad \psi\delta =\delta\psi.
\end{equation*}
对于任意的\(\alpha_i\in R(\lambda_i)\),\(\alpha_i\in {\rm Ker} (\phi-\lambda_i{\rm id}_V)^{e_i}\),有
\begin{equation*}
[g(\phi)- \lambda_i{\rm id}_V](\alpha_i) = q_i(\phi)(\phi-\lambda_i{\rm id}_V)^{e_i}(\alpha_i) =0,
\end{equation*}
所以
\begin{equation*}
\psi(\alpha_i)= g(\phi)(\alpha_i) = \lambda_i\alpha_i\ (i=1,\dots,t),
\end{equation*}
即每一个根子空间\(R(\lambda_i)\)都是\(\psi\)的特征子空间,所以\(\psi\)可对角化。
另一方面,
\begin{equation*}
\delta^{e_i}(\alpha_i) = h(\phi)^{e_i}(\alpha_i) = (\phi-g(\phi))^{e_i}(\alpha_i) = (\phi-\lambda_i{\rm id}_V)^{e_i}(\alpha_i) = 0.
\end{equation*}
取\(e =\max\{e_1,\dots,e_t\} \),则
\begin{equation*}
\delta^{e}(\alpha_i)= 0
\end{equation*}
对任意的\(i\)和任意的\(\alpha_i\in R(\lambda_i)\)都成立。
对任意的\(\alpha \in V =R(\lambda_1)\oplus\cdots\oplus R(\lambda_t) \),根据空间分解方式,存在\(\alpha_i\in R(\lambda_i) (i=1,\dots,t)\),使得
\begin{equation*}
\alpha = \alpha_1+\cdots+\alpha_t.
\end{equation*}
于是
\begin{equation*}
\delta^e(\alpha) = \delta^e(\alpha_1+\cdots+\alpha_t) = \delta^e(\alpha_1)+\cdots+\delta^e(\alpha_t)=0,
\end{equation*}
所以\(\delta\)是一个幂零线性变换。
下面证明分解的唯一性。设\(\phi\)有两个满足条件的分解式
\begin{equation*}
\phi = \psi+\delta = \psi_1+\delta_1.
\end{equation*}
因为\(\psi,\psi_1\)都是\(\phi\)的多项式,所以\(\psi,\psi_1\)可交换,从而可以同步对角化,进而可知\(\psi-\psi_1\)可对角化。
另一方面,设\(\delta^s=0\),\(\delta_1^k=0\),由于\(\delta,\delta_1\)可交换,
\begin{equation*}
(\delta -\delta_1)^{s+k} = \sum_{i=0}^{s+k} C_{s+k}^i \delta^{s+k-i}(-\delta_1)^i = 0.
\end{equation*}
于是可知\(\psi-\psi_1 = \delta_1-\delta\)既可对角且是幂零的线性变换,故只能是0线性变换,即
\begin{equation*}
\psi = \psi_1,\quad \delta = \delta_1,
\end{equation*}
所以分解式唯一。