设\(A\)与\(J(\lambda_0,n)\)可交换，即
\begin{equation*}
A\cdot\left(\lambda_0E_n+J(0,n)\right)=\left(\lambda_0E_n+J(0,n)\right)\cdot A,
\end{equation*}
则
\begin{equation*}
AJ(0,n)=J(0,n)A.
\end{equation*}
由于
\begin{equation*}
AJ(0,n)=\begin{pmatrix}
a_{12}&a_{13}&\cdots&a_{1n}&0\\
a_{22}&a_{23}&\cdots&a_{2n}&0\\
\vdots&\vdots&&\vdots&\vdots\\
a_{n-1,2}&a_{n-1,3}&\cdots&a_{n-1,n}&0\\
a_{n2}&a_{n3}&\cdots&a_{nn}&0
\end{pmatrix},
\end{equation*}
\begin{equation*}
J(0,n)A=\begin{pmatrix}
0&0&\cdots&0&0\\
a_{11}&a_{12}&\cdots&a_{1,n-1}&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2,n-1}&a_{2n}\\
\vdots&\vdots&&\vdots&\vdots\\
a_{n-1,1}&a_{n-1,2}&\cdots&a_{n-1,n-1}&a_{n-1,n}\\
\end{pmatrix},
\end{equation*}
比较系数得\(a_{ij}=0,\forall 1\leq i<j\leq n\)，且
\begin{equation*}
\begin{array}{c}
a_{11}=a_{22}=a_{33}=a_{44}=\cdots=a_{nn},\\
a_{21}=a_{32}=a_{43}=\cdots=a_{n,n-1},\\
a_{31}=a_{42}=\cdots=a_{n,n-2},\\
\vdots \\
a_{n-1,1}=a_{n2}
\end{array}
\end{equation*}
即
\begin{equation*}
A=\begin{pmatrix}
a_{1}&0&0&\cdots&0&0&0\\
a_2&a_1&0&\cdots&0&0&0\\
a_3&a_2&a_1&\cdots&0&0&0\\
a_4&a_3&a_2&\ddots&\vdots&\vdots&\vdots\\
a_5&a_4&\ddots&\ddots&a_1&0&0\\
\vdots&\ddots&\ddots&a_3&a_2&a_1&0\\
a_n&\cdots&a_5&a_4&a_3&a_2&a_1
\end{pmatrix},
\end{equation*}
所以
\begin{equation*}
\begin{array}{ccl}
A&=&a_1E_n+a_2J(0,n)+a_3J(0,n)^2+\cdots+a_nJ(0,n)^{n-1}\\
&=&a_1E_n+a_2\left(J(\lambda_0,n)-\lambda_0E_n\right)+a_3\left(J(\lambda_0,n)-\lambda_0E_n\right)^2\\
&&+\cdots+a_n\left(J(\lambda_0,n)-\lambda_0E_n\right)^{n-1}.
\end{array}
\end{equation*}
令
\begin{equation*}
f(\lambda)=a_1+a_2(\lambda-\lambda_0)+a_3(\lambda-\lambda_0)^2+\cdots+a_n(\lambda-\lambda_0)^{n-1},
\end{equation*}
则\(A\)可表示为\(J(\lambda_0,n)\)的多项式\(f\left(J(\lambda_0,n)\right)\)。
