设 $P=\left(\begin{array}{ccc}-1 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 1 & -1\end{array}\right), \Lambda=\left(\begin{array}{lll}1 & & \\ & 2 & \\ & & -3\end{array}\right)$ , $A P=P \Lambda$. 求 $\varphi(A)=A^3+2 A^2-3 A$.
【答案】 解: $|P|=\left|\begin{array}{ccc}-1 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 1 & -1\end{array}\right|=6 \neq 0$ ,所以矩阵 $P$ 可逆,于是有 $A=P \Lambda P^{-1}, \varphi(A)=P \varphi(\Lambda) P^{-1}$. 而 $\varphi(1)=0, \varphi(2)=10, \varphi(-3)=0$. 故 $\varphi(\Lambda)=\operatorname{diag}(0,10,0)$.
$$
\begin{aligned}
& \varphi(A)=P \varphi(\Lambda) P^{-1}=\left(\begin{array}{ccc}
-1 & 1 & 1 \\
1 & 0 & 2 \\
1 & 1 & -1
\end{array}\right)\left(\begin{array}{ccc}
0 & & \\
& 10 & \\
& & 0
\end{array}\right) \frac{1}{|P|} P^* \\
& =\frac{10}{6}\left(\begin{array}{lll}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 1 & 0
\end{array}\right)\left(\begin{array}{lll}
A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33}
\end{array}\right)=\frac{5}{3}\left(\begin{array}{ccc}
A_{12} & A_{22} & A_{32} \\
0 & 0 & 0 \\
A_{12} & A_{22} & A_{32}
\end{array}\right) \\
& \text { 而 } A_{12}=-\left|\begin{array}{cc}
1 & 2 \\
1 & -1
\end{array}\right|=3, A_{22}=\left|\begin{array}{cc}
-1 & 1 \\
1 & -1
\end{array}\right|=0 \text {, } \\
& A_{32}=-\left|\begin{array}{cc}
-1 & 1 \\
1 & 2
\end{array}\right|=3 \text {, } \\
& \text { 所以 } \varphi(A)=\left(\begin{array}{lll}
5 & 0 & 5 \\
0 & 0 & 0 \\
5 & 0 & 5
\end{array}\right) \text {. } \\
&
\end{aligned}
$$


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