若当标准型

证明. 计算特征多项式 \begin{align*} |\lambda I-A|=(\lambda-1) \lambda^2=0 \Rightarrow \lambda_1 & =1 \\ \lambda_2 & =\lambda_3=0 \end{align*} 得 $1$ 的几何重数为 $1$, $0$ 的几何重数也为 $1$, 那么 Jordan 标准型为 $$ J=\left[\begin{array}{cc:c} 0 & 1 & \\ & 0 & \\ \hdashline && 1 \end{array}\right] $$ 设 $P=\left(\alpha_1, \alpha_2, \alpha_3\right) ， \alpha_i \in \mathbb{R}^3$ 为过渡矩阵, 即 \begin{align*} P^{-1} A P=J & \Leftrightarrow A P=P J \\ & \Leftrightarrow A\left(\alpha_1, \alpha_2, \alpha_3\right)=\left(\alpha_1, \alpha_2, \alpha_3\right)\left(\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right) \\ & \Leftrightarrow\left\{\begin{array}{l} A \alpha_1=0 \\ A \alpha_2=\alpha_1 \\ A \alpha_3=\alpha_3 \end{array}\right. \end{align*} 其中 $\alpha_1, \alpha_2, \alpha_3$ 不唯一，可取 $$ P=\left(\alpha_1, \alpha_2, \alpha_3\right)=\left(\begin{array}{lll} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right) $$ 计算得 $$ P^{-1}=\left(\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{array}\right) $$ 则 $A=P J P^{-1}$.