## 答鄧勇──關於λ-矩陣的伴隨矩陣關係式

$B=A-\lambda I=\begin{bmatrix} a_{11}-\lambda&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}-\lambda&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}-\lambda \end{bmatrix}$

$(\mathrm{adj}(A-\lambda I))(A- \lambda I)=(\det(A-\lambda I))I$

$\mathrm{adj}(A-\lambda I)=\begin{bmatrix} d-\lambda&-b\\ -c&a-\lambda \end{bmatrix}=\mathrm{adj}A-\lambda I$

\begin{aligned} (\mathrm{adj}(A-\lambda I))(A-\lambda I)&=(\mathrm{adj}A-\lambda I)(A-\lambda I)\\ &=\lambda^2I-(\mathrm{adj}A+A)\lambda+(\mathrm{adj}A)A\\ &=\lambda^2I-(\mathrm{trace}A)\lambda I+(\det A)I\\ &=\left(\lambda^2-(\mathrm{trace}A)\lambda+\det A\right)I\\ &=(\det(A-\lambda I))I.\end{aligned}

$p(t)=\det(A-tI)=p_nt^n+\cdots+p_1t+p_0$

$q(t)=\det(B-tI)=q_nt^n+\cdots+q_1t+q_0$

$p(A)=p_nA^n+\cdots+p_1A+p_0I=0$

$\left(-p_nA^{n-1}-\cdots-p_2A-p_1I\right)A=(\det A)I$

$\mathrm{adj}A=-\left(p_nA^{n-1}+\cdots+p_2A+p_1I\right)$

$\mathrm{adj}B=-\left(q_nB^{n-1}+\cdots+q_2B+q_1I\right)$

$\mathrm{adj}(A-\lambda I)=-\left(q_n(A-\lambda I)^{n-1}+\cdots+q_2(A-\lambda I)+q_1I\right)$

\begin{aligned} p(t)&=\det(A-tI)=\det(B+\lambda I-tI)\\ &=\det(B-(t-\lambda)I)=q(t-\lambda). \end{aligned}

\begin{aligned} (\mathrm{adj}(A-\lambda I))(A-\lambda I)&=-\left(q_n(A-\lambda I)^{n-1}+\cdots+q_2(A-\lambda I)+q_1I\right)(A-\lambda I)\\ &=-\left(q_n(A-\lambda I)^n+\cdots+q_1(A-\lambda I)+q_0I\right)+q_0I\\ &=-q(A-\lambda I)+(\det(A-\lambda I))I\\ &=(\det(A-\lambda I))I, \end{aligned}

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