## 每週問題 May 6, 2013

Let $A$ be an $n\times n$ matrix.
(a) Show that $\mathrm{adj}(A-\lambda I)$ can be expressed in the form $\sum_{k=0}^{n-1}\lambda^k A_k$, where $A_k$’s are $n\times n$ matrices.
(b) For the $A_k$’s defined in (a), show that $A_kA-A_{k-1}$, $k=1,\ldots,n-1$, are scalar matrices, i.e., $A_kA-A_{k-1}=c_kI$ for some scalar $c_k$.

(a) 令 $A$ 的特徵多項式為

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

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

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

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

$A$ 替換為 $A-\lambda I$, 則有

$\mathrm{adj}(A-\lambda I)=-(p_n(A-\lambda I)^{n-1}+\cdots+p_2(A-\lambda I)+p_1I)$

(b) 因為 $\mathrm{adj}(A-\lambda I)(A-\lambda I)=\det(A-\lambda I)I$，將 $\mathrm{adj}(A-\lambda I)=\sum_{k=0}^{n-1}\lambda^k A_k$ 代入計算，

\displaystyle \begin{aligned} \left(\sum_{k=0}^{n-1}\lambda^kA_k\right)(A-\lambda I) &= \sum_{k=0}^{n-1}\lambda^k A_k A-\sum_{k=1}^n\lambda^kA_{k-1}\\ &= A_0A-\lambda^nA_{n-1}+\sum_{k=1}^{n-1}\lambda^k(A_kA-A_{k-1}),\end{aligned}

PowSol-May-6-13