每週問題 May 6, 2013

Posted on 05/06/2013 by ccjou

這是關於伴隨矩陣 \mathrm{adj}(A-\lambda I) 的表達式問題。

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.

 
參考解答：

(a) 令 A 的特徵多項式為

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

其中 p_0=p(0)=\det A。Cayley-Hamilton 定理表明 A 被 p(t) 消滅，即

p(A)=p_nA^n+\cdots+p_1A+p_0I=0。

將上式改寫成

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

因為 (\mathrm{adj}A)A=(\det A)I，可得

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

將 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)，

即證得所求。

(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}

上式等號右邊為一純量矩陣。由於 A_0=\mathrm{adj}A，即知 A_0A=(\det A)I 為一純量矩陣。另外，A_{n-1}=(-1)^np_nI 亦為純量矩陣，故可推論 A_kA-A_{k-1}，k=1,\ldots,n-1，必定也是純量矩陣。

PowSol-May-6-13

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