## 每週問題 October 19, 2015

Let $A$ and $B$ be $n\times n$ matrices such that $AB=BA$. If $A$ has $n$ distinct eigenvalues, show that $B$ can be expressed uniquely as a polynomial in $A$ with degree no more than $n-1$.

$n\times n$ 階矩陣 $A$ 的相異特徵值為 $\lambda_1,\ldots,\lambda_n$，對應線性獨立的特徵向量 $\mathbf{x}_1,\ldots,\mathbf{x}_n$，則 $A$ 可對角化為 $S^{-1}AS=\Lambda$，其中 $\Lambda=\hbox{diag}(\lambda_1,\ldots,\lambda_n)$$S=\begin{bmatrix} \mathbf{x}_1&\cdots&\mathbf{x}_n \end{bmatrix}$。令 $D=[d_{ij}]=S^{-1}BS$。因為 $AB=BA$

$\Lambda D=S^{-1}ASS^{-1}BS=S^{-1}ABS=S^{-1}BAS=S^{-1}BSS^{-1}AS=D\Lambda$

$p(t)=c_{n-1}t^{n-1}+\cdots+c_1t+c_0$

\begin{aligned} c_0+\lambda_1c_1+\cdots+\lambda_1^{n-1}c_{n-1}&=\mu_1\\ c_0+\lambda_2c_1+\cdots+\lambda_2^{n-1}c_{n-1}&=\mu_2\\ &\vdots\\ c_0+\lambda_nc_1+\cdots+\lambda_n^{n-1}c_{n-1}&=\mu_n. \end{aligned}

$\begin{bmatrix} 1&\lambda_1&\cdots&\lambda_1^{n-1}\\ 1&\lambda_2&\cdots&\lambda_2^{n-1}\\ \vdots&\vdots&\ddots&\vdots\\ 1&\lambda_n&\cdots&\lambda_n^{n-1} \end{bmatrix}$

(稱為 Vandermonde 矩陣) 可逆，線性方程組存在唯一解，換句話說，存在唯一的多項式 $p$ 滿足 $p(\Lambda)=\hbox{diag}(p(\lambda_1),\ldots,p(\lambda_n))=\hbox{diag}(\mu_1,\ldots,\mu_n)=D$。使用上式，可得

$p(A)=p(S\Lambda S^{-1})=Sp(\Lambda)S^{-1}=SDS^{-1}=B$