每週問題 June 26, 2017

對於秩-1方陣 A,證明 \det (A+I)=\hbox{trace}A+1

Let A be an n\times n matrix and \hbox{rank}A=1. Prove that \det (A+I)=\hbox{trace}A+1.

 
參考解答:

證明1. 使用 Schur 三角化定理。存在一個么正矩陣 (unitary matrix) UU^\ast=U^{-1},使得 A=UTU^\ast,其中 T 是上三角矩陣且 \hbox{rank}T=1,即

T=\begin{bmatrix} t_{11}&t_{12}&\cdots&t_{1n}\\ 0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0 \end{bmatrix}

使用跡數循環不變性,

\begin{aligned} \det(A+I)&=\det(UTU^\ast+UIU^\ast)=\det(U(T+I)U^\ast)\\ &=(\det U)\det(T+I)(\det U^\ast)=\det(T+I)=t_{11}+1\\ &=\hbox{trace}T+1=\hbox{trace}(U^\ast AU)+1=\hbox{trace} AUU^\ast)+1\\ &=\hbox{trace}A+1. \end{aligned}

 
證明2. 令 A=\mathbf{u}\mathbf{v}^T,其中 \mathbf{u}\mathbf{v} 為非零向量。使用 Sylvester 行列式定理,

\begin{aligned} \det(A+I)&=\det(\mathbf{u}\mathbf{v}^T+I)=\det (\mathbf{v}^T\mathbf{u}+1)=\mathbf{v}^T\mathbf{u}+1\\ &=\hbox{trace}(\mathbf{v}^T\mathbf{u})+1=\hbox{trace} (\mathbf{u}\mathbf{v}^T)+1\\ &=\hbox{trace}A+1. \end{aligned}

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2 Responses to 每週問題 June 26, 2017

  1. m010114 says:

    因為\mathrm{rank}(A)=1,所以A必有一個非零的eigenvalue,設此值為\lambda,且0必為A的eigenvalue,0的multiclipity為n-1。依此,A可以對角化。

    A=PDP^{-1}為A的eigen decomposition,則

    \mathrm{det}(A+I)=\mathrm{det}(P(D+I)P^{-1})=\mathrm{det}(D+I)=\lambda + 1 = \mathrm{trace}(A)+1

  2. 胡霖 says:

    秩1矩阵不一定能对角化吧!例如,A=[0 1; 0 0],A的两个特征值都是0,rank=1,但是不能对角化。

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