每週問題 February 25, 2013

這是關於半正定矩陣的判別法。

Let A be an n\times n real symmetric matrix. Show that A is positive semidefinite if and only if X^TAX is positive semidefinite for all n\times m real matrix X.

 
參考解答:

首先證明 X^TAX 是一 m\times m 階實對稱矩陣:(X^TAX)^T=X^TA^TX=X^TAX。對於任一 n 維實向量 \mathbf{x},若 \mathbf{x}^TA\mathbf{x}\ge 0,則對於所有 m 維實向量 \mathbf{y}

\mathbf{y}^T(X^TAX)\mathbf{y}=(X\mathbf{y})^TA(X\mathbf{y})\ge 0

因此證明 X^TAX 是半正定。相反的,假設對於所有 n\times m 階實矩陣 XX^TAX 是半正定。令

X=\begin{bmatrix}  \mathbf{x}&\mathbf{0}&\cdots&\mathbf{0}  \end{bmatrix}

其中 \mathbf{x} 是任一 n 維實向量,則

X^TAX=\begin{bmatrix}  \mathbf{x}^T\\  \mathbf{0}^T\\  \vdots\\  \mathbf{0}^T  \end{bmatrix}A\begin{bmatrix}  \mathbf{x}&\mathbf{0}&\cdots&\mathbf{0}  \end{bmatrix}=\begin{bmatrix}  \mathbf{x}^TA\mathbf{x}&0&\cdots&0\\  0&0&\cdots&0\\  \vdots&\vdots&\ddots&\vdots\\  0&0&\cdots&0  \end{bmatrix}

上式說明 X^TAX 有特徵值 \mathbf{x}^TA\mathbf{x},以及 m-1 個零特徵值,故可推論 \mathbf{x}^TA\mathbf{x}\ge 0

PowSol-Feb-25-13

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