每週問題 June 16, 2014

這是一個非常重要的命題:實對稱矩陣對應相異特徵值的特徵向量必定正交。

Let A be a real symmetric matrix. If \mathbf{x} and \mathbf{y} are eigenvectors of A, corresponding to distinct eigenvalues, show that \mathbf{x} and \mathbf{y} are orthogonal.

 
參考解答:

下面介紹一個快捷的證明。因為 A^T=A

\displaystyle  (A\mathbf{x})^T\mathbf{y}=\mathbf{x}^TA^T\mathbf{y}=\mathbf{x}^T(A\mathbf{y})

將特徵方程 A\mathbf{x}=\lambda\mathbf{x}A\mathbf{y}=\mu\mathbf{y} 代入上式,立得

\displaystyle  \lambda\mathbf{x}^T\mathbf{y}=\mu\mathbf{x}^T\mathbf{y}

\lambda\neq\mu,證得 \mathbf{x}^T\mathbf{y}=0

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