## 每週問題 April 1, 2013

A real matrix $A$ is said to be skew-symmetric if $A^T=-A$. If $A$ is a real skew-symmetric matrix, show that $A^2$ is symmetric negative semi-definite matrix.

$\mathbf{x}^TA^2\mathbf{x}=-\mathbf{x}^TA^TA\mathbf{x}=-(A\mathbf{x})^T(A\mathbf{x})=-\Vert A\mathbf{x}\Vert^2\le 0$

PowSol-April-1-13

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