## 每週問題 April 8, 2013

Let $A$ be an $n\times n$ real matrix. Show that $A$ is a skew-symmetric matrix, i.e., $A^T=-A$, if and only if $\mathbf{x}^TA\mathbf{x}=0$ for all $\mathbf{x}$.

$\displaystyle \mathbf{x}^TA\mathbf{x}=\sum_{i=1}^n\sum_{j=1}^na_{ij}x_ix_j=\sum_{i=1}^na_{ii}x_ix_i+\sum_{i

PowSol-April-8-13

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