每週問題 April 8, 2013

這是證明反對稱矩陣的二次型必為零。

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}.

 
參考解答:

寫出

\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<j}(a_{ij}+a_{ji})x_ix_j

若所有 \mathbf{x} 皆使上面的二次型為零,則每一 i, j 滿足 a_{ij}+a_{ji}=0,即 A^T=-A。具體地說,設 \mathbf{x}=\mathbf{e}_i,則 a_{ii}=0;設 \mathbf{x}=\mathbf{e}_i+\mathbf{e}_ji\neq j,則 a_{ij}+a_{ji}=0。相反陳述明顯成立。

PowSol-April-8-13

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