每週問題 January 30, 2017

Posted on 01/30/2017 by ccjou

證明反對稱矩陣的秩必為偶數。

Prove that the rank of a real skew-symmetric matrix is an even number.

 
參考解答：

假設 A^T=-A。若 A\mathbf{x}=\lambda\mathbf{x}，\mathbf{x}\neq\mathbf{0}，則

\overline{\lambda}\mathbf{x}^\ast\mathbf{x}=(\lambda\mathbf{x})^\ast\mathbf{x}=(A\mathbf{x})^\ast\mathbf{x}=\mathbf{x}^\ast A^T\mathbf{x}=-\mathbf{x}^\ast A\mathbf{x}=-\lambda\mathbf{x}^\ast\mathbf{x}。

因此，\overline{\lambda}=-\lambda，反對稱矩陣 A 的非零特徵值是成對的共軛純虛數。再者，A^TA=AA^T，反對稱矩陣 A 是正規矩陣 (normal matrix)，故可么正對角化 (unitarily diagonalizable) 為 U^\ast AU=D=\hbox{diag}(\lambda_1,\ldots,\lambda_n)，其中 U^\ast=U^{-1}。所以，\hbox{rank}A=\hbox{rank}D，也就是說 \hbox{rank}A 等於非零特徵值數，因此為偶數。

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