[再寄小读者之数学篇](2014-10-08 矩阵对称或反对称的一个充分条件)

设$A\in M_{n}(\mathbb F)$,且对任意的$\alpha,\beta\in\mathbb F^n$ 有 $$\alpha^TA\beta=0\Leftrightarrow\beta^TA\alpha=0$$ 且$A$不是对称矩阵,证明$A^T=-A$.

证明: [from 龙凤呈祥] 只需说明$a_{ii}=0$且 $$a_{ij}=-a_{ij},i\neq j$$ 由于不对称,不失一般性,不妨设$a_{12}\neq a_{21}$,那么二者不全为零,不妨设$a_{12}\neq0$. 首先说明$a_{11}=0$,否则 $$\beex \bea e_{1}^TAe_{2}-\frac{a_{12}}{a_{11}}e_{1}^TAe_{1} &=e_{1}^TA\left(e_{2}-\frac{a_{12}}{a_{11}}e_{1}\right)=0\\ \Rightarrow \left(e_2-\frac{a_{12}}{a_{11}}e_{1}\right)^TAe_{1}&=0\\ \Rightarrow a_{12}&=a_{21} \eea \eeex$$ 矛盾!所以$a_{11}=0$.类似可得$a_{22}=0$. 再说名一定有$a_{12}=-a_{21}$,这个注意到 $$\beex \bea 0&=a_{11}a_{12}+a_{12}a_{21}-a_{21}a_{12}-a_{22}a_{21}\\ &=\left(a_{21}e_{1}-a_{12}e_{2}\right)^TA(e_{1}+e_{2})\\ \Rightarrow (e_{1}+e_{2})^TA\left(a_{21}e_{1}-a_{12}e_{2}\right)&=0\\ \Rightarrow a_{12}^2&=a_{21}^2 \eea \eeex$$ 又二者不等,所以$a_{12}=-a_{21}\neq0$. 再来说明必有$a_{1j}=-a_{j1},j=3,\cdots,n$,如果$a_{1i}=a_{i1}=0$,那么显然成立.因此只需考虑某个$j$使得 $$a_{1j}\neq0$$ 的情形即可.那么 $$\beex \bea 0&=e_{1}^TAe_{2}-\frac{a_{12}}{a_{1j}}e_{1}^TAe_{j}\\ &=e_{1}^TA\left(e_{2}-\frac{a_{12}}{a_{1j}}e_{j}\right)\\ \Rightarrow \left(e_{2}-\frac{a_{12}}{a_{1j}}e_{j}\right)^TAe_{1}&=a_{21}-\frac{a_{12}}{a_{1j}}a_{j1}=0\\ \Rightarrow a_{1j}&=-a_{j1} \eea \eeex$$ 继续重复上面的步骤就可以说明$A$反对称. 

注记: [from torsor] 这是高代教材中的一个定理，你可以参考复旦高代第二版教材的定理10.3.1；如果没有复旦教材的话，可以参考Roman 的《Advanced Linear Algebra, 3rd ed.》第266页的Theorem 11.4.