设$Ain M_{n}(mathbb F)$,且对任意的$alpha,etainmathbb F^n$ 有$$ alpha^TAeta=0Leftrightarroweta^TAalpha=0 $$ 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$$ a_{ij}=-a_{ij},i eq j $$ 由于不对称,不失一般性,不妨设$a_{12} eq a_{21}$,那么二者不全为零,不妨设$a_{12} eq0$. 首先说明$a_{11}=0$,否则 $$eex ea e_{1}^TAe_{2}-frac{a_{12}}{a_{11}}e_{1}^TAe_{1} &=e_{1}^TAleft(e_{2}-frac{a_{12}}{a_{11}}e_{1} ight)=0\ Rightarrow left(e_2-frac{a_{12}}{a_{11}}e_{1} ight)^TAe_{1}&=0\ Rightarrow a_{12}&=a_{21} eea eeex$$ 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 $$eex ea 0&=a_{11}a_{12}+a_{12}a_{21}-a_{21}a_{12}-a_{22}a_{21}\ &=left(a_{21}e_{1}-a_{12}e_{2} ight)^TA(e_{1}+e_{2})\ Rightarrow (e_{1}+e_{2})^TAleft(a_{21}e_{1}-a_{12}e_{2} ight)&=0\ Rightarrow a_{12}^2&=a_{21}^2 eea eeex$$ 又二者不等,所以$a_{12}=-a_{21} eq0$. 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$$ a_{1j} eq0 $$ 的情形即可.那么 $$eex ea 0&=e_{1}^TAe_{2}-frac{a_{12}}{a_{1j}}e_{1}^TAe_{j}\ &=e_{1}^TAleft(e_{2}-frac{a_{12}}{a_{1j}}e_{j} ight)\ Rightarrow left(e_{2}-frac{a_{12}}{a_{1j}}e_{j} ight)^TAe_{1}&=a_{21}-frac{a_{12}}{a_{1j}}a_{j1}=0\ Rightarrow a_{1j}&=-a_{j1} eea eeex$$ 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.