# 张祖锦常用结论反对称矩阵的合同标准形
设 $\displaystyle A$ 是反对称矩阵, 则 $\displaystyle A$ 合同于下列形状的分块矩阵:
\begin\{aligned\} \mathrm\{diag\}(\underbrace\{S,\cdots,S\}\_r,0\_\{n-2r\}),\quad S=\left(\begin\{array\}\{cccccccccc\}0&1\\\\ -1&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 对矩阵的阶数 $\displaystyle n$ 作数学归纳法. 当 $\displaystyle n=1$ 时, $\displaystyle A=(0)$, 结论自明. 当 $\displaystyle n=2$ 时, 若 $\displaystyle A=0$, 也结论也成立. 若 $\displaystyle A\neq 0$, 则可设 $\displaystyle A=\left(\begin\{array\}\{cccccccccc\}0&a\\\\ -a&0\end\{array\}\right)$, $\displaystyle a\neq 0$. 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccc\}\frac\{1\}\{a\}&0\\\\ 0&1\end\{array\}\right)\left(\begin\{array\}\{cccccccccc\}0&a\\\\ -a&0\end\{array\}\right)\left(\begin\{array\}\{cccccccccc\}\frac\{1\}\{a\}&0\\\\ 0&1\end\{array\}\right)=\left(\begin\{array\}\{cccccccccc\}0&1\\\\ -1&0\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle A$ 合同于 $\displaystyle \left(\begin\{array\}\{cccccccccc\}0&1\\\\ -1&0\end\{array\}\right)$. 结论也得证. 假设对阶小于 $\displaystyle n$ 的矩阵结论成立, 现考虑 $\displaystyle n$ 阶反对称矩阵 $\displaystyle A$. 若 $\displaystyle A=0$, 则结论已成立. 若 $\displaystyle A\neq 0$, 则由反对称矩阵的主对角元为 $\displaystyle 0$ 知 $\displaystyle \exists\ i\neq j,\mathrm\{ s.t.\} a\_\{ij\}\neq 0$. 将 $\displaystyle A$ 的第 $\displaystyle 1$ 行与第 $\displaystyle i$ 行对调, 第 $\displaystyle 1$ 列与第 $\displaystyle i$ 列对调; 第 $\displaystyle 2$ 行与第 $\displaystyle j$ 行对调, 第 $\displaystyle 2$ 列与第 $\displaystyle j$ 列对调; 第 $\displaystyle 1$ 行乘以 $\displaystyle \frac\{1\}\{a\_\{ij\}\}$, 第 $\displaystyle 1$ 列乘以 $\displaystyle \frac\{1\}\{a\_\{ij\}\}$, 最后得到的矩阵合同于 $\displaystyle A$, 且具有下列形状:
\begin\{aligned\} M=\left(\begin\{array\}\{cccccccccc\}S&B\\\\ -B^\mathrm\{T\}&A\_\{n-2\}\end\{array\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle A\_\{n-2\}$ 是 $\displaystyle n-2$ 阶反对称矩阵. 由 $\displaystyle S$ 可逆及
\begin\{aligned\} &\left(\begin\{array\}\{cccccccccc\}E\_2&\\\\ B^\mathrm\{T\} S^\{-1\}&E\_\{n-2\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccc\}S&B\\\\ -B^\mathrm\{T\}&A\_\{n-2\}\end\{array\}\right)\left(\begin\{array\}\{cccccccccc\}E\_2&-S^\{-1\}B\\\\ &E\_\{n-2\}\end\{array\}\right)\\\\ =&\left(\begin\{array\}\{cccccccccc\}S&0\\\\ 0&A\_\{n-2\}+B^\mathrm\{T\} S^\{-1\} B\end\{array\}\right)\equiv N. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
易知 $\displaystyle A\_\{n-2\}+B^\mathrm\{T\} S^\{-1\} B$ 反对称, 而由归纳假设, 它合同于 $\displaystyle \mathrm\{diag\}(S,\cdots,S,0)$. 因此, 分块矩阵 $\displaystyle N$ 合同于
\begin\{aligned\} \mathrm\{diag\}(\underbrace\{S,\cdots,S\}\_r,0\_\{n-2r\}),\quad S=\left(\begin\{array\}\{cccccccccc\}0&1\\\\ -1&0\end\{array\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
结论得证. 跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/