65. 古伴的性质 (3)

本节, 我们再认识古伴的二个性质.

定理 65.1. 设 $A$ 是 $n$ 级阵 ($n⩾2$). 则 $\mathrm{adj}(\mathrm{adj}(A))=(\det (A){)}^{n-2}A$.

证. 设 $n=2$. 不难算出, $2$ 级阵 $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ 的古伴是 $\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$. 则 $\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ 的古伴是 $\left[\begin{array}{cc}a& -(-b)\\ -(-c)& d\end{array}\right]$, 即 $A$. 注意到, $(\det (A){)}^{2-2}A=1A=A$.

下设 $n>2$. 一方面, $$\begin{array}{r}\mathrm{adj}(A)\mathrm{adj}(\mathrm{adj}(A))=\det (\mathrm{adj}(A))I_n=(\det (A){)}^{n-1}{I}_n;\end{array}$$另一方面, $$\begin{array}{rl}\mathrm{adj}(A)((\det (A){)}^{n-2}A)=& (\det (A){)}^{n-2}(\mathrm{adj}(A)A)\\ =& (\det (A){)}^{n-2}(\det (A)I_n)\\ =& (\det (A){)}^{n-1}{I}_n.\end{array}$$故$$\begin{array}{r}\mathrm{adj}(A)\mathrm{adj}(\mathrm{adj}(A))=\mathrm{adj}(A)((\det (A){)}^{n-2}A).\end{array}$$若 $\det (A)\ne 0$, 则 $\det (\mathrm{adj}(A))=(\det (A){)}^{n-1}\ne 0$. 由消去律, 我们可在等式的二侧消去 $\mathrm{adj}(A)$, 即得结论.

若 $\det (A)=0$, 由上节的结论, $\mathrm{adj}(A)$ 的每一个 $n-1$ 级子阵的行列式都是 $0$. 故 $\mathrm{adj}(\mathrm{adj}(A))=0=0A=(\det (A){)}^{n-2}A$.

证毕.

定理 65.2. 设 $A$, $B$ 是 $n$ 级阵. 则$$\begin{array}{r}\mathrm{adj}(BA)=\mathrm{adj}(A)\mathrm{adj}(B).\end{array}$$

证. 注意到$$\begin{array}{rl}(BA)\mathrm{adj}(BA)=& \det (BA)I\\ =& (\det (B)\det (A))I\\ =& (\det (A)\det (B))I\\ =& \det (A)(\det (B)I)\\ =& \det (A)(B\mathrm{adj}(B))\\ =& B(\det (A)\mathrm{adj}(B))\\ =& B(\det (A)(I\mathrm{adj}(B)))\\ =& B((\det (A)I)\mathrm{adj}(B))\\ =& B((A\mathrm{adj}(A))\mathrm{adj}(B))\\ =& B(A(\mathrm{adj}(A)\mathrm{adj}(B)))\\ =& (BA)(\mathrm{adj}(A)\mathrm{adj}(B)),\end{array}$$若 $\det (BA)\ne 0$, 由消去律, 即得结论.

若 $\det (BA)=0$, 我们这么作. 由阵的积的定义, $$\begin{array}{rl}[\mathrm{adj}(A)\mathrm{adj}(B){]}_{i,j}=& \sum _{u=1}^n[\mathrm{adj}(A){]}_{i,u}[\mathrm{adj}(B){]}_{u,j}\\ =& \sum _{u=1}^n(-1{)}^{u+i}\det (A(u|i))(-1{)}^{j+u}\det (B(j|u))\\ =& \sum _{u=1}^n(-1{)}^{u+i}(-1{)}^{j+u}\det (A(u|i))\det (B(j|u))\\ =& \sum _{u=1}^n(-1{)}^{j+i}\det (B(j|u))\det (A(u|i))\\ =& (-1{)}^{j+i}\sum _{u=1}^n\det (B(j|u))\det (A(u|i))\\ =& (-1{)}^{j+i}\det ((BA)(j|i))\\ =& [\mathrm{adj}(BA){]}_{i,j}.\end{array}$$这里, 我们用了 Binet–Cauchy 公式的推广.

证毕.