62. 古伴的性质 (1)

本节, 我们认识古伴的三个性质.

一个 $n$ 级阵 $A$ 的古伴 $\mathrm{adj}(A)$ 是一个 $n$ 级阵, 其 $(i,j)$-元$$\begin{array}{r}[\mathrm{adj}(A){]}_{i,j}=(-1{)}^{j+i}\det (A(j|i)).\end{array}$$对任何方阵 $A$, 有$$\begin{array}{r}\mathrm{adj}(A)A=\det (A)I=A\mathrm{adj}(A).\end{array}$$

定理 62.1. 设 $A$ 是 $n$ 级阵. 则 $\mathrm{adj}(A^{\mathrm{T}})=(\mathrm{adj}(A){)}^{\mathrm{T}}$.

证. 注意, $A^{\mathrm{T}}(j|i)=(A(i|j){)}^{\mathrm{T}}$, 故$$\begin{array}{rl}[\mathrm{adj}(A^{\mathrm{T}}){]}_{i,j}=& (-1{)}^{j+i}\det (A^{\mathrm{T}}(j|i))\\ =& (-1{)}^{j+i}\det ((A(i|j){)}^{\mathrm{T}})\\ =& (-1{)}^{i+j}\det (A(i|j))\\ =& [\mathrm{adj}(A){]}_{j,i}\\ =& [(\mathrm{adj}(A){)}^{\mathrm{T}}{]}_{i,j}.\end{array}$$

证毕.

不难看出, 若 $B$ 是 $n$ 级阵, 且 $c$ 是数, 则 $\det (cB)=c^n\det (B)$.

定理 62.2. 设 $A$ 是 $n$ 级阵. 设 $c$ 是数. 则 $\mathrm{adj}(cA)=c^{n-1}\mathrm{adj}(A)$.

证. 注意, $(cA)(j|i)=c(A(j|i))$, 且 $A(j|i)$ 是 $n-1$ 级阵, 故$$\begin{array}{rl}[\mathrm{adj}(cA){]}_{i,j}=& (-1{)}^{j+i}\det ((cA)(j|i))\\ =& (-1{)}^{j+i}\det (c(A(j|i)))\\ =& (-1{)}^{j+i}{c}^{n-1}\det (A(j|i))\\ =& c^{n-1}(-1{)}^{j+i}\det (A(j|i))\\ =& c^{n-1}[\mathrm{adj}(A){]}_{i,j}\\ =& [c^{n-1}\mathrm{adj}(A){]}_{i,j}.\end{array}$$

证毕.

定理 62.3. 设 $A$ 是 $n$ 级阵 ($n⩾2$). 则 $\det (\mathrm{adj}(A))=(\det (A){)}^{n-1}$.

证. 注意, $$\begin{array}{rl}\det (A)\det (\mathrm{adj}(A))=& \det (A\mathrm{adj}(A))\\ =& \det (\det (A)I)\\ =& (\det (A){)}^n\det (I)\\ =& (\det (A){)}^n\\ =& \det (A)(\det (A){)}^{n-1}.\end{array}$$若 $\det (A)\ne 0$, 我们可在等式的二侧消去它, 得到结论.

若 $A=0$, 则 $\mathrm{adj}(A)=0$, 故 $\det (\mathrm{adj}(A))=0$.

若 $A\ne 0$, 且 $\det (A)=0$, 我们用反证法说明 $\det (\mathrm{adj}(A))=0$. 反设 $\det (\mathrm{adj}(A))\ne 0$. 则$$\begin{array}{r}\mathrm{adj}(A)A=\det (A)I=0=\mathrm{adj}(A)0.\end{array}$$因为 $\det (\mathrm{adj}(A))\ne 0$, 由消去律, 有 $A=0$. 这是矛盾.

证毕.