C.44. 杂例

例 C.44.1. 设 $A$ 是 $n$ 级阵. 设 $n$ 级阵 $D$ 适合$$\begin{array}{r}[D{]}_{i,j}=\{\begin{array}{ll}{d}_i,& i=j;\\ 0,& \text{其他}.\end{array}\end{array}$$则$$\begin{array}{rl}& \det (A+D)\\ =& \det (D)\\ & +\sum _{k=1}^{n-1}\sum _{1⩽j_1<j_2<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det (D(j_1,\dots ,j_k|j_1,\dots ,j_k))\\ & +\det (A).\end{array}$$

设 $A$ 的列 $1$, $2$, $\dots$, $n$ 分别是 $b_1^{(1)}$, $b_2^{(1)}$, $\dots$, $b_n^{(1)}$. 设 $D$ 的列 $1$, $2$, $\dots$, $n$ 分别是 $b_1^{(0)}$, $b_2^{(0)}$, $\dots$, $b_n^{(0)}$. 则 $A+D$ 的列 $j$ 是$$\begin{array}{r}{b}_j^{(1)}+b_j^{(0)}=b_j^{(0)}+b_j^{(1)}=\sum _{c_j=0}^1{b}_j^{(c_j)}.\end{array}$$则, 由多线性, $$\begin{array}{rl}& \det (A+D)\\ =& \det \left[\sum _{c_1=0}^1{b}_1^{(c_1)},\sum _{c_2=0}^1{b}_2^{(c_2)},\dots ,\sum _{c_n=0}^1{b}_n^{(c_n)}\right]\\ =& \sum _{c_1=0}^1\det \left[b_1^{(c_1)},\sum _{c_2=0}^1{b}_2^{(c_2)},\dots ,\sum _{c_n=0}^1{b}_n^{(c_n)}\right]\\ =& \sum _{c_1=0}^1\sum _{c_2=0}^1\det \left[b_1^{(c_1)},b_2^{(c_2)},\dots ,\sum _{c_n=0}^1{b}_n^{(c_n)}\right]\\ =& \dots \\ =& \sum _{c_1=0}^1\sum _{c_2=0}^1\cdots \sum _{c_n=0}^1\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \sum _{0⩽c_1,c_2,\dots ,c_n⩽1}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}].\end{array}$$由加法的结合律与交换律, 我们可按任何方式, 任何次序求这些$$\begin{array}{r}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\end{array}$$的和. 特别地, 我们可按 $c_1+c_2+\cdots +c_n$ 分这些数为若干组, 求出一组的组和 (即一组的元的和), 再求这些组的组和的和. 因为 $0⩽c_j⩽1$, 故 $0⩽c_1+c_2+\cdots +c_n⩽n$. 于是, 我们可分这些数为 $n+1$ 组: 适合 $c_1+c_2+\cdots +c_n=0$ 的项在一组; 适合 $c_1+c_2+\cdots +c_n=1$ 的项在一组; …… 适合 $c_1+c_2+\cdots +c_n=n$ 的项在一组. 不难看出, 每一项在某一组里, 且每一项不能在二个不同的组里. 则$$\begin{array}{rl}& \det (A+D)\\ =& \sum _{0⩽c_1,c_2,\dots ,c_n⩽1}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \sum _{k=0}^n\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=k\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=0\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ & +\sum _{k=1}^{n-1}\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=k\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ & +\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=n\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}].\end{array}$$

不难看出, $c_1+c_2+\cdots +c_n=0$ 时, $c_1=c_2=\cdots =c_n=0$. 则$$\begin{array}{r}\sum _{0⩽c_1,c_2,\dots ,c_n⩽1}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]=\det [b_1^{(0)},b_2^{(0)},\dots ,b_n^{(0)}]=\det (D).\end{array}$$

不难看出, $c_1+c_2+\cdots +c_n=n$ 时, $c_1=c_2=\cdots =c_n=1$. 则$$\begin{array}{r}\sum _{0⩽c_1,c_2,\dots ,c_n⩽1}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]=\det [b_1^{(1)},b_2^{(1)},\dots ,b_n^{(1)}]=\det (A).\end{array}$$

设正整数 $k<n$. 由 $c_1+c_2+\cdots +c_n=k$, 知在 $c_1$, $c_2$, $\dots$, $c_n$ 中, 有 $k$ 个 $1$ 与 $n-k$ 个 $0$. 设 $c_{j_1}=c_{j_2}=\cdots =c_{j_k}=1$ (其中, $1⩽j_1<j_2<\cdots <j_k⩽n$), 且 $c_{j_{k+1}}=\cdots =c_{j_n}=0$ (其中, $1⩽j_{k+1}<\cdots <j_n⩽n$). 记 $n$ 级阵$$\begin{array}{r}{B}_{c_1,\dots ,c_n}=[b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}].\end{array}$$注意, 当 $\ell >k$, 且 $i\ne j_{\ell }$ 时, $[B_{c_1,\dots ,c_n}{]}_{i,j_{\ell }}=0$, 且 $[B_{c_1,\dots ,c_n}{]}_{j_{\ell },j_{\ell }}=d_{j_{\ell }}$, 按列 $j_{k+1}$ 展开 $\det (B_{c_1,\dots ,c_n})$, 有$$\begin{array}{rl}\det (B_{c_1,\dots ,c_n})=& (-1{)}^{j_{k+1}+j_{k+1}}[B_{c_1,\dots ,c_n}{]}_{j_{k+1},j_{k+1}}\det (B_{c_1,\dots ,c_n}(j_{k+1}|j_{k+1}))\\ =& d_{j_{k+1}}\det (B_{c_1,\dots ,c_n}(j_{k+1}|j_{k+1})).\end{array}$$按列 $j_{k+2}-1$ 展开 $\det (B_{c_1,\dots ,c_n}(j_{k+1}|j_{k+1}))$, 有$$\begin{array}{rl}& \det (B_{c_1,\dots ,c_n}(j_{k+1}|j_{k+1}))\\ =& (-1{)}^{j_{k+2}-1+j_{k+2}-1}[B_{c_1,\dots ,c_n}{]}_{j_{k+2},j_{k+2}}\det (B_{c_1,\dots ,c_n}(j_{k+1},j_{k+2}|j_{k+1},j_{k+2}))\\ =& d_{j_{k+2}}\det (B_{c_1,\dots ,c_n}(j_{k+1},j_{k+2}|j_{k+1},j_{k+2})).\end{array}$$故$$\begin{array}{r}\det (B_{c_1,\dots ,c_n})=d_{j_{k+1}}{d}_{j_{k+2}}\det (B_{c_1,\dots ,c_n}(j_{k+1},j_{k+2}|j_{k+1},j_{k+2})).\end{array}$$…… 最后, 我们算出$$\begin{array}{rl}& \det (B_{c_1,\dots ,c_n})\\ =& d_{j_{k+1}}{d}_{j_{k+2}}\dots d_{j_n}\det (B_{c_1,\dots ,c_n}(j_{k+1},\dots ,j_n|j_{k+1},\dots ,j_n))\\ =& \det (B_{c_1,\dots ,c_n}(j_{k+1},\dots ,j_n|j_{k+1},\dots ,j_n))d_{j_{k+1}}{d}_{j_{k+2}}\dots d_{j_n}\\ =& \det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det (D(j_1,\dots ,j_k|j_1,\dots ,j_k)).\end{array}$$于是$$\begin{array}{rl}& \sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=k\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \sum _{1⩽j_1<j_2<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det (D(j_1,\dots ,j_k|j_1,\dots ,j_k)).\end{array}$$

综上, $$\begin{array}{rl}& \det (A+D)\\ =& \sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=0\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ & +\sum _{k=1}^{n-1}\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=k\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ & +\sum _{\begin{array}{c}0⩽c_1,c_2,\dots ,c_n⩽1\\ c_1+c_2+\cdots +c_n=n\end{array}}\det [b_1^{(c_1)},b_2^{(c_2)},\dots ,b_n^{(c_n)}]\\ =& \det (D)\\ & +\sum _{k=1}^{n-1}\sum _{1⩽j_1<j_2<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det (D(j_1,\dots ,j_k|j_1,\dots ,j_k))\\ & +\det (A).\end{array}$$

例 C.44.2. 设 $A$ 是 $n$ 级阵. 设 $x$ 是数. 则$$\begin{array}{r}\det (x{I}_n+A)=x^n+\sum _{k=1}^n{x}^{n-k}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right).\end{array}$$

注意, $$\begin{array}{r}[x{I}_n{]}_{i,j}=\{\begin{array}{ll}x,& i=j;\\ 0,& \text{其他}.\end{array}\end{array}$$故, 由上个例, $$\begin{array}{rl}& \det (x{I}_n+A)\\ =& \det (x{I}_n)\\ & +\sum _{k=1}^{n-1}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\det ((x{I}_n)(j_1,\dots ,j_k|j_1,\dots ,j_k))\\ & +\det (A)\\ =& x^n+\sum _{k=1}^{n-1}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)x^{n-k}\\ & +\det (A)\\ =& x^n+\sum _{k=1}^{n-1}{x}^{n-k}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ & +x^{n-n}\sum _{1⩽j_1<\cdots <j_n⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_n}{j_1,\dots ,j_n}\right)\right)\\ =& x^n+\sum _{k=1}^n{x}^{n-k}\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right).\end{array}$$

例 C.44.3. 设正整数 $m$, $n$ 适合 $m⩾n$. 设 $A$, $B$ 分别是 $m\times n$ 与 $n\times m$ 阵. 设正整数 $k⩽n$. 由 Binet–Cauchy 公式的推广, $$\begin{array}{rl}& \sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& \sum _{1⩽j_1<\cdots <j_k⩽m}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{i_1,\dots ,i_k}\right)\right)\det \left(B\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{j_1,\dots ,j_k}\right)\right)\\ =& \sum _{1⩽j_1<\cdots <j_k⩽m}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left(B\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{j_1,\dots ,j_k}\right)\right)\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{i_1,\dots ,i_k}\right)\right)\\ =& \sum _{1⩽i_1<\cdots <i_k⩽n}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left(B\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{j_1,\dots ,j_k}\right)\right)\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{i_1,\dots ,i_k}\right)\right)\\ =& \sum _{1⩽i_1<\cdots <i_k⩽n}\det \left((BA)\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{i_1,\dots ,i_k}\right)\right).\end{array}$$

例 C.44.4. 设正整数 $m$, $n$ 适合 $m⩾n$. 设 $A$, $B$ 分别是 $m\times n$ 与 $n\times m$ 阵. 设 $x$ 是数. 则$$\begin{array}{rl}& \det (x{I}_m+AB)\\ =& x^m+\sum _{k=1}^m{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& x^m+\sum _{k=1}^n{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ & +\sum _{k=n+1}^m{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& x^m+\sum _{k=1}^n{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ & +\sum _{k=n+1}^m{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}0\\ =& x^m+\sum _{k=1}^n{x}^{m-k}\sum _{1⩽j_1<\cdots <j_k⩽m}\det \left((AB)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& x^{m-n}{x}^n+\sum _{k=1}^n{x}^{m-n}{x}^{n-k}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left((BA)\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{i_1,\dots ,i_k}\right)\right)\\ =& x^{m-n}{x}^n+x^{m-n}\sum _{k=1}^n{x}^{n-k}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left((BA)\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{i_1,\dots ,i_k}\right)\right)\\ =& x^{m-n}\left(x^n+\sum _{k=1}^n{x}^{n-k}\sum _{1⩽i_1<\cdots <i_k⩽n}\det \left((BA)\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_k}{i_1,\dots ,i_k}\right)\right)\right)\\ =& x^{m-n}\det (x{I}_n+BA).\end{array}$$特别地, 代 $x$ 以数 $1$, 有$$\begin{array}{r}\det (I_m+AB)=\det (I_n+BA).\end{array}$$

例 C.44.5. 设 $a_1$, $b_1$, $a_2$, $b_2$, $\dots$, $a_m$, $b_m$ 是 $2m$ 个数. 作 $m$ 级阵 $C$ 如下: $$\begin{array}{r}[C{]}_{i,j}=\{\begin{array}{ll}1+a_i{b}_i,& i=j;\\ a_i{b}_j,& \text{其他}.\end{array}\end{array}$$我们计算 $\det (C)$.

设 $A=[a_1,a_2,\dots ,a_m{]}^{\mathrm{T}}$, 且 $B=[b_1,b_2,\dots ,b_m]$. 则$$\begin{array}{r}[AB{]}_{i,j}=[A{]}_{i,1}[B{]}_{1,j}=a_i{b}_j.\end{array}$$由此, 不难看出, $C=I_m+AB$. 故$$\begin{array}{rl}\det (C)=& \det (I_m+AB)\\ =& \det (I_1+BA)\\ =& [I_1+BA{]}_{1,1}\\ =& [I_1{]}_{1,1}+[BA{]}_{1,1}\\ =& 1+b_1{a}_1+b_2{a}_2+\cdots +b_m{a}_m.\end{array}$$

我们当然也可用别的方法. 作 $m+1$ 级阵 $G$ 如下: $$\begin{array}{r}[G{]}_{i,j}=\{\begin{array}{ll}1,& i=j=m+1;\\ -a_i,& i<j=m+1;\\ 0,& m+1=i>j;\\ [C{]}_{i,j},& \text{其他}.\end{array}\end{array}$$通俗地, $$\begin{array}{r}G=\left[\begin{array}{cccccc}1+a_1{b}_1& a_1{b}_2& \cdots & a_1{b}_{m-1}& a_1{b}_m& -a_1\\ a_2{b}_1& 1+a_2{b}_2& \cdots & a_2{b}_{m-1}& a_2{b}_m& -a_2\\ ⋮& ⋮& & ⋮& ⋮& ⋮\\ a_{m-1}{b}_1& a_{m-1}{b}_2& \cdots & 1+a_{m-1}{b}_{m-1}& a_{m-1}{b}_m& -a_{m-1}\\ a_m{b}_1& a_m{b}_2& \cdots & a_m{b}_{m-1}& 1+a_m{b}_m& -a_m\\ 0& 0& \cdots & 0& 0& 1\end{array}\right].\end{array}$$一方面, 按行 $m+1$ 展开, 有$$\begin{array}{r}\det (G)=(-1{)}^{m+1+m+1}1\det (G(1|1))=\det (C).\end{array}$$另一方面, 我们加列 $m+1$ 的 $b_1$ 倍于列 $1$, 加列 $m+1$ 的 $b_2$ 倍于列 $2$, …… 加列 $m+1$ 的 $b_m$ 倍于列 $m$, 得 $m+1$ 级阵$$\begin{array}{r}H=\left[\begin{array}{cccccc}1& 0& \cdots & 0& 0& -a_1\\ 0& 1& \cdots & 0& 0& -a_2\\ ⋮& ⋮& & ⋮& ⋮& ⋮\\ 0& 0& \cdots & 1& 0& -a_{m-1}\\ 0& 0& \cdots & 0& 1& -a_m\\ b_1& b_2& \cdots & b_{m-1}& b_m& 1\end{array}\right].\end{array}$$则 $\det (H)=\det (G)$. 故 $\det (C)=\det (H)$.

我们加列 $1$ 的 $a_1$ 倍于列 $m+1$, 加列 $1$ 的 $a_2$ 倍于列 $m+1$, …… 加列 $1$ 的 $a_m$ 倍于列 $m+1$, 得 $m+1$ 级阵$$\begin{array}{r}J=\left[\begin{array}{cccccc}1& 0& \cdots & 0& 0& 0\\ 0& 1& \cdots & 0& 0& 0\\ ⋮& ⋮& & ⋮& ⋮& ⋮\\ 0& 0& \cdots & 1& 0& 0\\ 0& 0& \cdots & 0& 1& 0\\ b_1& b_2& \cdots & b_{m-1}& b_m& 1+b_1{a}_1+b_2{a}_2+\cdots +b_m{a}_m\end{array}\right].\end{array}$$则 $\det (J)=\det (H)$. 故$$\begin{array}{r}\det (C)=\det (J)=1+b_1{a}_1+b_2{a}_2+\cdots +b_m{a}_m.\end{array}$$

例 C.44.6. 设 $A$ 是 $n$ 级阵. 则 $AA$ 是有意义的, 且也是方阵. 我们写 $A^2=AA$. 类似地, $A^{2}A$ 也是有意义的, 且也是方阵. 我们写 $A^3=A^{2}A$. 一般地, 对任何非负整数 $m$, 我们如下定义方阵 $A$ 的 $m$ 次方: $$\begin{array}{r}{A}^m=\{\begin{array}{ll}I,& m=0;\\ A^{m-1}A,& m⩾1.\end{array}\end{array}$$不难看出, $A^1=A^{0}A=IA=A$.

类似数的非负整数次方, 方阵的非负整数次方适合, 对任何方阵 $A$, 任何非负整数 $\ell$, $m$,

(1) $A^{\ell +m}=A^{\ell }{A}^m$;

(2) $(A^{\ell }{)}^m=A^{\ell m}$.

我们用数学归纳法证明 (1) 与 (2).

(1) 作命题 $P(m)$: $A^{\ell +m}=A^{\ell }{A}^m$. 我们的目标是, 对任何非负整数 $m$, $P(m)$ 是对的.

$P(0)$ 是对的, 因为 $A^{\ell +0}=A^{\ell }=A^{\ell }I=A^{\ell }{A}^0$.

假定 $P(m)$ 是对的. 我们由此证明, $P(m+1)$ 也是对的. 注意, $$\begin{array}{r}{A}^{\ell +(m+1)}=A^{(\ell +m)+1}=A^{\ell +m}A=(A^{\ell }{A}^m)A=A^{\ell }(A^{m}A)=A^{\ell }{A}^{m+1}.\end{array}$$所以, $P(m+1)$ 是对的. 由数学归纳法, 待证命题成立.

(2) 作命题 $Q(m)$: $(A^{\ell }{)}^m=A^{\ell m}$. 我们的目标是, 对任何非负整数 $m$, $Q(m)$ 是对的.

$Q(0)$ 是对的, 因为 $(A^{\ell }{)}^0=I=A^0=A^{\ell 0}$.

假定 $Q(m)$ 是对的. 我们由此证明, $Q(m+1)$ 也是对的. 注意, $$\begin{array}{r}(A^{\ell }{)}^{m+1}=(A^{\ell }{)}^m{A}^{\ell }=A^{\ell m}{A}^{\ell }=A^{\ell m+\ell }=A^{\ell (m+1)}.\end{array}$$所以, $Q(m+1)$ 是对的. 由数学归纳法, 待证命题成立.

例 C.44.7. 我们知道, 对任何数 $x$, $y$, 任何非负整数 $m$, 必 $(xy{)}^m=x^m{y}^m$. 可是, 存在二个同级的方阵 $A$, $B$, 存在非负整数 $m$, 使 $(AB{)}^m\ne A^m{B}^m$: 取 $m=2$, $$\begin{array}{r}A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right],B=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right].\end{array}$$不难算出$$\begin{array}{r}AB=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right],A^2=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right],B^2=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right].\end{array}$$则$$\begin{array}{r}(AB{)}^2=\left[\begin{array}{cc}5& 3\\ 3& 2\end{array}\right],A^2{B}^2=\left[\begin{array}{cc}5& 2\\ 2& 1\end{array}\right].\end{array}$$

不过, 若 $AB=BA$, 则我们仍有, 对任何非负整数 $m$, 必 $(AB{)}^m=A^m{B}^m$. 为证明此事, 我们要作准备.

作命题 $P(m)$: $B^{m}A=A{B}^m$. (注意, $A{B}^m$ 是 $A(B^m)$, 而不是 $(AB{)}^m$; 这就像 $x{y}^m$ 是 $x(y^m)$, 而不是 $(xy{)}^m$.) 我们用数学归纳法证明 $P(m)$.

$P(0)$ 是对的, 因为 $B^{0}A=IA=AI=A{B}^0$.

假定 $P(m)$ 是对的. 我们由此证明, $P(m+1)$ 也是对的. 注意, $$\begin{array}{rl}{B}^{m+1}A=& (B^{m}B)A=B^m(BA)=B^m(AB)\\ =& (B^{m}A)B=(A{B}^m)B=A(B^{m}B)=A{B}^{m+1}.\end{array}$$所以, $P(m+1)$ 是对的. 由数学归纳法, 待证命题成立.

好的. 我们作好了准备.

作命题 $Q(m)$: $(AB{)}^m=A^m{B}^m$. 我们的目标是, 对任何非负整数 $m$, $Q(m)$ 是对的.

$Q(0)$ 是对的, 因为 $(AB{)}^0=I=II=A^0{B}^0$.

假定 $Q(m)$ 是对的. 我们由此证明, $Q(m+1)$ 也是对的. 注意, $$\begin{array}{rl}(AB{)}^{m+1}=& (AB{)}^m(AB)=(A^m{B}^m)(AB)=((A^m{B}^m)A)B=(A^m(B^{m}A))B\\ =& (A^m(A{B}^m))B=((A^{m}A)B^m)B=A^{m+1}(B^{m}B)=A^{m+1}{B}^{m+1}.\end{array}$$所以, $Q(m+1)$ 是对的. 由数学归纳法, 待证命题成立.

例 C.44.8. 设 $A$ 是方阵. 设 $k$ 是数. 设 $m$ 是非负整数. 可以用数学归纳法证明:

(1) $(kA{)}^m=k^m{A}^m$. (用 $(kA)(\ell B)=(k\ell )(AB)$, 若 $A$, $B$ 都是 $n$ 级阵, 且 $k$, $\ell$ 是数.)

(2) $\det (A^m)=(\det (A){)}^m$. (用 $\det (AB)=\det (A)\det (B)$, 若 $A$, $B$ 都是 $n$ 级阵.)

(3) $(A^m{)}^{\mathrm{T}}=(A^{\mathrm{T}}{)}^m$. (用 $(BA{)}^{\mathrm{T}}=A^{\mathrm{T}}{B}^{\mathrm{T}}$, 若 $A$, $B$ 都是 $n$ 级阵; 见 “转置的性质”.)

(4) $\mathrm{adj}(A^m)=(\mathrm{adj}(A){)}^m$. (用 $\mathrm{adj}(BA)=\mathrm{adj}(A)\mathrm{adj}(B)$, 若 $A$, $B$ 都是 $n$ 级阵; 见 “古伴的性质 (3)”.)

请允许我留它们为您的习题.

例 C.44.9. 设 $A$, $B$ 是 $n$ 级阵. 则存在跟 $A$, $B$ 的元有关的数 $d_0$, $d_1$, $\dots$, $d_n$, 使对任何数 $x$, 有$$\begin{array}{r}\det (xA+B)=\sum _{k=0}^n{d}_k{x}^{n-k}=d_0{x}^n+d_1{x}^{n-1}+\cdots +d_n,\end{array}$$其中, $d_0=\det (A)$, 且 $d_n=\det (B)$.

作命题 $P(m)$: 对任何 $m$ 级阵 $A$, $B$, 存在跟 $A$, $B$ 的元有关的数 $d_0$, $d_1$, $\dots$, $d_m$, 使对任何数 $x$, 有$$\begin{array}{r}\det (xA+B)=\sum _{k=0}^m{d}_k{x}^{m-k}=d_0{x}^m+d_1{x}^{m-1}+\cdots +d_m,\end{array}$$其中, $d_0=\det (A)$, 且 $d_m=\det (B)$. 我们的目标是, 对任何非负整数 $m$, $P(m)$ 是对的.

$P(1)$ 是显然的: $$\begin{array}{r}\det (xA+B)=[xA+B{]}_{1,1}=x[A{]}_{1,1}+[B{]}_{1,1}=\det (A)x+\det (B).\end{array}$$

假定 $P(m-1)$ 是对的. 我们由此证明, $P(m)$ 也是对的.

设 $A$, $B$ 是 $m$ 级阵. 则$$\begin{array}{rl}\det (xA+B)=& \sum _{i=1}^m(-1{)}^{i+1}[xA+B{]}_{i,1}\det ((xA+B)(i|1))\\ =& \sum _{i=1}^m(-1{)}^{i+1}(x[A{]}_{i,1}+[B{]}_{i,1})\det (xA(i|1)+B(i|1)).\end{array}$$$A(i|1)$, $B(i|1)$ 是 $m-1$ 级阵. 由假定, 存在跟 $A(i|1)$, $B(i|1)$ 的元有关的数 (从而也是跟 $A$, $B$ 的元有关的数) $c_{i,0}$, $c_{i,1}$, $\dots$, $c_{i,m-1}$, 使对任何数 $x$, 有$$\begin{array}{r}\det (xA(i|1)+B(i|1))=\sum _{\ell =0}^{m-1}{c}_{i,\ell }{x}^{m-1-\ell },\end{array}$$其中, $c_{i,0}=\det (A(i|1))$, 且 $c_{i,m-1}=\det (B(i|1))$. 为方便, 记 $a_i=(-1{)}^{i+1}[A{]}_{i,1}$, 且 $b_i=(-1{)}^{i+1}[B{]}_{i,1}$. 则 $a_i$, $b_i$ 跟 $A$, $B$ 的元有关. 则$$\begin{array}{rl}& \det (xA+B)\\ =& \sum _{i=1}^m(a_{i}x+b_i)\sum _{\ell =0}^{m-1}{c}_{i,\ell }{x}^{m-1-\ell }\\ =& \sum _{i=1}^m\sum _{\ell =0}^{m-1}(a_{i}x+b_i)c_{i,\ell }{x}^{m-1-\ell }\\ =& \sum _{i=1}^m\sum _{\ell =0}^{m-1}(a_i{c}_{i,\ell }{x}^{m-\ell }+b_i{c}_{i,\ell }{x}^{m-1-\ell })\\ =& \sum _{\ell =0}^{m-1}\sum _{i=1}^m(a_i{c}_{i,\ell }{x}^{m-\ell }+b_i{c}_{i,\ell }{x}^{m-1-\ell })\\ =& \sum _{\ell =0}^{m-1}\sum _{i=1}^m{a}_i{c}_{i,\ell }{x}^{m-\ell }+\sum _{\ell =0}^{m-1}\sum _{i=1}^m{b}_i{c}_{i,\ell }{x}^{m-1-\ell }\\ =& \sum _{i=1}^m{a}_i{c}_{i,0}{x}^{m-0}+\sum _{\ell =1}^{m-1}\sum _{i=1}^m{a}_i{c}_{i,\ell }{x}^{m-\ell }\\ & +\sum _{\ell =0}^{m-2}\sum _{i=1}^m{b}_i{c}_{i,\ell }{x}^{m-1-\ell }+\sum _{i=1}^m{b}_i{c}_{i,m-1}{x}^{m-1-(m-1)}\\ =& \sum _{i=1}^m{a}_i{c}_{i,0}{x}^m+\sum _{\ell =1}^{m-1}\sum _{i=1}^m{a}_i{c}_{i,\ell }{x}^{m-\ell }\\ & +\sum _{\ell =0}^{m-2}\sum _{i=1}^m{b}_i{c}_{i,(\ell +1)-1}{x}^{m-(\ell +1)}+\sum _{i=1}^m{b}_i{c}_{i,m-1}\\ =& \sum _{i=1}^m{a}_i{c}_{i,0}{x}^m+\sum _{\ell =1}^{m-1}\sum _{i=1}^m{a}_i{c}_{i,\ell }{x}^{m-\ell }\\ & +\sum _{\ell =1}^{m-1}\sum _{i=1}^m{b}_i{c}_{i,\ell -1}{x}^{m-\ell }+\sum _{i=1}^m{b}_i{c}_{i,m-1}\\ =& \sum _{i=1}^m{a}_i{c}_{i,0}{x}^m+\sum _{\ell =1}^{m-1}\sum _{i=1}^m(a_i{c}_{i,\ell }+b_i{c}_{i,\ell -1})x^{m-\ell }+\sum _{i=1}^m{b}_i{c}_{i,m-1}.\end{array}$$记$$\begin{array}{r}{d}_k=\{\begin{array}{ll}\sum _{i=1}^m{a}_i{c}_{i,0},& k=0;\\ \sum _{i=1}^m(a_i{c}_{i,k}+b_i{c}_{i,k-1}),& 0<k<m;\\ \sum _{i=1}^m{b}_i{c}_{i,m-1},& k=m.\end{array}\end{array}$$则 $d_k$ 跟 $A$, $B$ 的元有关, 且对任何数 $x$, 有$$\begin{array}{r}\det (xA+B)=\sum _{k=0}^m{d}_k{x}^{m-k}=d_0{x}^m+d_1{x}^{m-1}+\cdots +d_m.\end{array}$$最后, 不难算出$$\begin{array}{rl}& d_0=\sum _{i=1}^m{a}_i{c}_{i,0}=\sum _{i=1}^m(-1{)}^{i+1}[A{]}_{i,1}\det (A(i|1))=\det (A),\\ & d_m=\sum _{i=1}^m{b}_i{c}_{i,m-1}=\sum _{i=1}^m(-1{)}^{i+1}[B{]}_{i,1}\det (B(i|1))=\det (B).\end{array}$$所以, $P(m)$ 是对的. 由数学归纳法, 待证命题成立.

例 C.44.10. 设 $A$ 是 $n$ 级阵. 则存在跟 $A$ 的元有关的数 $c_0$, $c_1$, $\dots$, $c_n$, 使对任何数 $x$, 有$$\begin{array}{r}\det (xI-A)=\det (xI+(-A))=\sum _{k=0}^n{c}_k{x}^{n-k}=c_0{x}^n+c_1{x}^{n-1}+\cdots +c_n,\end{array}$$其中, $c_0=\det (I)=1$, 且 $c_n=\det (-A)=(-1{)}^n\det (A)$.

我们考虑 $xI-A$ 的古伴 $\mathrm{adj}(xI-A)$. 它的 $(i,j)$-元$$\begin{array}{rl}[\mathrm{adj}(xI-A){]}_{i,j}=& (-1{)}^{j+i}\det ((xI-A)(j|i))\\ =& (-1{)}^{j+i}\det (xI(j|i)+(-A)(j|i)).\end{array}$$则有跟 $A$ 的元有关的数 $d_{i,j,k}$ ($k=0$, $1$, $\dots$, $n-1$), 使对任何数 $x$, 有$$\begin{array}{r}\det (xI(j|i)+(-A)(j|i))=d_{i,j,0}{x}^{n-1}+d_{i,j,1}{x}^{n-2}+\cdots +d_{i,j,n-1},\end{array}$$其中, $d_{i,j,0}=\det (I(j|i))$, 且 $d_{i,j,n-1}=\det ((-A)(j|i))$. 为方便, 记 $b_{i,j,k}=(-1{)}^{j+i}{d}_{i,j,k}$. 则有跟 $A$ 的元有关的数 $b_{i,j,k}$ ($k=0$, $1$, $\dots$, $n-1$), 使对任何数 $x$, 有$$\begin{array}{r}[\mathrm{adj}(xI-A){]}_{i,j}=b_{i,j,0}{x}^{n-1}+b_{i,j,1}{x}^{n-2}+\cdots +b_{i,j,n-1}.\end{array}$$作 $n$ 级阵 $B_k$, 使 $[B_k{]}_{i,j}=b_{i,j,k}$ ($k=0$, $1$, $\dots$, $n-1$). 则 $B_0$, $B_1$, $\dots$, $B_{n-1}$ 的元全跟 $A$ 的元有关, 且对任何数 $x$, 有$$\begin{array}{r}\mathrm{adj}(xI-A)=x^{n-1}{B}_0+x^{n-2}{B}_1+\cdots +B_{n-1}.\end{array}$$因为 $[B_{n-1}{]}_{i,j}=b_{i,j,n-1}=(-1{)}^{j+i}{d}_{i,j,n-1}=(-1{)}^{j+i}\det ((-A)(j|i))$, 我们有 $B_{n-1}=\mathrm{adj}(-A)$.

因为$$\begin{array}{rl}(\mathrm{adj}(xI-A))(xI-A)=& \det (xI-A)I\\ =& (c_0{x}^n+c_1{x}^{n-1}+\cdots +c_n)I\\ =& x^n(c_{0}I)+x^{n-1}(c_{1}I)+\cdots +(c_{n}I),\end{array}$$我们有$$\begin{array}{rl}& (x^{n-1}{B}_0+x^{n-2}{B}_1+\cdots +x{B}_{n-2}+B_{n-1})(xI-A)\\ =& x^n(c_{0}I)+x^{n-1}(c_{1}I)+\cdots +x(c_{n-1}I)+(c_{n}I).\end{array}$$上式的左侧是$$\begin{array}{rl}& (x^{n-1}{B}_0+x^{n-2}{B}_1+\cdots +x{B}_{n-2}+B_{n-1})(xI)\\ & +(x^{n-1}{B}_0+x^{n-2}{B}_1+\cdots +x{B}_{n-2}+B_{n-1})(-A)\\ =& x^n{B}_0+x^{n-1}{B}_1+\cdots +x^2{B}_{n-2}+x{B}_{n-1}\\ & -x^{n-1}(B_{0}A)-x^{n-2}(B_{1}A)-\cdots -x(B_{n-2}A)-B_{n-1}A\\ =& x^n{B}_0+x^{n-1}(B_1-B_{0}A)+x^{n-2}(B_2-B_{1}A)\\ & +\cdots +x(B_{n-1}-B_{n-2}A)+(-B_{n-1}A).\end{array}$$比较左侧与右侧, 有$$\begin{array}{rl}{B}_0& =c_{0}I,\\ B_1-B_{0}A& =c_{1}I,\\ B_2-B_{1}A& =c_{2}I,\\ & \dots ,\\ B_{n-1}-B_{n-2}A& =c_{n-1}I,\\ -B_{n-1}A& =c_{n}I.\end{array}$$则$$\begin{array}{rl}{B}_0{A}^{n-1}& =(c_{0}I)A^{n-1},\\ (B_1-B_{0}A)A^{n-2}& =(c_{1}I)A^{n-2},\\ (B_2-B_{1}A)A^{n-3}& =(c_{2}I)A^{n-3},\\ & \dots ,\\ (B_{n-1}-B_{n-2}A)A^0& =(c_{n-1}I)A^0.\end{array}$$则$$\begin{array}{rl}{B}_0{A}^{n-1}& =c_0{A}^{n-1},\\ B_1{A}^{n-2}-B_0{A}^{n-1}& =c_1{A}^{n-2},\\ B_2{A}^{n-3}-B_1{A}^{n-2}& =c_2{A}^{n-3},\\ & \dots ,\\ B_{n-1}-B_{n-2}A& =c_{n-1}I.\end{array}$$则$$\begin{array}{r}{B}_{n-1}=c_0{A}^{n-1}+c_1{A}^{n-2}+\cdots +c_{n-1}I.\end{array}$$

由 $B_{n-1}=\mathrm{adj}(-A)$, 知$$\begin{array}{r}\mathrm{adj}(-A)=c_0{A}^{n-1}+c_1{A}^{n-2}+\cdots +c_{n-1}I.\end{array}$$

由 $-B_{n-1}A=c_{n}I$, 知$$\begin{array}{rl}0=& B_{n-1}A+c_{n}I\\ =& (c_0{A}^{n-1}+c_1{A}^{n-2}+\cdots +c_{n-1}I)A+c_{n}I\\ =& c_0{A}^n+c_1{A}^{n-1}+\cdots +c_{n-1}A+c_{n}I.\end{array}$$

最后, 由前面的例, $1⩽k⩽n$ 时, $$\begin{array}{rl}{c}_k=& \sum _{1⩽j_1<\cdots <j_k⩽n}\det \left((-A)\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& \sum _{1⩽j_1<\cdots <j_k⩽n}(-1{)}^k\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right)\\ =& (-1{)}^k\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right);\end{array}$$特别地, $c_n=(-1{)}^n\det (A)$.

综上, 我们有:

定理 C.44.11. 设 $A$ 是 $n$ 级阵.

(1) 存在跟 $A$ 的元有关的数 $c_0$, $c_1$, $\dots$, $c_n$, 使对任何数 $x$, 有$$\begin{array}{r}\det (xI-A)=c_0{x}^n+c_1{x}^{n-1}+\cdots +c_n,\end{array}$$其中, $c_0=1$, 且 $c_n=(-1{)}^n\det (A)$. 更具体地, $1⩽k⩽n$ 时, $$\begin{array}{rl}{c}_k=& (-1{)}^k\sum _{1⩽j_1<\cdots <j_k⩽n}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1,\dots ,j_k}{j_1,\dots ,j_k}\right)\right).\end{array}$$

(2) 用 $A$ 的一些非负整数次方的数乘的和, 我们可如此表示 $-A$ 的古伴: $$\begin{array}{r}\mathrm{adj}(-A)=c_0{A}^{n-1}+c_1{A}^{n-2}+\cdots +c_{n-1}I.\end{array}$$则 $A$ 的古伴$$\begin{array}{r}\mathrm{adj}(A)=(-1{)}^{n-1}\mathrm{adj}(-A)=(-1{)}^{n-1}(c_0{A}^{n-1}+c_1{A}^{n-2}+\cdots +c_{n-1}I).\end{array}$$

(3) (Cayley–Hamilton (kēili–hāmérten) 定理) $$\begin{array}{r}{c}_0{A}^n+c_1{A}^{n-1}+\cdots +c_{n}I=0.\end{array}$$

例 C.44.12. 我们验证 $n=2$ 时的情形. 为方便, 设 $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$. 则 $xI-A=\left[\begin{array}{cc}x-a& -b\\ -c& x-d\end{array}\right]$. 直接计算, 有$$\begin{array}{r}\det (xI-A)=(x-a)(x-d)-(-c)(-b)=1x^2+(-(a+d))x+(ad-cb).\end{array}$$注意, $(-1{)}^2\det (A)=ad-cb$, 且$$\begin{array}{rl}(-1{)}^1\sum _{1⩽j_1⩽2}\det \left(A\left(\genfrac{}{}{0ex}{}{j_1}{j_1}\right)\right)=& (-1)\left(\det \left(A\left(\genfrac{}{}{0ex}{}{1}{1}\right)\right)+\det \left(A\left(\genfrac{}{}{0ex}{}{2}{2}\right)\right)\right)\\ =& -(a+d).\end{array}$$则 (1) 被验证.

因为 $-A=\left[\begin{array}{cc}-a& -b\\ -c& -d\end{array}\right]$, 我们有$$\begin{array}{r}\mathrm{adj}(-A)=\left[\begin{array}{cc}-d& b\\ c& -a\end{array}\right]=\left[\begin{array}{cc}a-(a+d)& b\\ c& d-(a+d)\end{array}\right]=1A+(-(a+d))I.\end{array}$$类似地, 可验证 $\mathrm{adj}(A)=(-1{)}^{2-1}\mathrm{adj}(-A)=-(1A+(-(a+d))I)$; 请允许我留它为您的习题. 则 (2) 被验证.

最后, $$\begin{array}{rl}& 1A^2+(-(a+d))A+(ad-cb)I\\ =& 1\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]+(-(a+d))\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]+(ad-cb)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ =& \left[\begin{array}{cc}{a}^2+bc& b(a+d)\\ c(a+d)& bc+d^2\end{array}\right]+\left[\begin{array}{cc}-a^2-ad& -b(a+d)\\ -c(a+d)& -ad-d^2\end{array}\right]+\left[\begin{array}{cc}ad-bc& 0\\ 0& ad-bc\end{array}\right]\\ =& \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\\ =& 0.\end{array}$$则 (3) 被验证.

例 C.44.13. 设 $A$ 是 $n$ 级阵. 设 $i$ 是不超过 $n$ 的正整数. 按行 $i$ 展开, 有$$\begin{array}{r}\det (A)=[A{]}_{i,i}\det (A(i|i))+\sum _{\begin{array}{c}1⩽\ell ⩽n\\ \ell \ne i\end{array}}(-1{)}^{i+\ell }[A{]}_{i,\ell }\det (A(i|\ell )).\end{array}$$注意, $A$ 的列 $i$ 对应 $A(i|\ell )$ 的列 $i-\rho (i,\ell )$. 则$$\begin{array}{r}\det (A(i|\ell ))=\sum _{\begin{array}{c}1⩽k⩽n\\ k\ne i\end{array}}(-1{)}^{k-\rho (k,i)+i-\rho (i,\ell )}[A{]}_{k,i}\det (A(i,k|\ell ,i)).\end{array}$$所以 (注意, 对整数 $p$, $q$, 有 $(-1{)}^{p+q}=(-1{)}^{p-q}$) $$\begin{array}{rl}& \det (A)\\ =& [A{]}_{i,i}\det (A(i|i))+\sum _{\begin{array}{c}1⩽k,\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell -\rho (k,i)+\rho (i,\ell )}[A{]}_{k,i}[A{]}_{i,\ell }\det (A(i,k|i,\ell )).\end{array}$$

设 $A$ 是反称阵. 则 $[A{]}_{i,i}=0$, 且 $[A{]}_{k,i}=-[A{]}_{i,k}$. 注意, $1-\rho (k,i)=\rho (i,k)$. 则$$\begin{array}{r}\det (A)=\sum _{\begin{array}{c}1⩽k,\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,k|i,\ell )).\end{array}$$则$$\begin{array}{rl}\det (A)=& \sum _{\begin{array}{c}1⩽k,\ell ⩽n\\ k,\ell \ne i\\ k=\ell \end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,k|i,\ell ))\\ & +\sum _{\begin{array}{c}1⩽k,\ell ⩽n\\ k,\ell \ne i\\ k<\ell \end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,k|i,\ell ))\\ & +\sum _{\begin{array}{c}1⩽k,\ell ⩽n\\ k,\ell \ne i\\ k>\ell \end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,k|i,\ell ))\\ =& \sum _{\begin{array}{c}1⩽k⩽n\\ k\ne i\end{array}}[A{]}_{i,k}^2\det (A(i,k|i,k))\\ & +\sum _{\begin{array}{c}1⩽k,\ell ⩽n\\ k,\ell \ne i\\ k<\ell \end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,k|i,\ell ))\\ & +\sum _{\begin{array}{c}1⩽k,\ell ⩽n\\ k,\ell \ne i\\ k<\ell \end{array}}(-1{)}^{\ell +k+\rho (i,\ell )+\rho (i,k)}[A{]}_{i,\ell }[A{]}_{i,k}\det (A(i,\ell |i,k))\\ =& \sum _{\begin{array}{c}1⩽k⩽n\\ k\ne i\end{array}}[A{]}_{i,k}^2\det (A(i,k|i,k))\\ & +\sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,k|i,\ell ))\\ & +\sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,\ell |i,k)).\end{array}$$注意, $(A(i,\ell |i,k){)}^{\mathrm{T}}=(A^{\mathrm{T}})(i,k|i,\ell )=(-A)(i,k|i,\ell )=-(A(i,k|i,\ell ))$. 则$$\begin{array}{rl}\det (A(i,\ell |i,k))=& \det (A(i,\ell |i,k){)}^{\mathrm{T}})\\ =& (-1{)}^{n-2}\det (A(i,k|i,\ell ))\\ =& (-1{)}^n\det (A(i,k|i,\ell )).\end{array}$$ndalign* 则$$\begin{array}{rl}& \det (A)\\ =& \sum _{\begin{array}{c}1⩽k⩽n\\ k\ne i\end{array}}[A{]}_{i,k}^2\det (A(i,k|i,k))\\ & +(1+(-1{)}^n)\sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,k|i,\ell )).\end{array}$$回想, 对反称阵 $A$, 有 $\det (A)=(\mathrm{pf}(A){)}^2$. 回想, $$\begin{array}{r}\mathrm{pf}(A)=\sum _{\begin{array}{c}1⩽j⩽n\\ j\ne i\end{array}}(-1{)}^{i-1+j+\rho (i,j)}[A{]}_{i,j}\mathrm{pf}(A(i,j|i,j)).\end{array}$$回想, $$\begin{array}{r}(x_1+x_2+\cdots +x_n{)}^2=(x_1^2+x_2^2+\cdots +x_n^2)+2\sum _{1⩽k<\ell ⩽n}{x}_k{x}_{\ell }.\end{array}$$则$$\begin{array}{rl}& (\mathrm{pf}(A){)}^2\\ =& \sum _{\begin{array}{c}1⩽k⩽n\\ k\ne i\end{array}}[A{]}_{i,k}^2(\mathrm{pf}(A(i,k|i,k)){)}^2\\ & +2\sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\mathrm{pf}(A(i,k|i,k))\mathrm{pf}(A(i,\ell |i,\ell )).\end{array}$$注意, $\det (A(i,k|i,k))=(\mathrm{pf}(A(i,k|i,k)){)}^2$. 则$$\begin{array}{rl}& (1+(-1{)}^n)\sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,k|i,\ell ))\\ =& 2\sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\mathrm{pf}(A(i,k|i,k))\mathrm{pf}(A(i,\ell |i,\ell )).\end{array}$$设 $n$ 是偶数. 则$$\begin{array}{rl}& \sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\det (A(i,k|i,\ell ))\\ =& \sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[A{]}_{i,k}[A{]}_{i,\ell }\mathrm{pf}(A(i,k|i,k))\mathrm{pf}(A(i,\ell |i,\ell )).\end{array}$$

设 $1⩽u<v⩽n$, $u\ne i$, 且 $v\ne i$. 作 $n$ 级反称阵 $B$, 其中, $$\begin{array}{r}[B{]}_{p,q}=\{\begin{array}{ll}1,& \text{p=i, 且 q=u, 或 p=i, 且 q=v};\\ -1,& \text{p=u, 且 q=i, 或 p=v, 且 q=i};\\ 0,& \text{p=i, 且 q=u,v, 或 q=i, 且 p=u,v};\\ [A{]}_{p,q},& \text{其他}.\end{array}\end{array}$$注意, $A(i|i)=B(i|i)$. 则 $A(i,k|i,k)=B(i,k|i,k)$, $A(i,\ell |i,\ell )=B(i,\ell |i,\ell )$, 且 $A(i,k|i,\ell )=B(i,k|i,\ell )$. 则, 由$$\begin{array}{rl}& \sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[B{]}_{i,k}[B{]}_{i,\ell }\det (B(i,k|i,\ell ))\\ =& \sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[B{]}_{i,k}[B{]}_{i,\ell }\mathrm{pf}(B(i,k|i,k))\mathrm{pf}(B(i,\ell |i,\ell )),\end{array}$$我们有$$\begin{array}{rl}& \sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[B{]}_{i,k}[B{]}_{i,\ell }\det (A(i,k|i,\ell ))\\ =& \sum _{\begin{array}{c}1⩽k<\ell ⩽n\\ k,\ell \ne i\end{array}}(-1{)}^{k+\ell +\rho (i,k)+\rho (i,\ell )}[B{]}_{i,k}[B{]}_{i,\ell }\mathrm{pf}(A(i,k|i,k))\mathrm{pf}(A(i,\ell |i,\ell )).\end{array}$$注意, 若 $k\ne u$, 或 $k\ne v$, 或 $\ell \ne u$, 或 $\ell \ne v$, 则 $[B{]}_{i,k}[B{]}_{i,\ell }=0$. 所以$$\begin{array}{rl}& (-1{)}^{u+v+\rho (i,u)+\rho (i,v)}1\cdot 1\cdot \det (A(i,u|i,v))\\ =& (-1{)}^{u+v+\rho (i,u)+\rho (i,v)}1\cdot 1\cdot \mathrm{pf}(A(i,u|i,u))\mathrm{pf}(A(i,v|i,v)),\end{array}$$即$$\begin{array}{r}\det (A(i,u|i,v))=\mathrm{pf}(A(i,u|i,u))\mathrm{pf}(A(i,v|i,v)).\end{array}$$

回想, $$\begin{array}{r}\det (A(i,u|i,v))=(-1{)}^n\det (A(i,v|i,u))=\det (A(i,v|i,u)).\end{array}$$则 $1⩽v<u⩽n$, $u\ne i$, 且 $v\ne i$ 时, $$\begin{array}{rl}\det (A(i,u|i,v))=& \det (A(i,v|i,u))\\ =& \mathrm{pf}(A(i,v|i,u))\mathrm{pf}(A(i,v|i,v))\\ =& \mathrm{pf}(A(i,u|i,u))\mathrm{pf}(A(i,v|i,v)).\end{array}$$回想, $\det (A(i,u|i,u))=(\mathrm{pf}(A(i,u|i,u)){)}^2$. 于是, 对不等于 $i$ 的, 且不超过 $n$ 的正整数 $u$, $v$, $$\begin{array}{r}\det (A(i,u|i,v))=\mathrm{pf}(A(i,u|i,u))\mathrm{pf}(A(i,v|i,v)).\end{array}$$所以, 若 $j\ne i$, $$\begin{array}{rl}& \det (A(i|j))\\ =& \sum _{\begin{array}{c}1⩽k⩽n\\ k\ne i\end{array}}(-1{)}^{k-\rho (k,i)+i-\rho (i,j)}[A{]}_{k,i}\det (A(i,k|i,j))\\ =& \sum _{\begin{array}{c}1⩽k⩽n\\ k\ne i\end{array}}(-1{)}^{k-\rho (k,i)+i-\rho (i,j)}[A{]}_{k,i}\mathrm{pf}(A(i,k|i,k))\mathrm{pf}(A(i,j|i,j))\\ =& \sum _{\begin{array}{c}1⩽k⩽n\\ k\ne i\end{array}}(-1{)}^{i+k+1-\rho (i,k)}(-1{)}^{1-\rho (i,j)}[A{]}_{i,k}\mathrm{pf}(A(i,k|i,k))\mathrm{pf}(A(i,j|i,j))\\ =& (-1{)}^{\rho (j,i)}\mathrm{pf}(A(i,j|i,j))\sum _{\begin{array}{c}1⩽k⩽n\\ k\ne i\end{array}}(-1{)}^{i-1+k+\rho (i,k)}[A{]}_{i,k}\mathrm{pf}(A(i,k|i,k))\\ =& (-1{)}^{\rho (j,i)}\mathrm{pf}(A(i,j|i,j))\mathrm{pf}(A).\end{array}$$

综上, 我们有:

定理 C.44.14. 设 $A$ 是 $2m$ 级反称阵. 设 $i$ 是不超过 $2m$ 的正整数. 设不超过 $2m$ 的正整数 $k$, $\ell$ 不等于 $i$. 则$$\begin{array}{r}\det (A(i,k|i,\ell ))=\mathrm{pf}(A(i,k|i,k))\mathrm{pf}(A(i,\ell |i,\ell )),\end{array}$$且$$\begin{array}{r}\det (A(i|\ell ))=(-1{)}^{\rho (\ell ,i)}\mathrm{pf}(A(i,\ell |i,\ell ))\mathrm{pf}(A).\end{array}$$则$$\begin{array}{r}[\mathrm{adj}(A){]}_{i,j}=\{\begin{array}{ll}(-1{)}^{\rho (i,j)}(-1{)}^{i+j}\mathrm{pf}(A(i,j|i,j))\mathrm{pf}(A),& i\ne j;\\ 0,& i=j.\end{array}\end{array}$$

例 C.44.15. 设 $A$ 是 $2m-1$ 级反称阵. 作 $2m$ 级反称阵 $B$, 其中, $$\begin{array}{r}[B{]}_{i,j}=\{\begin{array}{ll}0,& \text{i=2m 或 j=2m};\\ [A{]}_{i,j},& \text{其他}.\end{array}\end{array}$$则 $B(2m|2m)=A$. 设 $i$, $j$ 是不超过 $2m-1$ 的正整数. 则$$\begin{array}{rl}\det (A(j|i))=& \det (B(2m,j|2m,i))\\ =& \mathrm{pf}(B(2m,j|2m,j))\mathrm{pf}(B(2m,i|2m,i))\\ =& \mathrm{pf}(A(j|j))\mathrm{pf}(A(i|i)).\end{array}$$则$$\begin{array}{r}[\mathrm{adj}(A){]}_{i,j}=(-1{)}^i\mathrm{pf}(A(i|i))\cdot (-1{)}^j\mathrm{pf}(A(j|j)).\end{array}$$于是, 若 $(2m-1)\times 1$ 阵 $X$ 适合 $[X{]}_{i,1}=(-1{)}^i\mathrm{pf}(A(i|i))$, 则 $\mathrm{adj}(A)=X{X}^{\mathrm{T}}$.

例 C.44.16. 设 $A$ 是 $2m$ 级反称阵. 它的古伴 $\mathrm{adj}(A)$ 也是反称阵, 故 $\mathrm{adj}(A)$ 的 pfaffian 是有意义的. 则$$\begin{array}{r}\mathrm{pf}(A^{\mathrm{T}}\mathrm{adj}(A)A)=\mathrm{pf}(\mathrm{adj}(A))\det (A).\end{array}$$注意, $$\begin{array}{r}{A}^{\mathrm{T}}\mathrm{adj}(A)A=(-A)(\det (A)I)=(-\det (A))A,\end{array}$$故 $\mathrm{pf}(A^{\mathrm{T}}\mathrm{adj}(A)A)=(-\det (A){)}^m\mathrm{pf}(A)$. 于是, 若 $\det (A)\ne 0$, $$\begin{array}{r}\mathrm{pf}(\mathrm{adj}(A))=(-1{)}^m(\det (A){)}^{m-1}\mathrm{pf}(A)=(-1{)}^m(\mathrm{pf}(A){)}^{2m-1}.\end{array}$$若 $\det (A)=0$, 则 $\mathrm{pf}(A)=0$, 且由前面的结果, $\mathrm{adj}(A)=0$. 则 $\mathrm{pf}(\mathrm{adj}(A))=0$. 故上式仍是对的.

例 C.44.17. 设一元运算 $\mathrm{D}$ 适合, 对任何 $a$, $b$, 必$$\begin{array}{rl}\mathrm{D}(a+b)& =\mathrm{D}(a)+\mathrm{D}(b),\\ \mathrm{D}(a\cdot b)& =\mathrm{D}(a)\cdot b+a\cdot \mathrm{D}(b).\end{array}$$

由此, $\mathrm{D}(1)=\mathrm{D}(1\cdot 1)=\mathrm{D}(1)\cdot 1+1\cdot \mathrm{D}(1)=\mathrm{D}(1)+\mathrm{D}(1)$. 则 $\mathrm{D}(1)=0$.

注意, $\mathrm{D}(0)=\mathrm{D}(0+0)=\mathrm{D}(0)+\mathrm{D}(0)$. 则 $\mathrm{D}(0)=0$. 则 $\mathrm{D}(0\cdot x)=\mathrm{D}(0)=0=0\cdot \mathrm{D}(x)$.

注意, $\mathrm{D}(0)=\mathrm{D}(x+(-x))=\mathrm{D}(x)+\mathrm{D}(-x)$. 则 $\mathrm{D}(-x)=-\mathrm{D}(x)$. 则 $\mathrm{D}(\pm x)=\pm \mathrm{D}(x)$. 则 $\mathrm{D}(x\pm y)=\mathrm{D}(x)+\mathrm{D}(\pm y)=\mathrm{D}(x)\pm \mathrm{D}(y)$. 于是, 若 $s_1$, $s_2$, $\dots$, $s_n$ 的每一个是 $1$, $0$, 或 $-1$, 则$$\begin{array}{r}\mathrm{D}\left(\sum _{\ell =1}^n{s}_{\ell }{x}_{\ell }\right)=\sum _{\ell =1}^n{s}_{\ell }\mathrm{D}(x_{\ell }).\end{array}$$

设 $A$ 是 $n$ 级阵, 我们研究, $\mathrm{D}(\det (A))$ 会是什么.

设 $n=1$. 则 $\det (A)=[A{]}_{1,1}$. 则 $\mathrm{D}(\det (A))=\mathrm{D}([A{]}_{1,1})=\det [\mathrm{D}([A{]}_{1,1})]$.

设 $n=2$. 则 $\det (A)=[A{]}_{1,1}[A{]}_{2,2}-[A{]}_{2,1}[A{]}_{1,2}$. 则$$\begin{array}{rl}& \mathrm{D}(\det (A))\\ =& \mathrm{D}([A{]}_{1,1}[A{]}_{2,2}-[A{]}_{2,1}[A{]}_{1,2})\\ =& \mathrm{D}([A{]}_{1,1}[A{]}_{2,2})-\mathrm{D}([A{]}_{2,1}[A{]}_{1,2})\\ =& (\mathrm{D}([A{]}_{1,1})[A{]}_{2,2}+[A{]}_{1,1}\mathrm{D}([A{]}_{2,2}))-(\mathrm{D}([A{]}_{2,1})[A{]}_{1,2}+[A{]}_{2,1}\mathrm{D}([A{]}_{1,2}))\\ =& (\mathrm{D}([A{]}_{1,1})[A{]}_{2,2}-[A{]}_{2,1}\mathrm{D}([A{]}_{1,2}))+([A{]}_{1,1}\mathrm{D}([A{]}_{2,2})-\mathrm{D}([A{]}_{2,1})[A{]}_{1,2})\\ =& \det \left[\begin{array}{cc}\mathrm{D}([A{]}_{1,1})& [A{]}_{1,2}\\ \mathrm{D}([A{]}_{2,1})& [A{]}_{2,2}\end{array}\right]+\det \left[\begin{array}{cc}[A{]}_{1,1}& \mathrm{D}([A{]}_{1,2})\\ [A{]}_{2,1}& \mathrm{D}([A{]}_{2,2})\end{array}\right].\end{array}$$

设 $n=3$. 则$$\begin{array}{rl}\det (A)=& [A{]}_{1,1}\det \left[\begin{array}{cc}[A{]}_{2,2}& [A{]}_{2,3}\\ [A{]}_{3,2}& [A{]}_{3,3}\end{array}\right]-[A{]}_{2,1}\det \left[\begin{array}{cc}[A{]}_{1,2}& [A{]}_{1,3}\\ [A{]}_{3,2}& [A{]}_{3,3}\end{array}\right]\\ & +[A{]}_{3,1}\det \left[\begin{array}{cc}[A{]}_{1,2}& [A{]}_{1,3}\\ [A{]}_{2,2}& [A{]}_{2,3}\end{array}\right].\end{array}$$则$$\begin{array}{rl}& \mathrm{D}(\det (A))\\ =& \mathrm{D}([A{]}_{1,1})\det \left[\begin{array}{cc}[A{]}_{2,2}& [A{]}_{2,3}\\ [A{]}_{3,2}& [A{]}_{3,3}\end{array}\right]+[A{]}_{1,1}\mathrm{D}\left(\det \left[\begin{array}{cc}[A{]}_{2,2}& [A{]}_{2,3}\\ [A{]}_{3,2}& [A{]}_{3,3}\end{array}\right]\right)\\ & -\mathrm{D}([A{]}_{2,1})\det \left[\begin{array}{cc}[A{]}_{1,2}& [A{]}_{1,3}\\ [A{]}_{3,2}& [A{]}_{3,3}\end{array}\right]-[A{]}_{2,1}\mathrm{D}\left(\det \left[\begin{array}{cc}[A{]}_{1,2}& [A{]}_{1,3}\\ [A{]}_{3,2}& [A{]}_{3,3}\end{array}\right]\right)\\ & +\mathrm{D}([A{]}_{3,1})\det \left[\begin{array}{cc}[A{]}_{1,2}& [A{]}_{1,3}\\ [A{]}_{2,2}& [A{]}_{2,3}\end{array}\right]+[A{]}_{3,1}\mathrm{D}\left(\det \left[\begin{array}{cc}[A{]}_{1,2}& [A{]}_{1,3}\\ [A{]}_{2,2}& [A{]}_{2,3}\end{array}\right]\right)\\ =& \mathrm{D}([A{]}_{1,1})\det \left[\begin{array}{cc}[A{]}_{2,2}& [A{]}_{2,3}\\ [A{]}_{3,2}& [A{]}_{3,3}\end{array}\right]+[A{]}_{1,1}\det \left[\begin{array}{cc}\mathrm{D}([A{]}_{2,2})& [A{]}_{2,3}\\ \mathrm{D}([A{]}_{3,2})& [A{]}_{3,3}\end{array}\right]\\ & +[A{]}_{1,1}\det \left[\begin{array}{cc}[A{]}_{2,2}& \mathrm{D}([A{]}_{2,3})\\ [A{]}_{3,2}& \mathrm{D}([A{]}_{3,3})\end{array}\right]\\ & -\mathrm{D}([A{]}_{2,1})\det \left[\begin{array}{cc}[A{]}_{1,2}& [A{]}_{1,3}\\ [A{]}_{3,2}& [A{]}_{3,3}\end{array}\right]-[A{]}_{2,1}\det \left[\begin{array}{cc}\mathrm{D}([A{]}_{1,2})& [A{]}_{1,3}\\ \mathrm{D}([A{]}_{3,2})& [A{]}_{3,3}\end{array}\right]\\ & -[A{]}_{2,1}\det \left[\begin{array}{cc}[A{]}_{1,2}& \mathrm{D}([A{]}_{1,3})\\ [A{]}_{3,2}& \mathrm{D}([A{]}_{3,3})\end{array}\right]\\ & +\mathrm{D}([A{]}_{3,1})\det \left[\begin{array}{cc}[A{]}_{1,2}& [A{]}_{1,3}\\ [A{]}_{2,2}& [A{]}_{2,3}\end{array}\right]+[A{]}_{3,1}\det \left[\begin{array}{cc}\mathrm{D}([A{]}_{1,2})& [A{]}_{1,3}\\ \mathrm{D}([A{]}_{2,2})& [A{]}_{2,3}\end{array}\right]\\ & +[A{]}_{3,1}\det \left[\begin{array}{cc}[A{]}_{1,2}& \mathrm{D}([A{]}_{1,3})\\ [A{]}_{2,2}& \mathrm{D}([A{]}_{2,3})\end{array}\right]\\ =& \det \left[\begin{array}{ccc}\mathrm{D}([A{]}_{1,1})& [A{]}_{1,2}& [A{]}_{1,3}\\ \mathrm{D}([A{]}_{2,1})& [A{]}_{2,2}& [A{]}_{2,3}\\ \mathrm{D}([A{]}_{3,1})& [A{]}_{3,2}& [A{]}_{3,3}\end{array}\right]+\det \left[\begin{array}{ccc}[A{]}_{1,1}& \mathrm{D}([A{]}_{1,2})& [A{]}_{1,3}\\ [A{]}_{2,1}& \mathrm{D}([A{]}_{2,2})& [A{]}_{2,3}\\ [A{]}_{3,1}& \mathrm{D}([A{]}_{3,2})& [A{]}_{3,3}\end{array}\right]\\ & +\det \left[\begin{array}{ccc}[A{]}_{1,1}& [A{]}_{1,2}& \mathrm{D}([A{]}_{1,3})\\ [A{]}_{2,1}& [A{]}_{2,2}& \mathrm{D}([A{]}_{2,3})\\ [A{]}_{3,1}& [A{]}_{3,2}& \mathrm{D}([A{]}_{3,3})\end{array}\right].\end{array}$$

我们想, 我们或许知道 $\mathrm{D}(\det (A))$ 了. 为方便, 我们作这样的记号: 若 $B$ 是 $m\times n$ 阵, 且 $\ell$ 是正整数, 则 ${\mathrm{D}}_{\ell }(B)$ 也是 $m\times n$ 阵, 其中, $$\begin{array}{r}[{\mathrm{D}}_{\ell }(B){]}_{i,j}=\{\begin{array}{ll}\mathrm{D}([B{]}_{i,j}),& j=\ell ;\\ [B{]}_{i,j},& \text{其他}.\end{array}\end{array}$$

作命题 $P(n)$: 对任何 $n$ 级阵 $A$, $$\begin{array}{r}\mathrm{D}(\det (A))=\sum _{\ell =1}^n\det ({\mathrm{D}}_{\ell }(A)).\end{array}$$我们的目标是, 对任何正整数 $n$, $P(n)$ 是对的.

$P(1)$ 与 $P(2)$ 是对的.

假定 $P(n-1)$ 是对的. 我们由此证明, $P(n)$ 也是对的.

注意, $$\begin{array}{rl}& \mathrm{D}(\det (A))\\ =& \mathrm{D}\left(\sum _{i=1}^n(-1{)}^{i+1}[A{]}_{i,1}\det (A(i|1))\right)\\ =& \sum _{i=1}^n(-1{)}^{i+1}\mathrm{D}([A{]}_{i,1}\det (A(i|1)))\\ =& \sum _{i=1}^n(-1{)}^{i+1}(\mathrm{D}([A{]}_{i,1})\det (A(i|1))+[A{]}_{i,1}\mathrm{D}(\det (A(i|1))))\\ =& \sum _{i=1}^n(-1{)}^{i+1}\mathrm{D}([A{]}_{i,1})\det (A(i|1))+\sum _{i=1}^n(-1{)}^{i+1}[A{]}_{i,1}\mathrm{D}(\det (A(i|1))).\end{array}$$注意, 既然 $A$ 与 ${\mathrm{D}}_1(A)$ 的列 $j$ 相同 (若 $j\ne 1$), 则 $A(i|1)=({\mathrm{D}}_1(A))(i|1)$. 则$$\begin{array}{rl}& \sum _{i=1}^n(-1{)}^{i+1}\mathrm{D}([A{]}_{i,1})\det (A(i|1))\\ =& \sum _{i=1}^n(-1{)}^{i+1}\mathrm{D}([A{]}_{i,1})\det (({\mathrm{D}}_1(A))(i|1))\\ =& \det ({\mathrm{D}}_1(A)).\end{array}$$

由假定, $$\begin{array}{r}\mathrm{D}(\det (A(i|1)))=\sum _{v=1}^{n-1}\det ({\mathrm{D}}_v(A(i|1))).\end{array}$$注意, ${\mathrm{D}}_v(A(i|1))=({\mathrm{D}}_{v+1}(A))(i|1)$. 则$$\begin{array}{r}\mathrm{D}(\det (A(i|1)))=\sum _{v=1}^{n-1}\det (({\mathrm{D}}_{v+1}(A))(i|1))=\sum _{\ell =2}^n\det (({\mathrm{D}}_{\ell }(A))(i|1)).\end{array}$$注意, 当 $\ell \ne 1$ 时, $[{\mathrm{D}}_{\ell }(A){]}_{i,1}=[A{]}_{i,1}$. 则$$\begin{array}{rl}& \sum _{i=1}^n(-1{)}^{i+1}[A{]}_{i,1}\mathrm{D}(\det (A(i|1)))\\ =& \sum _{i=1}^n(-1{)}^{i+1}[{\mathrm{D}}_{\ell }(A){]}_{i,1}\sum _{\ell =2}^n\det (({\mathrm{D}}_{\ell }(A))(i|1))\\ =& \sum _{i=1}^n\sum _{\ell =2}^n(-1{)}^{i+1}[{\mathrm{D}}_{\ell }(A){]}_{i,1}\det (({\mathrm{D}}_{\ell }(A))(i|1))\\ =& \sum _{\ell =2}^n\sum _{i=1}^n(-1{)}^{i+1}[{\mathrm{D}}_{\ell }(A){]}_{i,1}\det (({\mathrm{D}}_{\ell }(A))(i|1))\\ =& \sum _{\ell =2}^n\det ({\mathrm{D}}_{\ell }(A)).\end{array}$$综上, $$\begin{array}{r}\mathrm{D}(\det (A))=\det ({\mathrm{D}}_1(A))+\sum _{\ell =2}^n\det ({\mathrm{D}}_{\ell }(A))=\sum _{\ell =1}^n\det ({\mathrm{D}}_{\ell }(A)).\end{array}$$所以, $P(n)$ 是对的. 由数学归纳法, 待证命题成立.

注意, 既然 $A$ 与 ${\mathrm{D}}_{\ell }(A)$ 的列 $j$ 相同 (若 $j\ne \ell$), 则 $A(i|\ell )=({\mathrm{D}}_{\ell }(A))(i|\ell )$. 并且, $[{\mathrm{D}}_{\ell }(A){]}_{i,\ell }=\mathrm{D}([A{]}_{i,\ell })$. 则$$\begin{array}{rl}\mathrm{D}(\det (A))=& \sum _{\ell =1}^n\det ({\mathrm{D}}_{\ell }(A))\\ =& \sum _{\ell =1}^n\sum _{i=1}^n(-1{)}^{i+\ell }[{\mathrm{D}}_{\ell }(A){]}_{i,\ell }\det (({\mathrm{D}}_{\ell }(A))(i|\ell ))\\ =& \sum _{\ell =1}^n\sum _{i=1}^n(-1{)}^{i+\ell }\mathrm{D}([A{]}_{i,\ell })\det (A(i|\ell ))\\ =& \sum _{\ell =1}^n\sum _{i=1}^n(-1{)}^{i+\ell }\det (A(i|\ell ))\mathrm{D}([A{]}_{i,\ell })\\ =& \sum _{\ell =1}^n\sum _{i=1}^n[\mathrm{adj}(A){]}_{\ell ,i}\mathrm{D}([A{]}_{i,\ell }).\end{array}$$为方便, 我们作这样的记号: 若 $B$ 是 $m\times n$ 阵, 则 ${\mathrm{D}}_{\mathrm{c}}(B)$ 也是 $m\times n$ 阵, 其中, $[{\mathrm{D}}_{\mathrm{c}}(B){]}_{i,j}=\mathrm{D}([B{]}_{i,j})$. 则$$\begin{array}{r}\mathrm{D}(\det (A))=\sum _{\ell =1}^n[\mathrm{adj}(A){\mathrm{D}}_{\mathrm{c}}(A){]}_{\ell ,\ell }.\end{array}$$

最后, 我们再引入方阵的 trace: 若 $A$ 是 $n$ 级阵, 则 $A$ 的 trace (chuìsi) $$\begin{array}{r}\mathrm{tr}(A)=\sum _{k=1}^n[A{]}_{k,k}.\end{array}$$则$$\begin{array}{r}\mathrm{D}(\det (A))=\mathrm{tr}(\mathrm{adj}(A){\mathrm{D}}_{\mathrm{c}}(A)).\end{array}$$

例 C.44.18. 设 $A$ 是 $n$ 级反称阵. 我们求 $\mathrm{D}(\mathrm{pf}(A))$.

设 $n=1$. 则 $\mathrm{pf}(A)=0$. 则 $\mathrm{D}(\mathrm{pf}(A))=0$.

设 $n=2$. 则 $\mathrm{pf}(A)=[A{]}_{1,2}$. 则 $\mathrm{D}(\mathrm{pf}(A))=\mathrm{D}([A{]}_{1,2})$.

设 $n=3$. 则 $\mathrm{pf}(A)=0$. 则 $\mathrm{D}(\mathrm{pf}(A))=0$.

设 $n=4$. 则 $\mathrm{pf}(A)=[A{]}_{1,2}[A{]}_{3,4}-[A{]}_{1,3}[A{]}_{2,4}+[A{]}_{1,4}[A{]}_{2,3}$. 则$$\begin{array}{rl}& \mathrm{D}(\mathrm{pf}(A))\\ =& \mathrm{D}([A{]}_{1,2})[A{]}_{3,4}+[A{]}_{1,2}\mathrm{D}([A{]}_{3,4})-\mathrm{D}([A{]}_{1,3})[A{]}_{2,4}-[A{]}_{1,3}\mathrm{D}([A{]}_{2,4})\\ & +\mathrm{D}([A{]}_{1,4})[A{]}_{2,3}+[A{]}_{1,4}\mathrm{D}([A{]}_{2,3})\\ =& \mathrm{D}([A{]}_{1,2})[A{]}_{3,4}-\mathrm{D}([A{]}_{1,3})[A{]}_{2,4}+\mathrm{D}([A{]}_{1,4})[A{]}_{2,3}\\ & +\mathrm{D}([A{]}_{2,3})[A{]}_{1,4}-\mathrm{D}([A{]}_{2,4})[A{]}_{1,3}+\mathrm{D}([A{]}_{3,4})[A{]}_{1,2}.\end{array}$$

作命题 $P(n)$: 对任何 $n$ 级反称阵 $A$, $$\begin{array}{r}\mathrm{D}(\mathrm{pf}(A))=\sum _{1⩽i<j⩽n}(-1{)}^{i+j-1}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(i,j|i,j)).\end{array}$$再作命题 $Q(n)$: $P(n-1)$ 与 $P(n)$ 是对的. 我们用数学归纳法证明, 对任何大于 $3$ 的正整数 $n$, $Q(n)$ 是对的. 则对不低于 $3$ 的正整数 $n$, $P(n)$ 是对的.

$P(3)$ 与 $P(4)$ 显然地是对的. 则 $Q(4)$ 是对的.

设 $Q(n-1)$ 是对的. 则 $P(n-2)$ 与 $P(n-1)$ 是对的. 我们要证, $Q(n)$ 是对的. 则我们要证, $P(n-1)$ 与 $P(n)$ 是对的. 于是, 若我们以 $P(n-2)$ 与 $P(n-1)$ 证, $P(n)$ 是对的, 则 $Q(n)$ 是对的.

设 $n>4$. 则$$\begin{array}{rl}& \mathrm{D}(\mathrm{pf}(A))\\ =& \mathrm{D}\left(\sum _{2⩽\ell ⩽n}(-1{)}^{\ell }[A{]}_{1,\ell }\mathrm{pf}(A(1,\ell |1,\ell ))\right)\\ =& \sum _{2⩽\ell ⩽n}(-1{)}^{\ell }\mathrm{D}([A{]}_{1,\ell }\mathrm{pf}(A(1,\ell |1,\ell )))\\ =& \sum _{2⩽\ell ⩽n}(-1{)}^{\ell }(\mathrm{D}([A{]}_{1,\ell })\mathrm{pf}(A(1,\ell |1,\ell ))+[A{]}_{1,\ell }\mathrm{D}(\mathrm{pf}(A(1,\ell |1,\ell ))))\\ =& \sum _{2⩽\ell ⩽n}(-1{)}^{\ell }\mathrm{D}([A{]}_{1,\ell })\mathrm{pf}(A(1,\ell |1,\ell ))\\ & +\sum _{2⩽\ell ⩽n}(-1{)}^{\ell }[A{]}_{1,\ell }\mathrm{D}(\mathrm{pf}(A(1,\ell |1,\ell ))).\end{array}$$注意, $[A{]}_{i,j}$ 在 $A(1,\ell |1,\ell )$ 的位置是行 $i-\rho (i,1)-\rho (i,\ell )$, 列 $j-\rho (j,1)-\rho (j,\ell )$. (回想, 若 $i⩾j$, 则 $\rho (i,j)=1$, 且若 $i<j$, 则 $\rho (i,j)=0$.) 由假定, $$\begin{array}{rl}& \mathrm{D}(\mathrm{pf}(A(1,\ell |1,\ell )))\\ =& \sum _{\begin{array}{c}2⩽i<j⩽n\\ i,j\ne \ell \end{array}}(-1{)}^{i-1-\rho (i,\ell )+j-1-\rho (j,\ell )-1}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(1,\ell ,i,j|1,\ell ,i,j))\\ =& \sum _{\begin{array}{c}2⩽i<j⩽n\\ i,j\ne \ell \end{array}}(-1{)}^{i-1+(\rho (\ell ,i)-1)+j-1+(\rho (\ell ,j)-1)-1}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(1,\ell ,i,j|1,\ell ,i,j))\\ =& \sum _{\begin{array}{c}2⩽i<j⩽n\\ i,j\ne \ell \end{array}}(-1{)}^{(i+j-1)-1-\rho (\ell ,i)+1-1-\rho (\ell ,j)+1}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(1,\ell ,i,j|1,\ell ,i,j))\\ =& \sum _{\begin{array}{c}2⩽i<j⩽n\\ i,j\ne \ell \end{array}}(-1{)}^{(i+j-1)-\rho (\ell ,i)-\rho (\ell ,j)}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(1,\ell ,i,j|1,\ell ,i,j)).\end{array}$$为方便, 我们简单地写 $A(1,\ell ,i,j|1,\ell ,i,j)$ 为 $A_{1,\ell ,i,j}$. 则$$\begin{array}{rl}& \sum _{2⩽\ell ⩽n}(-1{)}^{\ell }[A{]}_{1,\ell }\mathrm{D}(\mathrm{pf}(A(1,\ell |1,\ell )))\\ =& \sum _{2⩽\ell ⩽n}(-1{)}^{\ell }[A{]}_{1,\ell }\sum _{\begin{array}{c}2⩽i<j⩽n\\ i,j\ne \ell \end{array}}(-1{)}^{(i+j-1)-\rho (\ell ,i)-\rho (\ell ,j)}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A_{1,\ell ,i,j})\\ =& \sum _{2⩽\ell ⩽n}\sum _{\begin{array}{c}2⩽i<j⩽n\\ i,j\ne \ell \end{array}}(-1{)}^{\ell }[A{]}_{1,\ell }(-1{)}^{(i+j-1)-\rho (\ell ,i)-\rho (\ell ,j)}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A_{1,\ell ,i,j})\\ =& \sum _{2⩽\ell ⩽n}\sum _{\begin{array}{c}2⩽i<j⩽n\\ i,j\ne \ell \end{array}}(-1{)}^{i+j-1}\mathrm{D}([A{]}_{i,j})(-1{)}^{\ell -\rho (\ell ,i)-\rho (\ell ,j)}[A{]}_{1,\ell }\mathrm{pf}(A_{1,\ell ,i,j})\\ =& \sum _{2⩽i<j⩽n}\sum _{\begin{array}{c}2⩽\ell ⩽n\\ \ell \ne i,j\end{array}}(-1{)}^{i+j-1}\mathrm{D}([A{]}_{i,j})(-1{)}^{\ell -\rho (\ell ,i)-\rho (\ell ,j)}[A{]}_{1,\ell }\mathrm{pf}(A_{1,\ell ,i,j})\\ =& \sum _{2⩽i<j⩽n}(-1{)}^{i+j-1}\mathrm{D}([A{]}_{i,j})\sum _{\begin{array}{c}2⩽\ell ⩽n\\ \ell \ne i,j\end{array}}(-1{)}^{\ell -\rho (\ell ,i)-\rho (\ell ,j)}[A{]}_{1,\ell }\mathrm{pf}(A_{1,\ell ,i,j})\\ =& \sum _{\begin{array}{c}1⩽i<j⩽n\\ i⩾2\end{array}}(-1{)}^{i+j-1}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(i,j|i,j)).\end{array}$$注意, $$\begin{array}{rl}& \sum _{2⩽\ell ⩽n}(-1{)}^{\ell }\mathrm{D}([A{]}_{1,\ell })\mathrm{pf}(A(1,\ell |1,\ell ))\\ =& \sum _{\begin{array}{c}1⩽i<j⩽n\\ i=1\end{array}}(-1{)}^{i+j-1}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(i,j|i,j)).\end{array}$$综上, $$\begin{array}{rl}\mathrm{D}(\mathrm{pf}(A))=& \sum _{\begin{array}{c}1⩽i<j⩽n\\ i=1\end{array}}(-1{)}^{i+j-1}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(i,j|i,j))\\ & +\sum _{\begin{array}{c}1⩽i<j⩽n\\ i⩾2\end{array}}(-1{)}^{i+j-1}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(i,j|i,j))\\ =& \sum _{1⩽i<j⩽n}(-1{)}^{i+j-1}\mathrm{D}([A{]}_{i,j})\mathrm{pf}(A(i,j|i,j)).\end{array}$$所以, $P(n)$ 是对的. 则 $Q(n)$ 是对的. 由数学归纳法, 待证命题成立.

例 C.44.19. 回想, 若 $A$ 是 $n$ 级阵, 则 $A$ 的 trace $$\begin{array}{r}\mathrm{tr}(A)=\sum _{k=1}^n[A{]}_{k,k}.\end{array}$$我们说, trace 也是方阵的一个属性, 就像行列式是方阵的一个属性那样.

设 $A$, $B$ 是 $n$ 级阵. 设 $s$ 是数. 设 $C$ 是 $n\times m$ 阵, 且 $D$ 是 $m\times n$ 阵. 则:

(1) $\mathrm{tr}(A+B)=\mathrm{tr}(A)+\mathrm{tr}(B)$. 注意, $$\begin{array}{rl}\mathrm{tr}(A+B)=& \sum _{k=1}^n[A+B{]}_{k,k}\\ =& \sum _{k=1}^n([A{]}_{k,k}+[B{]}_{k,k})\\ =& \sum _{k=1}^n[A{]}_{k,k}+\sum _{k=1}^n[B{]}_{k,k}\\ =& \mathrm{tr}(A)+\mathrm{tr}(B).\end{array}$$

(2) $\mathrm{tr}(sA)=s\mathrm{tr}(A)$. 注意, $$\begin{array}{r}\mathrm{tr}(sA)=\sum _{k=1}^n[sA{]}_{k,k}=\sum _{k=1}^{n}s[A{]}_{k,k}=s\sum _{k=1}^n[A{]}_{k,k}=s\mathrm{tr}(A).\end{array}$$

(3) $\mathrm{tr}(A^{\mathrm{T}})=\mathrm{tr}(A)$. 注意, $$\begin{array}{r}\mathrm{tr}(A^{\mathrm{T}})=\sum _{k=1}^n[A^{\mathrm{T}}{]}_{k,k}=\sum _{k=1}^n[A{]}_{k,k}=\mathrm{tr}(A).\end{array}$$

(4) $\mathrm{tr}(CD)=\mathrm{tr}(DC)$. 注意, $$\begin{array}{rl}\mathrm{tr}(CD)=& \sum _{k=1}^n[CD{]}_{k,k}\\ =& \sum _{k=1}^n\sum _{\ell =1}^m[C{]}_{k,\ell }[D{]}_{\ell ,k}\\ =& \sum _{\ell =1}^m\sum _{k=1}^n[C{]}_{k,\ell }[D{]}_{\ell ,k}\\ =& \sum _{\ell =1}^m\sum _{k=1}^n[D{]}_{\ell ,k}[C{]}_{k,\ell }\\ =& \sum _{\ell =1}^m[DC{]}_{\ell ,\ell }\\ =& \mathrm{tr}(DC).\end{array}$$(其实, 此事是例 C.44.3 的一个特别的情形.)

值得提, 当 $n⩾2$ 时, $\mathrm{tr}(AB)=\mathrm{tr}(A)\mathrm{tr}(B)$ 一般不是对的. 为此, 设 $A$, $B$ 是这样的 $n$ 级阵: $$\begin{array}{r}[A{]}_{i,j}=\{\begin{array}{ll}1,& i=j=1;\\ 0,& \text{其他},\end{array}[B{]}_{i,j}=\{\begin{array}{ll}1,& i=j=2;\\ 0,& \text{其他}.\end{array}\end{array}$$可以验证, $AB=0$. 则 $\mathrm{tr}(AB)=0$. 但是, $\mathrm{tr}(A)=1=\mathrm{tr}(B)$.

例 C.44.20. 设 $A$ 是 $n$ 级阵. 则存在 $n$ 级阵 $B$, 使 $AB=\det (A)I_n=BA$.

我们知道, $A\mathrm{adj}(A)=\det (A)I_n=\mathrm{adj}(A)A$. 若 $\mathrm{adj}(A)\ne 0$, 我们取 $B$ 为 $\mathrm{adj}(A)$.

以下, 我们设 $\mathrm{adj}(A)=0$. 注意, 此时, $n>1$ (注意, 若 $n=1$, 则 $A$ 的古伴总为 $I_1$), 且 $\det (A)=0$.

若 $A=0$, 我们可取 $B$ 为任何非零的 $n$ 级阵 (如 $I_n$).

以下, 我们设 $\mathrm{adj}(A)=0$, 且 $A\ne 0$. 注意, 此时, $n>2$ (注意, 若 $n=2$, 则 $A$ 的古伴不为 $0$, 除非 $A=0$).

我们说, 存在小于 $n-1$ 的正整数 $r$, 使 $A$ 有行列式非零的 $r$ 级子阵, 且 $A$ 没有行列式非零的 $r+1$ 级子阵. 设 $C=A\left(\genfrac{}{}{0ex}{}{i_1,i_2,\dots ,i_{r+1}}{j_1,j_2,\dots ,j_{r+1}}\right)$ (其中, $i_1<i_2<\cdots <i_{r+1}$, 且 $j_1<j_2<\cdots <j_{r+1}$; 则 $[C{]}_{s,t}=[A{]}_{i_s,j_t}$) 是 $A$ 的一个 $r+1$ 级子阵, 且 $C$ 有行列式非零的 $r$ 级子阵 (此 $r$ 级子阵当然是 $A$ 的行列式非零的 $r$ 级子阵). 则 $\mathrm{adj}(C)\ne 0$, 且 $C\mathrm{adj}(C)=0=\mathrm{adj}(C)C$.

作 $n$ 级阵 $B$, 使$$\begin{array}{r}[B{]}_{i,j}=\{\begin{array}{ll}[\mathrm{adj}(C){]}_{t,s},& \text{若 i 等于某 jt, 且 j 等于某 is};\\ 0,& \text{其他}.\end{array}\end{array}$$则 $B\ne 0$ (因为 $\mathrm{adj}(C)$ 的元都是 $B$ 的元, 且 $\mathrm{adj}(C)\ne 0$). 我们说明, $AB=0=\det (A)I_n$, 且 $BA=0=\det (A)I_n$.

注意, $$\begin{array}{rl}[AB{]}_{i,j}=& \sum _{1⩽\ell ⩽n}[A{]}_{i,\ell }[B{]}_{\ell ,j}\\ =& \sum _{\begin{array}{c}1⩽\ell ⩽n\\ \ell \text{等于某}{j}_t\end{array}}[A{]}_{i,\ell }[B{]}_{\ell ,j}+\sum _{\begin{array}{c}1⩽\ell ⩽n\\ \ell \text{不等于任何}{j}_t\end{array}}[A{]}_{i,\ell }[B{]}_{\ell ,j}\\ =& \sum _{1⩽t⩽r+1}[A{]}_{i,j_t}[B{]}_{j_t,j}+\sum _{\begin{array}{c}1⩽\ell ⩽n\\ \ell \text{不等于任何}{j}_t\end{array}}[A{]}_{i,\ell }0\\ =& \sum _{1⩽t⩽r+1}[A{]}_{i,j_t}[B{]}_{j_t,j}.\end{array}$$若 $j$ 不等于任何 $i_s$, 则 $[B{]}_{j_t,j}=0$. 则 $[AB{]}_{i,j}=0$. 若 $j$ 等于某 $i_s$, 则$$\begin{array}{rl}[AB{]}_{i,i_s}=& \sum _{1⩽t⩽r+1}[A{]}_{i,j_t}[B{]}_{j_t,i_s}\\ =& \sum _{1⩽t⩽r+1}[A{]}_{i,j_t}[\mathrm{adj}(C){]}_{t,s}.\end{array}$$若 $i$ 等于某 $i_k$, 则$$\begin{array}{rl}[AB{]}_{i_k,i_s}=& \sum _{1⩽t⩽r+1}[A{]}_{i_k,j_t}[\mathrm{adj}(C){]}_{t,s}\\ =& \sum _{1⩽t⩽r+1}[C{]}_{k,t}[\mathrm{adj}(C){]}_{t,s}\\ =& [C\mathrm{adj}(C){]}_{k,s}\\ =& [0{]}_{k,s}\\ =& 0.\end{array}$$若 $i$ 不等于任何 $i_k$, 则$$\begin{array}{rl}[AB{]}_{i,i_s}=& \sum _{1⩽t⩽r+1}[A{]}_{i,j_t}[\mathrm{adj}(C){]}_{t,s}\\ =& \sum _{1⩽t⩽r+1}[A{]}_{i,j_t}(-1{)}^{s+t}\det (C(s|t)).\end{array}$$作 $r+1$ 级阵 $D$, 使$$\begin{array}{r}[D{]}_{u,v}=\{\begin{array}{ll}[A{]}_{i,j_v},& u=s;\\ [C{]}_{u,v},& \text{其他}.\end{array}\end{array}$$则 $C$ 的行 $u$ 与 $D$ 的行 $u$ 相等 (若 $u\ne s$). 则 $C(s|t)=D(s|t)$, 且 $[A{]}_{i,j_t}=[D{]}_{s,t}$. 则 $[AB{]}_{i,i_s}=\det (D)$.

设 $i$ 在 $i_1$, $\dots$, $i_{s-1}$, $i$, $i_{s+1}$, $\dots$, $i_{r+1}$ 中是第 $z$ 小的数. 若 $z=s$, 则 $D$ 就是 $A$ 的 $r+1$ 级子阵 $F=A\left(\genfrac{}{}{0ex}{}{i_1,\dots ,i_{s-1},i,i_{s+1},\dots ,i_{r+1}}{j_1,j_2,\dots ,j_{r+1}}\right)$. 注意, $\det (F)=0$. 则 $\det (D)=0$. 若 $z>s$, 则我们分别地使 $D$ 的行 $s$ 与在行 $s$ 的下面的 $z-s$ 行, 行 $i_{s+1}$, $\dots$, 行 $i_z$, 交换位置, 变 $D$ 为 $F$. 则 $\det (D)=(-1{)}^{z-s}\det (F)=0$. 若 $z<s$, 则我们分别地使 $D$ 的行 $s$ 与在行 $s$ 的上面的 $s-z$ 行, 行 $i_{s-1}$, $\dots$, 行 $i_z$, 交换位置, 变 $D$ 为 $F$. 则 $\det (D)=(-1{)}^{s-z}\det (F)=0$. 则 $[AB{]}_{i,i_s}=\det (D)=0$.

综上, 既然 $AB$ 的每一个元都是 $0$, 则 $AB=0$.

我们可几乎类似地证, $BA=0$. 当然, 我们也可这样作.

$BA=0$ 相当于 $A^{\mathrm{T}}{B}^{\mathrm{T}}=0^{\mathrm{T}}=0$. 且 $A^{\mathrm{T}}$ 也没有行列式非零的 $r+1$ 级子阵. 注意, $C^{\mathrm{T}}=A^{\mathrm{T}}\left(\genfrac{}{}{0ex}{}{j_1,j_2,\dots ,j_{r+1}}{i_1,i_2,\dots ,i_{r+1}}\right)$ 是 $A^{\mathrm{T}}$ 的一个 $r+1$ 级子阵, 且 $C^{\mathrm{T}}$ 有行列式非零的 $r$ 级子阵 (此 $r$ 级子阵当然是 $A^{\mathrm{T}}$ 的行列式非零的 $r$ 级子阵). 则 $\mathrm{adj}(C^{\mathrm{T}})\ne 0$, 且 $C^{\mathrm{T}}\mathrm{adj}(C^{\mathrm{T}})=0=\mathrm{adj}(C^{\mathrm{T}})C^{\mathrm{T}}$.

我们类似地作 $n$ 级阵 $G$, 使$$\begin{array}{r}[G{]}_{i,j}=\{\begin{array}{ll}[\mathrm{adj}(C^{\mathrm{T}}){]}_{s,t},& \text{若 i 等于某 is, 且 j 等于某 jt};\\ 0,& \text{其他}.\end{array}\end{array}$$则 $G=B^{\mathrm{T}}$. (注意, $[\mathrm{adj}(C^{\mathrm{T}}){]}_{s,t}=[\mathrm{adj}(C){]}_{t,s}$, 故 $[G{]}_{i_s,j_t}=[B{]}_{j_t,i_s}$. 在其他的情形, $[G{]}_{i,j}=0=[B{]}_{j,i}$.) 并且, 当然, $G\ne 0$. 那么, 由前面的讨论, $A^{\mathrm{T}}{B}^{\mathrm{T}}=0$.