58. 行列式为 0 的阵的性质

本节, 我们讨论行列式为 $0$ 的阵的几个性质.

定理 58.1. 设 $A$ 是一个 $n$ 级阵. 设 $\det (A)=0$. 则存在不超过 $n$ 的正整数 $i$, 使$$|[A{]}_{i,i}|⩽\sum _{\begin{array}{c}1⩽u⩽n\\ u\ne i\end{array}}|[A{]}_{i,u}|.$$(58.1)

证. 因为 $\det (A)=0$, 故有非零的 $n\times 1$ 阵 $x$, 使 $Ax=0$ (见第一章, 节 23 或节 25).

考虑不全为 $0$ 的非负实数 $|[x{]}_{1,1}|$, $|[x{]}_{2,1}|$, $\dots$, $|[x{]}_{n,1}|$. 则有不超过 $n$ 的正整数 $i$, 使对任何不超过 $n$ 的正整数 $u$, 有 $|[x{]}_{i,1}|⩾|[x{]}_{u,1}|$, 且 $|[x{]}_{i,1}|>0$. 则$$\begin{array}{rl}0=& [0{]}_{i,1}\\ =& [Ax{]}_{i,1}\\ =& \sum _{1⩽u⩽n}[A{]}_{i,u}[x{]}_{u,1}\\ =& \sum _{\begin{array}{c}1⩽u⩽n\\ u\ne i\end{array}}[A{]}_{i,u}[x{]}_{u,1}+[A{]}_{i,i}[x{]}_{i,1}.\end{array}$$则$$\begin{array}{rl}|[A{]}_{i,i}||[x{]}_{i,1}|=& |-1||[A{]}_{i,i}[x{]}_{i,1}|\\ =& |-[A{]}_{i,i}[x{]}_{i,1}|\\ =& |\sum _{\begin{array}{c}1⩽u⩽n\\ u\ne i\end{array}}[A{]}_{i,u}[x{]}_{u,1}|\\ ⩽& \sum _{\begin{array}{c}1⩽u⩽n\\ u\ne i\end{array}}|[A{]}_{i,u}[x{]}_{u,1}|\\ =& \sum _{\begin{array}{c}1⩽u⩽n\\ u\ne i\end{array}}|[A{]}_{i,u}||[x{]}_{u,1}|\\ ⩽& \sum _{\begin{array}{c}1⩽u⩽n\\ u\ne i\end{array}}|[A{]}_{i,u}||[x{]}_{i,1}|\\ =& (\sum _{\begin{array}{c}1⩽u⩽n\\ u\ne i\end{array}}|[A{]}_{i,u}|)|[x{]}_{i,1}|.\end{array}$$因为 $|[x{]}_{i,1}|>0$, 故式 (58.1) 是对的.

证毕.

不过, 此事反过来不一定是对的.

例 58.2. 设 $A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$ 是一个 $2$ 级阵. 取 $i=1$ 或 $i=2$, 即有$$\begin{array}{r}|[A{]}_{i,1}|=0⩽1=\sum _{\begin{array}{c}1⩽u⩽2\\ u\ne i\end{array}}|[A{]}_{i,u}|.\end{array}$$可是, $\det (A)=1\ne 0$.

定理 58.3. 设 $A$ 是一个 $n$ 级阵 ($n⩾2$). 设 $\det (A)=0$. 则存在不超过 $n$, 且不等的正整数 $j$, $k$, 使$$|[A{]}_{j,j}||[A{]}_{k,k}|⩽(\sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}|\left)\right(\sum _{\begin{array}{c}1⩽q⩽n\\ q\ne k\end{array}}|[A{]}_{k,q}|).$$(58.2)

证. 因为 $\det (A)=0$, 故有非零的 $n\times 1$ 阵 $x$, 使 $Ax=0$ (见第一章, 节 23 或节 25).

考虑不全为 $0$ 的非负实数 $|[x{]}_{1,1}|$, $|[x{]}_{2,1}|$, $\dots$, $|[x{]}_{n,1}|$. 则有不超过 $n$, 且不等的正整数 $j$, $k$, 使对任何不超过 $n$, 且不等于 $j$ 的正整数 $p$, 有 $|[x{]}_{j,1}|⩾|[x{]}_{k,1}|⩾|[x{]}_{p,1}|$, 且 $|[x{]}_{j,1}|>0$.

注意, 对任何不超过 $n$ 的正整数 $u$, 有$$\begin{array}{rl}0=& [0{]}_{u,1}\\ =& [Ax{]}_{u,1}\\ =& \sum _{1⩽v⩽n}[A{]}_{u,v}[x{]}_{v,1}\\ =& \sum _{\begin{array}{c}1⩽v⩽n\\ v\ne u\end{array}}[A{]}_{u,v}[x{]}_{v,1}+[A{]}_{u,u}[x{]}_{u,1}.\end{array}$$故$$\begin{array}{r}-[A{]}_{u,u}[x{]}_{u,1}=\sum _{\begin{array}{c}1⩽v⩽n\\ v\ne u\end{array}}[A{]}_{u,v}[x{]}_{v,1}.\end{array}$$则$$\begin{array}{rl}|[A{]}_{u,u}||[x{]}_{u,1}|=& |-1||[A{]}_{u,u}[x{]}_{u,1}|\\ =& |-[A{]}_{u,u}[x{]}_{u,1}|\\ =& |\sum _{\begin{array}{c}1⩽v⩽n\\ v\ne u\end{array}}[A{]}_{u,v}[x{]}_{v,1}|\\ ⩽& \sum _{\begin{array}{c}1⩽v⩽n\\ v\ne u\end{array}}|[A{]}_{u,v}[x{]}_{v,1}|\\ =& \sum _{\begin{array}{c}1⩽v⩽n\\ v\ne u\end{array}}|[A{]}_{u,v}||[x{]}_{v,1}|.\end{array}$$则 (取 $u=j$, 并注意, 对任何不超过 $n$, 且不等于 $j$ 的正整数 $p$, 有 $|[x{]}_{p,1}|⩽|[x{]}_{k,1}|$) $$\begin{array}{rl}|[A{]}_{j,j}||[x{]}_{j,1}|⩽& \sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}||[x{]}_{p,1}|\\ ⩽& \sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}||[x{]}_{k,1}|\\ =& (\sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}|)|[x{]}_{k,1}|.\end{array}$$因为 $|[x{]}_{j,1}|>0$, $$\begin{array}{r}|[A{]}_{j,j}|⩽(\sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}|)|[x{]}_{k,1}||[x{]}_{j,1}{|}^{-1}.\end{array}$$另一方面 (取 $u=k$, 并注意, 对任何不超过 $n$ 的正整数 $q$, 有 $|[x{]}_{q,1}|⩽|[x{]}_{j,1}|$), $$\begin{array}{rl}|[A{]}_{k,k}||[x{]}_{k,1}|⩽& \sum _{\begin{array}{c}1⩽q⩽n\\ q\ne k\end{array}}|[A{]}_{k,q}||[x{]}_{q,1}|\\ ⩽& \sum _{\begin{array}{c}1⩽q⩽n\\ q\ne k\end{array}}|[A{]}_{k,q}||[x{]}_{j,1}|\\ =& (\sum _{\begin{array}{c}1⩽q⩽n\\ q\ne k\end{array}}|[A{]}_{k,q}|)|[x{]}_{j,1}|.\end{array}$$故$$\begin{array}{r}(|[A{]}_{j,j}||[A{]}_{k,k}|)|[x{]}_{k,1}|⩽(\sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}|)(\sum _{\begin{array}{c}1⩽q⩽n\\ q\ne k\end{array}}|[A{]}_{k,q}|)|[x{]}_{k,1}|.\end{array}$$

若 $|[x{]}_{k,1}|>0$, 则式 (58.2) 当然是对的. 若 $|[x{]}_{k,1}|=0$, 则$$\begin{array}{r}0⩽|[A{]}_{j,j}|⩽(\sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}|)0|[x{]}_{j,1}{|}^{-1}=0.\end{array}$$故 $|[A{]}_{j,j}|=0$. 则式 (58.2) 的左侧是 $0$.

证毕.

此事反过来不一定是对的: 考虑上个例的 $A$ 即可.

我们说, 若存在不超过 $n$, 且不等的正整数 $j$, $k$, 使式 (58.2) 是对的, 则存在不超过 $n$ 的正整数 $i$, 使式 (58.1) 是对的. 用反证法, 不难看出, 若式 (58.2) 是对的, 则$$\begin{array}{r}|[A{]}_{j,j}|>\sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}|\end{array}$$与$$\begin{array}{r}|[A{]}_{k,k}|>\sum _{\begin{array}{c}1⩽q⩽n\\ q\ne k\end{array}}|[A{]}_{k,q}|\end{array}$$不能全是对的. 故$$\begin{array}{r}|[A{]}_{j,j}|⩽\sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}|\end{array}$$与$$\begin{array}{r}|[A{]}_{k,k}|⩽\sum _{\begin{array}{c}1⩽q⩽n\\ q\ne k\end{array}}|[A{]}_{k,q}|\end{array}$$的至少一个是对的. 取 $i$ 为 $j$ 或 $k$ 即可.

或许, 您会想:

设 $A$ 是一个 $n$ 级阵 ($n⩾3$). 设 $\det (A)=0$. 则存在不超过 $n$, 且互不相同的正整数 $j$, $k$, $\ell$, 使$$\begin{array}{rl}& |[A{]}_{j,j}||[A{]}_{k,k}||[A{]}_{\ell ,\ell }|\\ ⩽& (\sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{j,p}|\left)\right(\sum _{\begin{array}{c}1⩽q⩽n\\ q\ne k\end{array}}|[A{]}_{k,q}|)(\sum _{\begin{array}{c}1⩽r⩽n\\ r\ne \ell \end{array}}|[A{]}_{\ell ,r}|).\end{array}$$

不过, 这不是对的.

例 58.4. 设 $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 10\end{array}\right]$ 是一个 $3$ 级阵. 不难算出, $\det (A)=0$. 可是 (此处 $j$, $k$, $\ell$ 分别是 $1$, $2$, $3$) $$\begin{array}{rl}& |[A{]}_{j,j}||[A{]}_{k,k}||[A{]}_{\ell ,\ell }|\\ =& 1\cdot 1\cdot 10\\ >& 2\cdot 2\cdot 2\\ =& (\sum _{\begin{array}{c}1⩽p⩽3\\ p\ne j\end{array}}|[A{]}_{j,p}|\left)\right(\sum _{\begin{array}{c}1⩽q⩽3\\ q\ne k\end{array}}|[A{]}_{k,q}|)(\sum _{\begin{array}{c}1⩽r⩽3\\ r\ne \ell \end{array}}|[A{]}_{\ell ,r}|).\end{array}$$$j$, $k$, $\ell$ 取其他的数时, 也有类似的结果.

最后, 注意, 一个阵与其转置的行列式相等: 若 $\det (A)=0$, 则 $\det (A^{\mathrm{T}})=\det (A)=0$. 应用前二个定理于 $A^{\mathrm{T}}$, 立得

定理 58.5. 设 $A$ 是一个 $n$ 级阵. 设 $\det (A)=0$. 则存在不超过 $n$ 的正整数 $i$, 使$$\begin{array}{r}|[A{]}_{i,i}|⩽\sum _{\begin{array}{c}1⩽u⩽n\\ u\ne i\end{array}}|[A{]}_{u,i}|.\end{array}$$

定理 58.6. 设 $A$ 是一个 $n$ 级阵 ($n⩾2$). 设 $\det (A)=0$. 则存在不超过 $n$, 且不等的正整数 $j$, $k$, 使$$\begin{array}{r}|[A{]}_{j,j}||[A{]}_{k,k}|⩽(\sum _{\begin{array}{c}1⩽p⩽n\\ p\ne j\end{array}}|[A{]}_{p,j}|\left)\right(\sum _{\begin{array}{c}1⩽q⩽n\\ q\ne k\end{array}}|[A{]}_{q,k}|).\end{array}$$