拼接矩阵会对秩产生什么样的影响？

一、题目

已知 $\boldsymbol{A}, \boldsymbol{B}$ 均为 $n$ 阶非零矩阵, 且秩 $r(\boldsymbol{A})=r(\boldsymbol{B})$, 则以下说法中，正确的是哪个？

(A) $r(\boldsymbol{A}, \boldsymbol{B})=r(\boldsymbol{A})$.

(B) $r(\boldsymbol{A}, \boldsymbol{B})=2 r(\boldsymbol{B})$.

(C) $r(\boldsymbol{A}, \boldsymbol{B}) \leqslant 2 r(\boldsymbol{B})$.

(D) $r(\boldsymbol{A}-\boldsymbol{B})=0$.

难度评级：

二、解析

解法一：举特例

若：

$$
A=\left[\begin{array}{ll}0 & 1 \ 0 & 0\end{array}\right], \quad B=\left[\begin{array}{ll}0 & 0 \ 1 & 0\end{array}\right].
$$

则：

$$
r(A, B)=r\left[\begin{array}{llll}0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0\end{array}\right]=2.
$$

若：

$$
A=\left[\begin{array}{ll}0 & 1 \ 0 & 0\end{array}\right], \quad B=\left[\begin{array}{ll}0 & 0 \ 0 & 0\end{array}\right].
$$

则：

$$
r(A, B)=r\left[\begin{array}{llll}0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0\end{array}\right]=1
$$

综上可知，只有 $C$ 选项正确。

解法二：利用极大无关组

设：

$$
A=\left(\alpha_{1}, \alpha_{2}, \ldots \alpha_{n}\right) \quad B=\left(\beta_{1}, \beta_{2}, \ldots, \beta_{n}\right)
$$

则由题可知：

$A$ 中存在极大无关组 $\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{r}\right)$ 使得 $r(A)=r$.

$B$ 中存在极大无关组 $\left(\beta_{1}, \beta_{2}, \ldots, \beta_{r}\right)$ 使得 $r(B)=r$.

且 $\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}, \beta_{1}, \beta_{2}, \ldots, \beta_{n}\right)$ 可以被 $\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{r}\right)$ 和 $\left(\beta_{1}, \beta_{2}, \ldots, \beta_{r}\right)$ 线性表出。

于是：

$$
r \left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}, \beta_{1}, \beta_{2}, \ldots, \beta_{n}\right) \leqslant r\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{r}, \beta_{1}, \beta_{2}, \ldots, \beta_{r}\right).
$$

又：

$$
r\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{r}, \beta_{1}, \beta_{2}, \ldots, \beta_{r}\right) \leqslant 2 r.
$$

因此：

$$
r\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}, \beta_{1}, \beta_{2}, \ldots, \beta_{n}\right) \leqslant 2 r = 2 r(A) = 2 r(B).
$$

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