南京师范大学 2016年高等代数第0题

假设 $V = \mathscr{A}V \oplus \mathscr{A}^{-1}(0)$。对任意 $v \in V$，存在 $u \in \mathscr{A}V$ 和 $w \in \mathscr{A}^{-1}(0)$ 使得 $v = u + w$。则 $\mathscr{A}v = \mathscr{A}u + \mathscr{A}w = \mathscr{A}u$。由于 $u \in \mathscr{A}V$，存在 $x \in V$ 使得 $u = \mathscr{A}x$，故 $\mathscr{A}v = \mathscr{A}^2 x \in \mathscr{A}^2 V$。因此 $\mathscr{A}V \subseteq \mathscr{A}^2 V$。显然 $\mathscr{A}^2 V \subseteq \mathscr{A}V$，所以 $\mathscr{A}^2 V = \mathscr{A}V$。

假设 $\mathscr{A}^2 V = \mathscr{A}V$。首先证明 $\mathscr{A}V \cap \mathscr{A}^{-1}(0) = \{0\}$。取 $y \in \mathscr{A}V \cap \mathscr{A}^{-1}(0)$，则存在 $x \in V$ 使得 $y = \mathscr{A}x$，且 $\mathscr{A}y = 0$。于是 $\mathscr{A}^2 x = 0$。由 $\mathscr{A}^2 V = \mathscr{A}V$，存在 $z \in V$ 使得 $\mathscr{A}x = \mathscr{A}^2 z$，即 $y = \mathscr{A}^2 z$。那么 $\mathscr{A}y = \mathscr{A}^3 z = 0$。由于 $\mathscr{A}^2 V = \mathscr{A}V$，$\mathscr{A}$ 在 $\mathscr{A}V$ 上的限制是满射，且有限维线性空间上满射必为单射，故 $\mathscr{A}$ 在 $\mathscr{A}V$ 上可逆。因此由 $\mathscr{A}y = 0$ 且 $y \in \mathscr{A}V$ 得 $y = 0$。所以交为零。

其次证明 $V = \mathscr{A}V + \mathscr{A}^{-1}(0)$。对任意 $v \in V$，由 $\mathscr{A}^2 V = \mathscr{A}V$，存在 $u \in V$ 使得 $\mathscr{A}v = \mathscr{A}^2 u$。于是 $\mathscr{A}(v - \mathscr{A}u) = 0$，即 $v - \mathscr{A}u \in \mathscr{A}^{-1}(0)$。因此 $v = \mathscr{A}u + (v - \mathscr{A}u) \in \mathscr{A}V + \mathscr{A}^{-1}(0)$。

由前两步，$\mathscr{A}V \cap \mathscr{A}^{-1}(0) = \{0\}$ 且 $V = \mathscr{A}V + \mathscr{A}^{-1}(0)$，根据直和的定义，$V = \mathscr{A}V \oplus \mathscr{A}^{-1}(0)$。

必要性已证，充分性已证，故 $V = \mathscr{A}V \oplus \mathscr{A}^{-1}(0)$ 当且仅当 $\mathscr{A}^2 V = \mathscr{A}V$。