大连理工大学 2024年高等代数第0题
📝 题目
4．设 $\mathscr{A}$ 是线性空间 $V$ 上的线性变换， $\operatorname{Im} \mathscr{A}=\{\mathscr{A} \xi \mid \xi \in V\}, \operatorname{Ker} \mathscr{A}=\{\xi \mid \mathscr{A} \xi=0, \xi \in V\}$ 。证明： $\operatorname{Im} \mathscr{A}^{2}=\operatorname{Im} \mathscr{A}$ 当且仅当 $\operatorname{Ker} \mathscr{A}^{2}=\operatorname{Ker} \mathscr{A}$ ．
💡 答案解析
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📋 详细解题步骤
步骤 1/4
目标：明确已知包含关系
首先，注意到对于任意线性变换 $\mathscr{A}$，总有 $\operatorname{Im} \mathscr{A}^2 \subseteq \operatorname{Im} \mathscr{A}$ 和 $\operatorname{Ker} \mathscr{A} \subseteq \operatorname{Ker} \mathscr{A}^2$。这是因为：若 $\xi \in V$，则 $\mathscr{A}^2 \xi = \mathscr{A}(\mathscr{A} \xi) \in \operatorname{Im} \mathscr{A}$；若 $\mathscr{A} \xi = 0$，则 $\mathscr{A}^2 \xi = \mathscr{A}(0) = 0$。因此，要证明的等式等价于证明反向包含关系成立。
提示：注意包含关系的方向，不要混淆。
步骤 2/4
目标：必要性：由像相等推出核相等
假设 $\operatorname{Im} \mathscr{A}^2 = \operatorname{Im} \mathscr{A}$。要证 $\operatorname{Ker} \mathscr{A}^2 = \operatorname{Ker} \mathscr{A}$，只需证 $\operatorname{Ker} \mathscr{A}^2 \subseteq \operatorname{Ker} \mathscr{A}$。取 $\xi \in \operatorname{Ker} \mathscr{A}^2$，则 $\mathscr{A}^2 \xi = 0$。于是 $\mathscr{A} \xi \in \operatorname{Im} \mathscr{A} \cap \operatorname{Ker} \mathscr{A}$。由 $\operatorname{Im} \mathscr{A}^2 = \operatorname{Im} \mathscr{A}$，存在 $\eta \in V$ 使得 $\mathscr{A} \xi = \mathscr{A}^2 \eta$。则 $\mathscr{A} \xi = \mathscr{A}(\mathscr{A} \eta)$，所以 $\mathscr{A}(\xi - \mathscr{A} \eta) = 0$，即 $\xi - \mathscr{A} \eta \in \operatorname{Ker} \mathscr{A}$。又因为 $\mathscr{A}^2 \xi = 0$，有 $\mathscr{A}(\mathscr{A} \xi) = 0$，所以 $\mathscr{A} \xi \in \operatorname{Ker} \mathscr{A}$。于是 $\mathscr{A} \xi = 0$，从而 $\xi \in \operatorname{Ker} \mathscr{A}$。因此 $\operatorname{Ker} \mathscr{A}^2 \subseteq \operatorname{Ker} \mathscr{A}$，故 $\operatorname{Ker} \mathscr{A}^2 = \operatorname{Ker} \mathscr{A}$。
公式：$\mathscr{A}^2 \xi = 0$ 推出 $\mathscr{A} \xi \in \operatorname{Ker} \mathscr{A}$
提示：注意 $\mathscr{A} \xi$ 同时属于像和核，利用像相等条件构造 $\eta$。
步骤 3/4
目标：充分性：由核相等推出像相等（利用维数公式）
假设 $\operatorname{Ker} \mathscr{A}^2 = \operatorname{Ker} \mathscr{A}$。要证 $\operatorname{Im} \mathscr{A}^2 = \operatorname{Im} \mathscr{A}$，只需证 $\operatorname{Im} \mathscr{A} \subseteq \operatorname{Im} \mathscr{A}^2$。由维数公式：$\dim V = \dim \operatorname{Im} \mathscr{A} + \dim \operatorname{Ker} \mathscr{A}$，且 $\dim V = \dim \operatorname{Im} \mathscr{A}^2 + \dim \operatorname{Ker} \mathscr{A}^2$。由于 $\operatorname{Ker} \mathscr{A}^2 = \operatorname{Ker} \mathscr{A}$，得 $\dim \operatorname{Im} \mathscr{A}^2 = \dim \operatorname{Im} \mathscr{A}$。又 $\operatorname{Im} \mathscr{A}^2 \subseteq \operatorname{Im} \mathscr{A}$，故 $\operatorname{Im} \mathscr{A}^2 = \operatorname{Im} \mathscr{A}$。
公式：$\dim V = \dim \operatorname{Im} \mathscr{A} + \dim \operatorname{Ker} \mathscr{A}$
提示：维数公式要求 $V$ 是有限维线性空间，题目中未明确，但通常高等代数中默认有限维。若无限维，需用其他方法。
步骤 4/4
目标：总结结论
综上，我们证明了 $\operatorname{Im} \mathscr{A}^2 = \operatorname{Im} \mathscr{A}$ 当且仅当 $\operatorname{Ker} \mathscr{A}^2 = \operatorname{Ker} \mathscr{A}$。
提示：注意充分性证明中使用了维数公式，若空间无限维，则需用其他方法（如直接构造）。
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