合肥工业大学 2025年高等代数第11题
📝 题目
11、已知 $\displaystyle \mathscr{A}$ 是 $V$ 上的线性变化，试证 $\displaystyle \mathscr{A}^{2}=\mathscr{A} \Leftrightarrow V=\mathscr{N} \otimes \operatorname{ker} \mathscr{A}$ 。
💡 答案解析
暂无答案解析
📋 详细解题步骤
步骤 1/7
目标：明确要证明的等价关系
设 $\mathscr{A}$ 是 $V$ 上的线性变换，$\mathscr{A}^2 = \mathscr{A}$ 当且仅当 $V = \operatorname{Im}\mathscr{A} \oplus \ker\mathscr{A}$。
提示：注意直和符号 $\oplus$ 表示子空间的和是直和，即交为 $\{0\}$。
步骤 2/7
目标：必要性：证明 $\operatorname{Im}\mathscr{A} \cap \ker\mathscr{A} = \{0\}$
假设 $\mathscr{A}^2 = \mathscr{A}$。任取 $\alpha \in \operatorname{Im}\mathscr{A} \cap \ker\mathscr{A}$，则存在 $\beta \in V$ 使得 $\alpha = \mathscr{A}\beta$，且 $\mathscr{A}\alpha = 0$。于是 $0 = \mathscr{A}\alpha = \mathscr{A}(\mathscr{A}\beta) = \mathscr{A}^2\beta = \mathscr{A}\beta = \alpha$，故 $\alpha = 0$。
公式：$\mathscr{A}^2 = \mathscr{A}$
提示：注意 $\alpha \in \ker\mathscr{A}$ 意味着 $\mathscr{A}\alpha = 0$。
步骤 3/7
目标：必要性：证明 $V = \operatorname{Im}\mathscr{A} + \ker\mathscr{A}$
任取 $\alpha \in V$，令 $\beta = \mathscr{A}\alpha$，$\gamma = \alpha - \mathscr{A}\alpha$。则 $\beta \in \operatorname{Im}\mathscr{A}$，且 $\mathscr{A}\gamma = \mathscr{A}(\alpha - \mathscr{A}\alpha) = \mathscr{A}\alpha - \mathscr{A}^2\alpha = \mathscr{A}\alpha - \mathscr{A}\alpha = 0$，故 $\gamma \in \ker\mathscr{A}$。于是 $\alpha = \beta + \gamma \in \operatorname{Im}\mathscr{A} + \ker\mathscr{A}$。
公式：$\mathscr{A}^2 = \mathscr{A}$
提示：构造 $\gamma = \alpha - \mathscr{A}\alpha$ 是关键技巧。
步骤 4/7
目标：必要性结论：$V = \operatorname{Im}\mathscr{A} \oplus \ker\mathscr{A}$
由前两步，$\operatorname{Im}\mathscr{A} \cap \ker\mathscr{A} = \{0\}$ 且 $V = \operatorname{Im}\mathscr{A} + \ker\mathscr{A}$，故 $V = \operatorname{Im}\mathscr{A} \oplus \ker\mathscr{A}$。
提示：直和需要同时满足和与交的条件。
步骤 5/7
目标：充分性：利用直和分解表示任意向量
假设 $V = \operatorname{Im}\mathscr{A} \oplus \ker\mathscr{A}$。任取 $\alpha \in V$，则存在唯一的 $\beta \in \operatorname{Im}\mathscr{A}$ 和 $\gamma \in \ker\mathscr{A}$ 使得 $\alpha = \beta + \gamma$。由于 $\beta \in \operatorname{Im}\mathscr{A}$，存在 $\delta \in V$ 使得 $\beta = \mathscr{A}\delta$。
提示：注意直和分解的唯一性。
步骤 6/7
目标：充分性：证明 $\mathscr{A}\beta = \beta$
考虑 $\mathscr{A}\beta - \beta$。因为 $\mathscr{A}(\mathscr{A}\beta - \beta) = \mathscr{A}^2\beta - \mathscr{A}\beta = \mathscr{A}\beta - \mathscr{A}\beta = 0$，所以 $\mathscr{A}\beta - \beta \in \ker\mathscr{A}$。又 $\mathscr{A}\beta \in \operatorname{Im}\mathscr{A}$，$\beta \in \operatorname{Im}\mathscr{A}$，故 $\mathscr{A}\beta - \beta \in \operatorname{Im}\mathscr{A}$。由于 $\operatorname{Im}\mathscr{A} \cap \ker\mathscr{A} = \{0\}$，得 $\mathscr{A}\beta - \beta = 0$，即 $\mathscr{A}\beta = \beta$。
公式：$\mathscr{A}^2\beta = \mathscr{A}\beta$ 由幂等性推出，但此处尚未证明，需用直和性质。
提示：这里 $\mathscr{A}^2\beta$ 中的 $\mathscr{A}^2$ 是复合，但 $\beta$ 是像，需注意 $\mathscr{A}\beta$ 不一定等于 $\beta$，需证明。
步骤 7/7
目标：充分性：证明 $\mathscr{A}^2\alpha = \mathscr{A}\alpha$
由 $\alpha = \beta + \gamma$，$\beta = \mathscr{A}\delta$，且 $\mathscr{A}\gamma = 0$，得 $\mathscr{A}\alpha = \mathscr{A}\beta + \mathscr{A}\gamma = \mathscr{A}\beta$。又 $\mathscr{A}^2\alpha = \mathscr{A}(\mathscr{A}\alpha) = \mathscr{A}(\mathscr{A}\beta) = \mathscr{A}^2\beta$。由 $\mathscr{A}\beta = \beta$，有 $\mathscr{A}^2\beta = \mathscr{A}(\mathscr{A}\beta) = \mathscr{A}\beta = \beta$。而 $\mathscr{A}\alpha = \mathscr{A}\beta = \beta$，故 $\mathscr{A}^2\alpha = \beta = \mathscr{A}\alpha$。由 $\alpha$ 任意性，$\mathscr{A}^2 = \mathscr{A}$。
公式：$\mathscr{A}^2\alpha = \mathscr{A}\alpha$
提示：注意 $\mathscr{A}^2\beta = \mathscr{A}(\mathscr{A}\beta)$，利用上一步 $\mathscr{A}\beta = \beta$。
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