东华大学 2026年高等代数第0题
📝 题目
1．（8 分）若 $\mathscr{A}, \mathscr{B}$ 是幕等变换，即 $\mathscr{A}^{2}=\mathscr{A}, \mathscr{B}^{2}=\mathscr{B}$ ，证明： $\mathscr{A}+\mathscr{B}$ 是幂等变换的充要条件是
$$
\mathscr{A} \mathscr{B}=\mathscr{B} \mathscr{A}=\mathscr{O} .
$$
💡 答案解析
暂无答案解析
📋 详细解题步骤
步骤 1/7
目标：明确已知条件
设 $\mathscr{A}, \mathscr{B}$ 是幂等变换，即 $\mathscr{A}^2 = \mathscr{A}$, $\mathscr{B}^2 = \mathscr{B}$。
提示：注意幂等变换的定义：变换的平方等于自身。
步骤 2/7
目标：必要性：假设 $\mathscr{A}+\mathscr{B}$ 是幂等变换
若 $\mathscr{A}+\mathscr{B}$ 是幂等变换，则 $(\mathscr{A}+\mathscr{B})^2 = \mathscr{A}+\mathscr{B}$。
提示：幂等变换的定义直接应用。
步骤 3/7
目标：展开平方并化简
计算左边：$(\mathscr{A}+\mathscr{B})^2 = \mathscr{A}^2 + \mathscr{A}\mathscr{B} + \mathscr{B}\mathscr{A} + \mathscr{B}^2 = \mathscr{A} + \mathscr{A}\mathscr{B} + \mathscr{B}\mathscr{A} + \mathscr{B}$。由幂等性得：$\mathscr{A} + \mathscr{A}\mathscr{B} + \mathscr{B}\mathscr{A} + \mathscr{B} = \mathscr{A} + \mathscr{B}$。两边消去 $\mathscr{A}+\mathscr{B}$，得：$\mathscr{A}\mathscr{B} + \mathscr{B}\mathscr{A} = \mathscr{O}$。
公式：$(\mathscr{A}+\mathscr{B})^2 = \mathscr{A}^2 + \mathscr{A}\mathscr{B} + \mathscr{B}\mathscr{A} + \mathscr{B}^2$
提示：注意展开时顺序不能交换，因为变换不一定可交换。
步骤 4/7
目标：推导 $\mathscr{A}\mathscr{B} = \mathscr{B}\mathscr{A}$
将 $\mathscr{A}\mathscr{B} + \mathscr{B}\mathscr{A} = \mathscr{O}$ 左乘 $\mathscr{A}$：$\mathscr{A}^2\mathscr{B} + \mathscr{A}\mathscr{B}\mathscr{A} = \mathscr{A}\mathscr{O} = \mathscr{O} \Rightarrow \mathscr{A}\mathscr{B} + \mathscr{A}\mathscr{B}\mathscr{A} = \mathscr{O}$。右乘 $\mathscr{A}$：$\mathscr{A}\mathscr{B}\mathscr{A} + \mathscr{B}\mathscr{A}^2 = \mathscr{O} \Rightarrow \mathscr{A}\mathscr{B}\mathscr{A} + \mathscr{B}\mathscr{A} = \mathscr{O}$。两式相减得：$(\mathscr{A}\mathscr{B} + \mathscr{A}\mathscr{B}\mathscr{A}) - (\mathscr{A}\mathscr{B}\mathscr{A} + \mathscr{B}\mathscr{A}) = \mathscr{O} \Rightarrow \mathscr{A}\mathscr{B} - \mathscr{B}\mathscr{A} = \mathscr{O}$，所以 $\mathscr{A}\mathscr{B} = \mathscr{B}\mathscr{A}$。
提示：左乘和右乘时要小心，不能随意交换顺序。
步骤 5/7
目标：得到 $\mathscr{A}\mathscr{B} = \mathscr{O}$ 和 $\mathscr{B}\mathscr{A} = \mathscr{O}$
由 $\mathscr{A}\mathscr{B} = \mathscr{B}\mathscr{A}$ 代入 $\mathscr{A}\mathscr{B} + \mathscr{B}\mathscr{A} = \mathscr{O}$ 得 $2\mathscr{A}\mathscr{B} = \mathscr{O}$，即 $\mathscr{A}\mathscr{B} = \mathscr{O}$，从而 $\mathscr{B}\mathscr{A} = \mathscr{O}$。
提示：注意零变换乘以任何变换仍为零变换。
步骤 6/7
目标：充分性：假设 $\mathscr{A}\mathscr{B} = \mathscr{B}\mathscr{A} = \mathscr{O}$
若 $\mathscr{A}\mathscr{B} = \mathscr{B}\mathscr{A} = \mathscr{O}$，则 $(\mathscr{A}+\mathscr{B})^2 = \mathscr{A}^2 + \mathscr{A}\mathscr{B} + \mathscr{B}\mathscr{A} + \mathscr{B}^2 = \mathscr{A} + \mathscr{O} + \mathscr{O} + \mathscr{B} = \mathscr{A}+\mathscr{B}$。
提示：直接代入计算即可。
步骤 7/7
目标：得出结论
故 $\mathscr{A}+\mathscr{B}$ 是幂等变换。综上，$\mathscr{A}+\mathscr{B}$ 是幂等变换的充要条件是 $\mathscr{A}\mathscr{B} = \mathscr{B}\mathscr{A} = \mathscr{O}$。
提示：注意充要条件需要证明两个方向。
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