comp_1_数学类_决赛_8

【参考证明】：记 $E_{j}=\operatorname{Ker} f_{j}, j=1,2$ 。由 $f_{j} \neq 0$ 知 $\operatorname{dim} E_{j}=n-1, j=1,2$ 。不失一般性，可令 $$ \begin{aligned} & V=\mathbb{C}^{n}=\left\{\alpha=\left(x_{1}, \ldots, x_{n}\right): x_{1}, x_{2}, \ldots, x_{n} \in \mathbb{C}\right\} \\ & f_{j}(\alpha)=a_{j 1} x_{1}+a_{j 2} x_{2}+\ldots+a_{j n} x_{n}, j=1,2 \end{aligned} $$ 由 $f_{1} \neq 0, f_{2} \neq 0, f_{1} \neq c f_{2}, \forall c \in \mathbb{C}$ ，知 $$ \left\{\begin{array}{l} a_{11} x_{1}+a_{12} x_{2}+\ldots+a_{1 n} x_{n}=0 \\ a_{21} x_{1}+a_{22} x_{2}+\ldots+a_{2 n} x_{n}=0 \end{array}\right. $$ 的系数矩阵之秩为 2 。因此其解空间维数为 $n-2$ ，即 $\operatorname{dim}\left(E_{1} \cap E_{2}\right)=n-2$ 。但 $$ \operatorname{dim} E_{1}+\operatorname{dim} E_{2}=\operatorname{dim}\left(E_{1}+E_{2}\right)+\operatorname{dim}\left(E_{1} \cap E_{2}\right), $$ 故有 $\operatorname{dim}\left(E_{1}+E_{2}\right)=n$ ，即 $E_{1}+E_{2}=V$ 。 现在，任 意 的 $\alpha \in V$ 都 可 表为 $\alpha=\alpha_{1}+\alpha_{2}$ ，其 中 $\alpha_{1} \in E_{1}, \alpha_{2} \in E_{2}$ 。注 意 到 $f_{1}\left(\alpha_{1}\right)=0, f_{2}\left(\alpha_{2}\right)=0$ ，因此 $$ f_{1}(\alpha)=f_{1}\left(\alpha_{2}\right), f_{2}(\alpha)=f_{2}\left(\alpha_{1}\right) $$