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习题二23:
设$M_1,M_2,\cdots,M_k$为数域$K$上线性空间$V$的子空间。证明和$\sum\limits_{i=1}^kM_i$为直和的充分必要条件是
$$M_i \cap \left(\sum\limits_{j=1}^{i-1}M_j\right)=\left\{0\right\}(i=2,3,\cdots,k)$$
解:
必要性
由于$\sum\limits_{i=1}^kM_i$为直和,故
$$M_i \cap \left(\sum\limits_{j \ne i}M_j\right)=\left\{0\right\}(i=1,2,\cdots,k)$$
显然有
$$M_i \cap \left(\sum\limits_{j=1}^{i-1}M_j\right)=\left\{0\right\}(i=2,3,\cdots,k)$$
充分性
归纳假设对于$\sum\limits_{i=1}^{k-1}M_i$为直和
反证,若和$\sum\limits_{i=1}^kM_i$不是直和,则
$$M_i \cap \left(\sum\limits_{j \ne i}M_j\right) \ne \left\{0\right\}(i=1,2,\cdots,k)$$
故$\exists \alpha_i \in M_i$
$$\alpha_i=\sum\limits_{j \ne i}\alpha_j(i=1,2,\cdots,k)$$
于是
$$\alpha_k=\alpha_1+\alpha_2+\cdots+\alpha_{i-1}-\alpha_i+\alpha_{i+1}+\cdots+\alpha_{k-1}$$
但是
$$M_i \cap \left(\sum\limits_{j=1}^{i-1}M_j\right)=\left\{0\right\}(i=2,3,\cdots,k)$$
所以
$$\alpha_k=0$$
根据归纳假设
$$\alpha_i=0(i=1,2,\cdots,k)$$
那么
$$M_i \cap \left(\sum\limits_{j \ne i}M_j\right) = \left\{0\right\}(i=1,2,\cdots,k)$$
矛盾。 |
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