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Suppose that $W_1$ and $W_2$ are both four-dimensional subspaces of a vector space $V$ of dimension seven. Explain why $W_1 \cap W_2 \neq \{0\}$.

Suppose $W_1\cap W_2 = \{0\}$, since $\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1\cap W_2)$ and $\dim(W_1)=\dim(W_2)=4$, $\dim(V)=7$,

$$\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(\{0\})=4+4-0= 8 > \dim(V)=7$$ which does not make sense since both $W_1$, $W_2$ are subspaces of $V$. Therefore, $W_1 \cap W_2 \neq \{0\}$

This is how I solved it. Is it right?

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    Can you please add the work you've done on this problem? Otherwise you give the impression that you expect the community to do it for you. – 2017-02-10
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    Looks perfectly fine to me. – 2017-02-10

2 Answers 2

1

Your proof is fine, but there's no need to use contradiction.

By Grassmann’s formula $$ 7=\dim V\ge\dim(W_1+W_2)=\dim W_1+\dim W_2-\dim(W_1\cap W_2)= 4+4-\dim(W_1\cap W_2) $$ Therefore $$ \dim(W_1\cap W_2)\ge 4+4-7=1 $$

2

Towards a contradiction, if $W_1\cap W_2=\{0\}$, then the subspace $W_1+W_2$ is in fact a direct sum. What is its dimension?