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Let $W$ be a 7-dimensional space. Prove that for any 5-dimensional subspaces $W_1,W_2,W_3 \subset W$ the intersection $W_1 \cap W_2 \cap W_3$ is non-zero.

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2 Answers

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Hint: $2+2+2<7$.

It might help to consider the complement.

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up vote 3 down vote

Proof by contradiction. Suppose not, then $W_1$ and $W_2 \cap W_3$ are disjoint. Take a basis of each. Because they are disjoint you can combine these two bases to get a linearly independent set in $W$. This proves that:

$$\dim W_1 + \dim(W_2 \cap W_3) \leq \dim W$$

hence

$$\dim(W_2 \cap W_3) \leq 2.$$

Is this possible if both $W_2$ and $W_3$ are $5$-dimensional? To see take a basis for $W_2 \cap W_3$ and then extend this to a basis for $W_2$ and a basis for $W_3$. These vectors are all linearly independent so there can't be more than $7$ of them. Count them up and see how many you actually get.

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Don't you mean there can't be more than 5 for them to still be linearly independent since they have dimension 5? I think I am still a little stuck. Let $\{u_1,u_2\}$ be a basis for $W_2 \cap W_3$. Now extend this basis for $W_2$ and $W_3$, $\{u_1,u_2, w_1, w_2, w_3\}$ and $\{u_1,u_2, x_1, x_2, x_3\}$ respectively. I could extend both of these even further to get a basis for $W$, but $W$ can have multiple bases, so I guess I don't see the contradiction yet. – user4593 Feb 7 at 3:14
If $\{u_1, u_2, w_1, w_2, w_3\}$ and $\{u_1, u_2, x_1, x_2, x_3\}$ are bases for $W_2$ and $W_3$ respectively then try and prove that $\{u_1, u_2, w_1, w_2, w_3, x_1, x_2, x_3\}$ is a linearly independent set. – Jim Feb 7 at 7:40
Well that's just because you can't have $n+1$ linearly independent vectors in a vector space of dimension $n$. How can we try to claim that $\{u_1, u_2, w_1, w_2, w_3, x_1, x_2, x_3\}$ is linearly independent? – user4593 Feb 7 at 20:20
Take a linear combination of the vectors and rewrite it as $a_1u_1 + a_2u_2 + b_1w_1 + b_2w_2 + b_3w_3 = c_1x_1 + c_2x_2 + c_3x_3$. We want to prove that all the $a_i, b_i, c_i$ are zero. Which subspaces contain the element $c_1x_1 + c_2x_2 + c_3x_3$? – Jim Feb 7 at 21:37

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