Let $W_1, W_2, W_3$ be $3$ distinct subspaces of $\Bbb{R}$$^{10}$ such that each $W_i$ has dimension 9.
Let $W = W_1 \cap W_2 \cap W_3$. Then which of the following can we conclude?
- $W$ may not be a subspace of $\Bbb{R}$$^{10}$
- $\dim W \le 8$
- $\dim W \ge 7$
- $\dim W \le 3$
I know that first and fourth options are incorrect. I have even find examples where dimension of W is $7$ and $8$.
But still from here, how can I conclude that the third option is correct. I know that dimension of $W$ should be less than or equal to $9$. It can't be $9$ since those subspaces are distinct. So second option will be correct. But how do I know that the dimension of $W$ Will be greater than or equal to $7$?
Best Answer
For any two subspaces $X$ and $Y$ of a vector space $V$ you have $$ \dim(X+Y) = \dim(X) + \dim(Y) - \dim(X\cap Y). $$ Starting with $W_1$ and $W_2$ this gives you $$ \dim(W_1\cap W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1+W_2) = 9 + 9 - 10 = 8, $$ where $\dim(W_1+W_2)$ is $10$ since the two $9$-dimensional subspaces of $\mathbb R^{10}$ are distinct.
Since $W_1\cap W_2\cap W_3$ is a subspace of $W_1\cap W_2$ we immediately conclude $$ \dim(W_1\cap W_2\cap W_3) \le 8. $$ Finally $$ \dim(W_1\cap W_2\cap W_3) = \underbrace{\dim(W_1\cap W_2)}_{{}=8} + \underbrace{\dim(W_3)}_{{}=9} - \dim((W_1\cap W_2) + W_3). $$ Since $\dim((W_1\cap W_2) + W_3)\le 10$ we can conclude $$ \dim(W_1\cap W_2\cap W_3) \ge 7. $$