Let V be a vector space, U, W, Z subspaces. Given their dimensions, prove that there exists a vector v .v != 0 ,v ∈ U ∩ W ∩ Z

Question

As the title says, the question gives me: dim U = dim W = dim Z = 10 ,dim V = 14

I've tried using the dimension theorem, and figured out that each dimension of an intersection between two of the subspaces is blocked by 6 and 10. The previous questions in this exercise are directing me to use this in the answer, so if it's possible I would love a hint towards that direction. Else, any help would do. Thanks!

Answer 1

We know that $\dim(U+W)=\dim(U)+\dim(W)-\dim(U\cap W)$. Since $\dim(U)+\dim(W)=20$, but $\dim(U+W)\leq \dim(V)=14$, it follows that $\dim(U\cap W)$ is at least $6$ (and at most $10$, but that is immaterial here).

Now look at $H=U\cap W$. Again we have that $\dim(H+Z)=\dim(H)+\dim(Z)-\dim(H\cap Z)$. The left hand side is at most $14$ and at least $10$. And $\dim(H)+\dim(Z)$ is at least $16$. So $\dim(H\cap Z)$ is at least $2$.