Dimension of the intersection of $2$ subspaces: $\dim(W \cap X) \leq \min \{\dim(W), \dim(X)\}$ Let $V$ be an $n$-dimensional vector space, $W$ a subspace of $V$, and $X$ another subspace of $V$.
Prove that
$$
\dim(W \cap X) \leq \min \{\dim(W), \dim(X)\}.
$$
I'm not sure the best way to prove it. Any idea? Thanks!

$W \cap X \subseteq W$ and $W \cap X \subseteq X$. So, any basis for $W$ spans $W \cap X$. Hence, the dimension of $W \cap X$ is at most the dimension of $W$.
The same applies for $X$, and the result follows immediately.
