Dimension of $U\cap V$ using informations about other dimensions

Question

$X$ is linear space, $U$ and $V$ are subspaces of $X$, such that $\dim(U)=\dim⁡(V)=5$. Then $\dim(X)=8\Rightarrow \dim(U\cap V)\ge 2$. Is true implication ? For me it is true, but I am not sure about proof. Can you help me?

Answer 1

Recall that $\dim (U+V) = \dim(U) + \dim(V) - \dim(U \cap V)$ and that
$\dim (U+V)\le \dim(X)$.

Plug in the values you have and the result follows.