Linear algebra in Hilbert space Let $M,N$ be closed subspaces of a separable Hilbert space.
How to prove rigorously the following:
$\operatorname{dim} M >\operatorname{dim} N => \exists u\neq0 \in M, u\in N^\perp$
Intuitively it seems clear since $\operatorname{dim} M >\operatorname{dim} N$ would imply that there is a vector in $M$ which is linearly independent from a given basis of $N$. So by the decomposition $H=N \oplus N^\perp$ that vector would be in $N^\perp$. But I can't manage to write it down in a rigorous way.
I'd appreciate some direction to get started, not necessarily a complete solution.
Thanks for the help.

*

*We can assume that $M$ is finite dimensional.

*Let $\pi\colon M\to N$ be the orthogonal projection on $N$. If the conclusion is not satisfied, then $\pi$ is injective, hence $\dim M\leqslant \dim N$.
