While studying differential geometry I have formulated a result in linear algebra which I haven't been able to prove:
Let $V$ be a real vector space with an inner product, and let $\{e_1,\dots,e_k\}\subset V$ be an orthonormal basis. Let $W\subset V$ be a subspace of dimension $n\geq 1$. There exist $i_1,\dots,i_n$ such that $P_S|_W:W\to S$ is an isomorphism, where $S=\operatorname{span}(e_{i_1},\dots,e_{i_n})$ and $P_S:V\to S$ is the orthogonal projection onto $S$.
Here's (what I think is) the most fruitful approach I've tried.
I have geometrically visualized that "almost always" any $(i_1,\dots,i_n)$ will do, except when some $e_i\in W$. So, I assert that we can pick $(i_1,\dots,i_n)$ such that $W$ is not perpendicular to $\operatorname{span}(e_{i_j})$ for all $j=1,\dots,n$.
In that case, I try to prove that $P_S|_W$ is surjective. Let $v=\sum_j \langle v,e_{i_j}\rangle e_{i_j}\in S$. Define $w_{i_j}=P_W(e_{i_j})$ and $w=\sum_j \langle v,e_{i_j}\rangle w_{i_j}$. A computation gives $$P_S(w)=\sum_k \langle v,e_{i_k}\rangle P_S P_W(e_{i_k})$$
If I could prove that $P_S P_W(e_{i_k})\in \operatorname{span}(e_{i_k})$ (which I geometrically believe, although I have a feeling it might fail in higher-than-imaginable dimensions), then I could, redefining $w_{i_j}$ by a scalar multiple, get that $P_SP_W(e_{i_k})=e_{i_k}$ thus obtaining $P_S(w)=v$ and finishing the proof. But I haven't been able to obtain a useful expression of $P_SP_W(e_{i_k})$.
I'd appreciate any proof, but I'd especially appreciate any comments on this approach (I've been at it for some hours now).
Best Answer
Induct on the dimension of $V$: The statement is clear when $n = 1$. First consider the projection $PW$ onto the span of $e_2,\ldots,e_n$. By induction, there are some $\{e_i\}_{i \in I}$ such that the projection $P'$ of $PW$ onto the span of the $e_i$ for $i \in I$ is an isomorphism. ($I \subseteq \{2,3,\ldots,n\}$). The kernel of $P$ when restricted to $W$ is either zero-dimensional or one-dimensional. In the first case we are done: $P'P$ is an isomorphism of $W$ onto the span of the $\{e_i\}$ for $i \in I$. In the second case, we must have $e_1 \in W$, and so we simply need to add $e_1$ to the list of vectors.