Consider the following statement:
Let $H$ be an Hilbert Space and let $K$ be a closed, non-empty subspace of $H$ (so that it inherits the Hilbert Space structure). Let $\{\underline{v}_a\}_{a \in A}$ be an orthonormal basis of $H$ and let $\hat P_K$ be the orthogonal projection operator on $K$, then a maximal orthogonal set of vectors in $\{\hat P_K \underline{v}_a\}_{a \in A}$ is an orthogonal basis of $K$ (which may then be normalized).
I have thought about some geometrical examples in $\Bbb R^3$ first and then I moved on to $\Bbb R^n$ and in all the examples it seems to be possible. However I am not sure on how to proceed more generally. Any help would be much appreciated.
The reason why I am asking this question is to prove that the restriction of an observable in quantum mechanics to an invariant subspace admits an orthonormal basis of eigenvectors.
In $\Bbb{R}^3$ $$e_1=(1,0,0)\quad e_2=(0,1,0)\quad e_3=(0,0,1)$$ and the unit vector $$v=\frac{1}{\sqrt{3}}(1,1,1)$$ Let $u_1, u_2, u_3$ be the projections of $e_1, e_2, e_3$ on the plane perpendicular to $v$ $$\begin{align} u_1=e_1-(e_1\cdot v) v&=(\frac{2}{3}, -\frac{1}{3}, -\frac{1}{3})\\ u_2=e_2-(e_2\cdot v) v&=(-\frac{1}{3}, \frac{2}{3}, -\frac{1}{3})\\ u_3=e_3-(e_3\cdot v) v&=(-\frac{1}{3}, -\frac{1}{3}, \frac{2}{3}) \end{align}$$ and $$u_1\cdot u_2=u_2\cdot u_3=u_3\cdot u_1=-\frac{1}{3}$$ So a maximal orthogonal set among $u_1, u_2, u_3$ has just one vector. They span a space which is two dimensional.