"Finite-Dimensional-Type-Spectral-Theorem" for Orthogonal Projections Let $H$ be a Hilbert space, not assumed separable, and $p$ and $q$ (bounded - not sure if that is important) orthogonal projections.

Question 1: Is it the case that $p$ has an orthonormal eigenbasis for $H$/is diagonalisable?
Question 2: Is it the case that $p$ and $q$ commute if and only if they share an eigenbasis/are simultaneously diagonalisable?

If the answer to question 1 is no, what are some reasonable assumptions on $H$ that guarantee such an eigenbasis exists (it seems to me that separability is enough. I don't think assumptions (such as compactness) on $p$ and $q$ are much good to me.).
If the answer to question 1 is no, do these assumptions give a positive answer to question 2?
Thanks for your help.

I am a little concerned when I work in infinite dimensions. For example I would just say, OK,
$$H=\operatorname{ran }p \oplus \ker{p},$$
each are closed and so each are Hilbert spaces and so each have onb... and just union those, bingo-bango, jobs a good 'un... but I am concerned there is an error there.

Yes, there's nothing tricky about this.  As you say, $H$ is the orthogonal direct sum of the image of $p$ and the kernel of $p$.  Both of these are closed (the image is closed since it is the kernel of $1-p$), and so can pick orthonormal bases for each of them and their union is an orthonormal basis for $H$ which diagonalizes $p$.
If $p$ and $q$ commute, then $pq$, $p(1-q)$, $(1-p)q$, and $(1-p)(1-q)$ are all orthogonal projections which are pairwise orthogonal and their sum is $1$.  So again, $H$ is the orthogonal direct sum of their ranges, and picking an orthonormal basis for each of their ranges, you get an orthonormal basis for $H$ which simultaneously diagonalizes $p$ and $q$.
