This question is Exercise 2.10 in Rordam's K-Theory for C$^{*}$-Algebras book. Let $p,q$ be projections in an infinite-dimensional separable Hilbert space.
Show that $p$ and $q$ are unitarily equivalent (i.e., $\exists$ unitary $u\in H$ with $q=upu^{*}$) if and only if $\dim(p(H))=\dim(q(H))$ and $\dim(p(H)^{\perp})=\dim(q(H)^{\perp})$.
This question was asked once before here, but no answer was given.
I have solved the first part of the problem, which asks to show that $p$ and $q$ are equivalent in the Murray von-Neumann sense iff $\dim(p(H))=\dim(q(H))$. Since in a C$^{*}$-algebra unitary equivalence implies Murray von-Neumann equivalence, the forward direction reduces to showing $\dim(p(H)^{\perp})=\dim(q(H)^{\perp})$. But I am stuck in excluding the case where $\dim(p(H))$ and $\dim(q(H))$ are co-finite with $\dim(H)\setminus\dim(p(H))\not=\dim(H)\setminus\dim(q(H))$ I have thought about this question for hours, but I am at a loss as to how to proceed. I would really appreciate any help with completing the forward implication and with obtaining the reverse implication.
Thank you.
Let $\{e_i\mid i\in I_1\}$ be a Hilbert basis of $p(H)$ and $\{b_i\mid i\in I_1\}$ be a Hilbert basis of $q(H)$, note that the index set is the same, this follows from $\mathrm{dim}\,p(H)=\mathrm{dim}\,q(H)$.
Extend these to Hilbert basises $\{e_i\mid i\in I_2\}$ and $\{b_i\mid i\in I_2\}$. Again the index sets are the same, which follows from $\mathrm{dim}\,(1-q)(H) = \mathrm{dim}\,q(H)^\perp=\mathrm{dim}\,p(H)^\perp=\mathrm{dim}\,(1-p)(H)$.
The map $$U:H\to H,\quad \sum_{i\in I_2} x_i e_i\mapsto \sum_{i\in I_2} x_i b_i$$ is unitary. Note that: $$Up\left(\sum_{i\in I_2} x_i e_i\right) = U\left(\sum_{i\in I_1} x_i e_i\right)=\sum_{i\in I_1}x_ib_i = q\left(\sum_{i\in I_2}x_i b_i\right)=qU\left(\sum_{i\in I_2}x_i e_i\right) $$
so $Up=qU$.