Why are the ranges of these projections isomorphic?

Question

For a separable, infinite dimensional Hilbert space $H$ and two projections $p,q\in B(H)$ (i.e. $p^2=p^*=p$) such that there exists an operator $v\in B(H)$ satisfying $p=v^*v$ and $q=vv^*$ I want to show that $\operatorname{im}(p)=pH$ and $\operatorname{im}(q)=qH$ are isomorphic Hilbert spaces.

I read that the isomorphism $pH\to qH$ is given by $v$, but I'm not sure about the details.

Presumably if $x\in pH$ (and $qH$ respectively) it is $px=x$ and $v$ looks like $v:pH\to qH\; x\mapsto x$, or, more precisely $v(px)=qx$. Why is this map surjective? (the other properties are clear)

Answer 1

The first thing to check is that $v$ actually maps $pH$ into $qH$: let $px\in pH$, then

$$v(px) = vv^*vx=q(vx)\in qH.$$

Now let us check surjectivity:

Let $qx\in qH$. We have $qx=q^2 x = vv^*vv^*x = v(pv^*x)$ and $p(v^*x)\in pH$.

For the sake of completeness let us also check injectivity:

Let $v(px)=0$. Then $0=\langle v(px),vx\rangle=\langle px, v^* vx\rangle = \|px\|^2$, so $px=0$.

Answer 2

Generality In addition this is, indeed, a general fact. Take $v\,:\,V\rightarrow W$ and $w\,:\,W\rightarrow V$ such that $p=wv$ and $q=vw$ are projectors (resp. of $V$ and $W$) then $pV$ and $qW$ are isomorphic as it can be checked easily that $$ v\,:\, pV\rightarrow qW\ ;\ w\,:\, qW\rightarrow pV\ $$ are reciprocal isomorphisms.

Case of Hilbert spaces This shows that, for Hilbert spaces, you do not need the hypothesis of separability. Only with $v\in B(H)$, you get $v^*,v^*v,vv^*\in B(H)$ so your projectors have closed images (the kernels of their complement) and hence $pH,qH$ are Hilbert spaces.