I have to prove the following statement:
Let $H$ be a Hilbertspace and $M,N$ closed subspaces. Then the following holds:
If $M \sim N $ and $N$ is finite, then $M$ is finite.
I think it should say finite dimensional, since a finite subspace should only be the trivial subspace.
Here $M\sim N$ means, that there is a partial isometry $U$ such that $P=U^* U$ und $Q=U U^*$ with $M=P(H)$ and $N=Q(H)$.
The statement seems very obvious, but I don't know how to prove ist. Probably I have to use some relations betwenn kernels and images of bounded operators but I don't know where to start.
Take a basis $\{v_i\}$ of $N$. In particular $v_i=Qv_i=UU^* v_i$.
We can show that $U^*v_i$ is a basis of $M$.
In fact, $U^*v_i=U^*Qv_i=U^*UU^*v_i=PU^*v_i$, so $U^*v_i\in M$.
If $0=\sum a_iU^*v_i$. Then $0=\sum a_iUU^*v_i=\sum a_iv_i$. Therefore $a_i=0$, i.e. $U^*v_i$ are linearly independent.
If $x\in M$, then $x=Px=U^*Ux$, then $Ux=UU^*Ux=QUx$. This means that $Ux\in N$. Therefore there are $a_i$ such that $Ux=\sum a_iv_i$. Therefore $x=U^*Ux=\sum a_iU^*v_i$. Hence $U^*v_i$ generates $M$.
To finish notice that the bases $v_i$ and $U^*v_i$ of $N$ and $M$ have the same number of elements.