Connection between isometries and projectors in QM I realize this question is technically a mathematical one but I think it is seen often enough in quantum information so I ask it here. The following is the definition of an isometry in Mark Wilde's book

Let $\mathcal{H}$ and $\mathcal{H}^{\prime}$ be Hilbert spaces such
  that $\operatorname{dim}(\mathcal{H}) \leq$
$\operatorname{dim}\left(\mathcal{H}^{\prime}\right)$ An isometry $V$
  is a linear map from $\mathcal{H}$ to $\mathcal{H}^{\prime}$ such that
  $V^{\dagger} V=I_{\mathcal{H}}$. Equivalently, an isometry $V$ is a
  linear, norm-preserving operator, in the sense that
  $\||\psi\rangle\left\|_{2}=\right\| V|\psi\rangle \|_{2}$ for all
  $|\psi\rangle \in \mathcal{H}$.

He also points out that $V V^{\dagger}=\Pi_{\mathcal{H}^{\prime}}$ which is a projection onto $\mathcal{H'}$. 
My questions are about $V^\dagger$. 


*

*By the definition, it is not an isometry but it is a linear map from $\mathcal{H'}$ to $\mathcal{H}$. Is $V^\dagger$ itself a projector from $\mathcal{H'}$ to a subspace of $\mathcal{H'}$ of dimension $\text{dim}(\mathcal{H})$ followed by a unitary from this subspace to $\mathcal{H}$? 

*Does every projector have a corresponding isometry? That is, suppose I am given a projector $\Pi_{\mathcal{H}}$ onto a subspace of $\mathcal{H}$ called $\mathcal{K}$. Then does every isometry $V$ from $\mathcal{K}$ to $\mathcal{H}$ satisfy $VV^\dagger = \Pi_{\mathcal{H}}$?
 A: A partial isometry is mapping a sub-vector space $K$ of a Hilbert space $H$ onto another sub-vector space $K'$ of the same dimension isometrically, that is
$$(V\psi, V\phi) = (\psi,\phi)$$
for any two vectors in the initial domain of the isometry, that is $K=V^*VH$. The fact that $V^*VH$ is the initial domain of $V$ can be proved by showing that $E=V^*V$ is precisely the projection onto $K$. Similarly, one can show that $F=VV^*$ is the projection onto $K'$, so that $K' = FH$. To get an idea of what a partial isometry is, observe that every unitary is a partial isometry, but not every partial isometry is a unitary because $V^*VH$ and $VV^*H$ are generally not the whole of $H$ (although they could be isomorphic to it). Indeed, when $K$ is all of $H$, one talks about isometries. An important example of an isometry is the adjoint of the shift operator $S$ on a separable infinite dimensional Hilbert space $H$ with ONB $\{e_0,e_1,\ldots\}$,
$$Se_0=0,\qquad Se_k=e_{k-1}.$$
Note how $S^*$ maps the whole of $H$ onto the orthogonal complement of $e_0$ isometrically.
It is also easy to prove the following identities that characterise partial isometries:
$$VV^*V = V\qquad V^*VV^* = V^*.$$
Finally, one sees that projections are (rather trivial) examples of partial isometries.
A: You can characterise isometries as those linear maps that can be written in the form
$$V = \sum_{k=1}^d |u_k'\rangle\!\langle u_k| \in \operatorname{Lin}(\mathcal H,\mathcal H'),$$
where $\{|u_k\rangle\}_k$ is an orthonormal basis for $\mathcal H$, $\{|u_k'\rangle\}_k$ is an orthonormal set in $\mathcal H'$ (but not a basis if $\operatorname{dim}(\mathcal H)<\operatorname{dim}(\mathcal H')$), and $d\equiv\operatorname{dim}(\mathcal H)$.
In this notation, $V^\dagger$ is obtained by simply switching $|u_k\rangle$ and $|u_k'\rangle$:
$$V^\dagger = \sum_{k=1}^d |u_k\rangle\!\langle u_k'| \in \operatorname{Lin}(\mathcal H',\mathcal H).$$
Now to address your questions:


*

*Indeed, $V^\dagger$ is not an isometry if $\operatorname{dim}\mathcal H<\operatorname{dim}\mathcal H'$. You can write it as
$$V^\dagger = \left( \sum_{j=1}^d |u_j\rangle\!\langle u_j'| \right) \left( \sum_{k=1}^d |u_k'\rangle\!\langle u_k'| \right)=V^\dagger \left( \sum_{k=1}^d |u_k'\rangle\!\langle u_k'| \right).$$
This amounts to simply multiplying to the right with a projector onto the support of $V^\dagger$, which you can always do freely.
This clearly is not a very insightful statement.
However, you could think of $V^\dagger$ as a unitary operation when restricting its domain to its support. In other words, $V^\dagger|_{\operatorname{supp}(V^\dagger)}$ is unitary. That's probably how close you can get to your statement.

*Let $W:\mathcal K\to\mathcal H$ be an isometry, with $d'\equiv \operatorname{dim}\mathcal K\le d$. Then we can write it as
$$ W = \sum_{k=1}^{d'} |v_k\rangle\!\langle v_k'|,$$
with $|v_k'\rangle$ orthonormal basis for $\mathcal K$ and $|v_k\rangle$ orthonormal set in $\mathcal H$. Then,
$$ W W^\dagger = \sum_{k=1}^{d'} |v_k\rangle\!\langle v_k|.$$
This is therefore a projector onto a subset of $\mathcal H$ of dimension $d'$ (though not necessarily a projector onto $\mathcal K$).
