I've read a couple of times about an alternative way of expressing the trace norm of a complex matrix $A$, mapping from some finite Hilbert space $\mathcal{H}_1$ to some other finite Hilbert space $\mathcal{H}_2$ (for simplicity, I set $m:=|\mathcal{H}_2|$ and $n:= \mathcal{H}_1$). The alternative expression reads as follows (second equality sign), $||A||_1 := Tr[\sqrt{A^*A}] = \max_{W:\mathcal{H}_2 \rightarrow \mathcal{H}_1}(Tr[WA])$, where we take the maximum over all partial isometries from $W$, mapping from $\mathcal{H}_2$ to $\mathcal{H}_1$.
I have never seen any proof of this statement, so I tried to prove it on my own, but I was only partially successful. Here's my attempt:
Assuming that $m=n$, I was able to show $||A||_1 \leq\max_{W:\mathcal{H}_2 \rightarrow \mathcal{H}_1}(Tr[WA])$ by using the singular value decomposition of $A$.
For $A$ the singular value decomposition reads $A = U \Sigma V^*$, where $U$ and $W$ are unitaries. Then we find that $\sqrt{A^*A} = \sqrt{(U\Sigma V)^*(U\Sigma V)} = \sqrt{V \Sigma^{\top}\Sigma V^*} = V\Sigma V^*$.
Hence, we deduce $||A||_1 = Tr[\sqrt{A^*A}] = Tr[V \Sigma V^*] = Tr[VU^*U\Sigma V^*] = Tr[W A] \leq \max_{W:\mathcal{H}_2 \rightarrow > \mathcal{H}_1}(Tr[WA])$ , where we set $W := VU^*$.
Unfortunately, I have serious problems when I try to prove the general case, where $m \neq n$, similiarly, as, for example, $\sqrt{V \Sigma^{\top}\Sigma V^*} = V \Sigma V^*$ doesn't make sense any more because the dimensions don't fit. The same applies to several equalities the second line of argumentation.
I wondered if I just can "extend" the dimensions of the matrices such that the matrix multiplications are defined by adding zeros (so "enforcing" square matrices), but I am not sure whether this changes the singular value decomposition. Furthermore, the statement asks explicitly for partial isometries from $\mathcal{H}_2$ to $\mathcal{H}_1$, so I would manipulate the statement.
Does anyone have an idea how I can prove this statement (so this direction, and/or the reverse one, where I haven't found a promising approach)?