Suppose $V$ and $W$ be two partial isometries in a von Neumann algebra $\mathscr{R}$ such that $V^*V = E_1;~VV^*=F_1$ and $W^*W=E_2; ~ WW^*=F_2$. I am thinking about
When is $V + W$ a partial isometry?
My guess (based on $M_n(\Bbb{C})$) is that their initial and final spaces has to be orthogonal, i.e., $E_1E_2=0=F_1F_2$....! So I thought to claim:
$E_1E_2=0=F_1F_2\iff V+W$ is a partial isometry $~~~~~~~~~~~~~~~~~~~~$ (1)
Proof($\implies$) Assume $E_1E_2=0=F_1F_2$ and $U:=V + W$. Then $$U^*U=(V+W)^*(V+W)=E_1+V^*W + W^*V+E_2=E_1 + 2 \text{Re}(W^*V) + E_2=E_1 +2 \text{Re}(W^*F_2F_1V)+E_2=E_1+E_2$$ which is a projection and hence $U$ is a partial isometry.
But I cannot prove the other direction. Any help on this? Thanks.
Here is a partial answer, adressing only the converse you try to prove (and not the actual question):
You can not prove the other direction because it is false. In fact, there are very easy counterexamples: Let $V$ be a non-zero partial isometry. Then $$(-V)(-V)^* (-V) = -VV^*V = -V$$ so $-V$ is also a partial isometry. With $W= -V$, $$E_1 = V^*V = (-V)^*(-V) = E_2, \quad F_1 = VV^* = (-V)(-V)^* = F_2.$$ We then have $F_1F_2 = VV^*VV^* = VV^* = F_1$ which is the projection onto the closure of the image of $V$, which is non-zero because $V$ is chosen to be non-zero. Hence, $F_1F_2 =0$ is not true, so your claim is false.