Let $A\in M_n(\Bbb{C})$, show that there exists unique partial isometry $V$ such that $A=VP$ for some positive $P$ and rank$(A)=$rank$(V)$ First I have shown the existence. Define $P:=\sqrt{A^*A}$. Then it's easy to verify that $\lVert Pv\rVert=\lVert Av\rVert$ for all $v\in\Bbb{C}^n$.
We have $\Bbb{C}^n=Ran(P)\oplus Ran(P)^{\perp}=Ran(P)\oplus Ker(P)$ (as $P^*=P$), define $V$ as follows $$V(x+y)=Tx\ \text{where } x\in Ran(P),y\in Ker(P)$$
$V$ is well defined follows from the fact $\lVert Pv\rVert=\lVert Av\rVert$ for all $v\in\Bbb{C}^n$. It can be shown that $Ker(V)=Ker(P)$ and $V$ is partial isometry.
Now, $Ker(P)=Ker(\sqrt{A^*A})=Ker(A^*A)=Ker(A)$. This proves the existence.
But I am stuck with the uniqueness of $V$ part. If we have $A=WQ$ where $W$ is partial isometry and $Q$ is positive such that rank$(A)=$rank$(W)$ then we need to prove $W=V$. As we have $Ran(A)\subseteq Ran(W)$ and their dimensions are given to be same then $Ran(A)=Ran(W)$.
I can prove the uniqueness if $Ker(Q)=Ker(W)$, which is exactly the statement of Polar Decomposition Theorem. But how to prove the existance of partial isometry $V$ in this case rank$(A)=$rank$(W)$.
Can anyone give me a wayout? Thanks for your help in advance.
 A: The answer is no.
Consider $\mathbb{C}^2.$ For fixed unit vectors $u,v$ let $Wx=\langle x,u\rangle v.$ Then $W$ is a partial isometry. For an arbitrary nonnegative operator $Q,$ such that $Qu\neq 0,$ let $A=WQ$. Hence
$$Ax=WQx=\langle Qx,u\rangle v=\langle x, Qu\rangle v$$
Now we determine the standard polar decomposition of $A,$ i.e. $A=V|A|,$ where $V$ is the partial isometry from ${\rm Im}\,|A|$ onto ${\rm Im}\,A.$ We have
$$A^*x=\langle x,v\rangle Qu$$
Hence $$A^*Ax=\langle x, Qu\rangle Qu,\quad |A|x={1\over \|Qu\|}\langle x,Qu\rangle Qu$$
The partial isometry $V$ satisfies
$V|A|x=Ax .$ For $x=Qu$ we get $|A|Qu=\|Qu\|\,Qu$ and $AQu=\|Qu\|^2\,v.$ Hence
$$ VQu=\|Qu\|\,v $$ On the other hand
$$WQu= \langle Qu,u\rangle v$$
Thus $W\neq V$ provided that $$\langle Qu,u\rangle \neq \|Qu\|$$
Now we are ready to construct an adequate example.
Let $$u=v=\begin{pmatrix} {\sqrt{3}\over 2}\\ {1\over 2}\end{pmatrix},\qquad Q=\begin{pmatrix}
1 & 0\\
0 & 0
\end{pmatrix}$$ Then $$W=\begin{pmatrix} {3\over 4} &{\sqrt{3}\over 4}\\
{\sqrt{3}\over 4} &{1\over 4}
\end{pmatrix}\qquad
A=\begin{pmatrix} {3\over 4} & 0\\
{\sqrt{3}\over 4}& 0
\end{pmatrix}$$
Moreover
$$V=\begin{pmatrix}
{\sqrt{3}\over 2} & 0\\
{1\over 2} & 0
\end{pmatrix}\qquad |A|=\begin{pmatrix}
{\sqrt{3}\over 2} & 0\\
0 & 0
\end{pmatrix}$$
All the operators showing up in the example have rank $1.$
Remark
If we assume that $Q=|A|$ then $W$ is uniquely determined.
