Polar decomposition with a unitary The following is a problem from V. S. Sunder. Assume we know   the regular polar decomposition of Hilbert space operators. Show that an operator $T \in L(H,K)$ can be expressed in the form $T = WA$ where $A$ is a positive operator and  $W\in L(H,K)$ is unitary if and only if $$\dim\ker T = \dim\ker T^{\star}.$$ 
 A: So you have that $T=UA$ with $A\in L(H)_+$, and $U\in L(H,K)$ is a partial isometry with $\ker T=\ker U=\ker A$. 
The operator $U^*U\in L(H)$ is a projection with 
$$
\ker U^*U=\ker T=\ker A.
$$
So $\operatorname{ran}U^*U=(\ker U^*U)^\perp=(\ker A)^\perp=\operatorname{ran}A$. This implies that $U^*UA=A$. Thus 
$$
T^*T=A^*U^*UA=A^*A=A^2 
$$
(which Sunder mentions in the hint, but I don't think is necessary)
Also
$$\tag1
\ker U^*=(\operatorname{ran}U)^\perp=(\operatorname{ran}T)^\perp=\ker T^*. 
$$
Suppose that $\dim \ker T=\dim\ker T^*$. By mapping an orthonormal basis to another, we can construct a partial isometry $V:\ker T\to\ker T^*$. Define $$W=U+V.$$ By $(1)$ we have that $U^*V=0$. Then 
$$
W^*W=U^*U+V^*V+2\operatorname{Re} U^*V=U^*U+V^*V=I_H
$$
The equality with the identity is due to $U^*U$ being the projection onto $\operatorname{ran}A$, and $V^*V$ being the projection onto $\ker T=\ker A=(\operatorname{ran}A)^\perp$. Similarly, since $\operatorname{ran} U^*=(\ker U)^\perp=(\ker T)^\perp$, we have $VU^*=0$. Then
$$
WW^*=UU^*+VV^*=I_K.
$$
The last equality now holds because $UU^*$ is the projection onto $$\operatorname{ran}UU^*=(\ker UU^*)^\perp=(\ker U^*)^\perp=(\ker T^*)^\perp,$$ and $VV^*$ is the projection onto $\ker T^*$. 
Finally, we have $VA=VU^*UA=0$, so $WA=(U+V)A=UV$. 

For the converse, if $T=WA$ with $W$ a unitary, as $W$ maps $\operatorname{ran} A$ onto $\operatorname{ran} T$, it also maps $(\operatorname{ran} A)^\perp=\ker T$ onto $(\operatorname{ran} T)^\perp=\ker T^*$. 
