On polar decomposition

Question

In the paper "Isometries of non-commutative $L_p$-spaces" by Yeadon the author states the following: Let $H$ is separable Hilbert space, $\mathcal B(H)$ is $\ast-$algebra of all bounded linear operators on $H$. Then for operators $A,B\in\mathcal B(H)$ the condition $AB^*=A^*B=0$ is equivalent to the condition $W_AW_B^*=W_A^*W_B=0$, where $X = W_X |X|$ polar decomposition of an operator $X\in\mathcal B(H)$. Unfortunately the proof is not given I couldn't proof it myself. Does anybody know literature where the proof of the above state is given.
Thank's.

Answer 1

First note that $AB^\ast=A^\ast B=0$ is equivalent to $\operatorname{ran} B^\ast\subset \ker A$ and $\operatorname{ran} B\subset \ker A^\ast$. The kernel of a bounded linear operator is always closed and $\overline{\operatorname{ran}}X=(\ker X^\ast)^\perp$. Hence the two condition above are equivalent to $(\ker B)^\perp\subset \ker A$ and $\operatorname{ran} B\subset (\operatorname{ran} A)^\perp$.

The partial isometry in the polar decomposition has the property $\ker W_X=\ker X$ and $\operatorname{ran} W_X=\operatorname{ran}X$. Now you can read the proof from the first paragraph backwards to see that $(\ker W_B)^\perp\subset \ker W_A$ and $\operatorname{ran} W_B\subset (\operatorname{ran} W_A)^\perp$ is equivalent to $W_A W_B^\ast=W_A^\ast W_B=0$.