2
$\begingroup$

Let $T$ be a bounded operator on a Hilbert space $H$ and $U,V$ be two isometries such that $T=USV^\ast$ for some unitary $S$. Then $TT^\ast=USV^\ast VS^\ast U^\ast=UU^\ast$ and $T^\ast T=VS^\ast U^\ast USV^\ast=VV^\ast$.

But on converse if $TT^\ast=UU^\ast$ and $T^\ast T=VV^\ast$. Does that imply $T=USV^\ast$ for some unitary $S$?

For $h\in H$,

$\langle (T-USV^\ast)h,(T-USV^\ast)h\rangle$

$=\lVert Th\rVert^2-\langle Th, USV^\ast h\rangle-\langle USV^\ast h, Th\rangle+\langle USV^\ast h,USV^\ast h\rangle$

$=\lVert Th\rVert^2-\langle Th, USV^\ast h\rangle-\langle USV^\ast h, Th\rangle+\langle VV^\ast h, h\rangle$ (as $U, S$ are isometry)

$=\lVert Th\rVert^2-\langle Th, USV^\ast h\rangle-\langle USV^\ast h, Th\rangle+\langle T^\ast T h, h\rangle$

$=2\lVert Th\rVert^2-\langle Th, USV^\ast h\rangle-\langle USV^\ast h, Th\rangle$ (as $T^\ast T=VV^\ast$)

I need to find the unitary $S$ such that the above expression becomes $0$. I think that I need to use polar decomposition somewhere and I am yet to use $TT^\ast =UU^\ast$.

Can anyone help me to find a wayout? Thanks for your help in advance.

$\endgroup$

1 Answer 1

3
$\begingroup$

The answer seems to be yes.

Since $U$ and $V$ are isometries, we have $U^*U = V^*V = I$.

Consider $S = U^*TV$ and note that $S$ is unitary: $$S^*S = V^*T^*(UU^*)TV = V^*(T^*T)^2V = V^*(VV^*)^2V = (V^*V)^3 = I^3 = I,$$ $$SS^* = U^*T(VV^*)T^*U = U^*(TT^*)^2U = U^*(UU^*)^2U = (U^*U)^3 = I^3 = I.$$

Furthermore, we have $$USV^* = (UU^*) T (VV^*) = T(T^*T)VV^* = TV(V^*V)V^* = T(VV^*) = TT^*T = T.$$ To see that the last equality holds, note that $$(T^*T)^2 = V(V^*V)V^* = VV^* = T^*T$$ and therefore $$(T^*TT-T)^*(T^*TT-T) = (T^*T)^3 - 2(T^*T)^2 + T^*T = 0$$ which implies $TT^*T-T=0$ by the $C^*$-property.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .