# If $T\in B(H)$ and let $U,V\in B(H)$ be two isomtries such that $TT^*=UU^*$ and $T^*T=VV^*$. Is it true that $T=USV^*$ for some unitary S?

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.

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.