A question about trace class operatos and partial isomertry

Question

I'm a beginner in operator theory and I'm trying to understand just one step in a proof,which claims that if $H$ is a Hilbert space, $A\in B_1(H)$(that is , $A$ is a trace-class operator), and $U$ is a partial isometry such that $UA=|A|$, then $A=U^*UA=U^*|A|$. I have tried to prove the equation "$A=U^*UA$" but all my attempts have failed.

For example, I can prove that $U^*U$ is the projection onto the initial sapce. And I have tried to set $z:=U^*UA-A$ and attempted to prove that $z^*z=0$. But I still can't prove the equation $A=U^*UA$.

Can anyone help? And sorry for the language barrier.

Answer 1

You're done if you can show that the range of $A$ is in the initial space of $U$. Based on the first condition in the proof here, it's sufficient to show for any $x, y$ that $\langle UAx, UAy\rangle = \langle Ax, Ay \rangle$.

We have: \begin{align*} \langle UAx, UAy \rangle &= \langle |A|x, |A|y \rangle \\ &= \langle |A|^2x, y \rangle \\ &= \langle A^* A x, y \rangle \\ &= \langle Ax, Ay \rangle \\ \end{align*}

So the initial space of $U$ contains the range of $A$. You're right that $P = U^*U$ is the projection onto $U$'s initial space, but it's critical that this space necessarily includes the range of $A$, which is why $PA = A$.

You can in general take $U$'s initial space to be exactly the (closure of the) range of $A$.

Answer 2

If $P$ is any (self-adjoint) projection then $\langle Px, x\rangle = \langle P^2 x, x\rangle = \langle P^* P x, x\rangle = \langle Px, Px \rangle = \|Px\|^2$.

$U^* U$ is a projection on the initial space and $(I-U^*U)$ a projection on the orthogonal complement of the initial space. So $\|(I-U^* U)Ax\|^2 = \langle (I-U^*U)Ax, Ax \rangle = \langle A^*(I-U^*U)Ax, x\rangle$.

But $$\begin{align}A^*(I-U^*U)A &= A^*A - A^* U^* UA = A^* A - (UA)^* UA \\&= A^* A - |A|^* |A|= A^* A - |A|^2 = 0\end{align}$$ (since $UA = |A|$ by assumption.)

Combining we get $\|(I-U^* U)Ax\|^2 = 0$ for all $x$. So $0=(I-U^*U)A = A - U^*U A\implies A = U^*UA$.

Answer 3

If $U$ is a partial isometry, then by definition the operator $U$ is an isometry on $(\ker U)^\perp={\rm Im}\,U^*$ and $U^*U$ is the projection onto ${\rm Im}\,U^*.$

The condition $UA=|A|$ and the fact that $\|Ax\|=\||A|x\|$ imply that $U$ is an isometry on ${\rm Im}\, A.$ Therefore ${\rm Im}\, A\subset {\rm Im}\,U^*.$ Thus $U^*U(Ax)=Ax.$ Hence $U^*|A|=A.$