I've come across the following while solving a specific problem. To conclude that original problem I need to (dis)prove the following:
Suppose $\mathcal{H}$ is a complex Hilbert space and $T \in \mathcal{B}(\mathcal{H})$. Then is it true that $$\text{rank}((T^*T)^{\frac{1}{2}})=\text{rank}(T)$$
I haven't done any satisfactory progress here. Any help is appreciated. Thank you.
Edit. Let me mention how did I come to this problem:
Initially wanted to find all $T \in \mathcal{B}(\mathcal{H})$ such that $T$ is Hilbert-Schmidt operator and $||T||_{op}=||T||_{2}$, where $||T||_{2}$ denotes the Hilbert-Schmidt norm of $T$.
The 'rank one' operators are such operators. But I wonder whether these are the only such operators.
So I assumed $T\in \mathcal{B}(\mathcal{H})$ such that $||T||_{op}=||T||_{2}<\infty$ and then my claim is rank$(T)=1$.
I've managed to prove this under the additional assumption that '$T$ is self-adjoint': i.e., if $T^*=T$ and $||T||_{op}=||T||_{2}<\infty$ then rank$(T)=1$.
For the general case, I considered $A:=(T^*T)^{1/2}$. Then by the assumption on $T$ (i.e., $||T||_{op}=||T||_{2}<\infty$) have $$||A||_{op}=||T||_{op}=||T||_2=||A||_2$$. Then I concluded that $\text{rank}((T^*T)^{1/2})=\text{rank}(A)=1$ (since $A$ is self-adjoint and satisfies $||A||_{op}=||A||_2<\infty$). But does this imply $\text{rank}(T)=1$?
It's true. Let $\operatorname{Dom}(T) = \operatorname{Ker}(T)^\perp$. Then any bounded operator has a polar decomposition $T=U|T|$ where $|T|=\sqrt{T^* T}$ is positive semi-definite and $\operatorname{Dom}(|T|)=\operatorname{Dom}(T)$. $U$ is a partial isometry from $\operatorname{Dom}(T)$ to $\operatorname{closure}(\operatorname{Range}(T))$.
Edit: The part you need is simple to show. For any vector $x$ $$||Tx||^2 = \langle Tx,Tx\rangle=\langle T^*Tx,x\rangle=\langle |T|^2x,x\rangle=\langle |T|x,|T|x \rangle=|| |T|x||^2$$ which proves that $\operatorname{Ker}(T)=\operatorname{Ker}(|T|)$ so $T$ and $|T|$ have the same domain. You can also see from this identity that if a $U$ exists that maps $|T|x$ to $Tx$, it has to be an isometry because they have the same norm. The rank you are taking about is the dimension of the domain or the dimension of the closure of the range. They are the same dimension, even if infinite.