Zero diagonal operators and Riesz bases 
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*We shall concern ourselves only with separable complex Hilbert spaces and bounded linear operators on them. An operator $A: \mathcal H\to \mathcal H$ is said to be a zero diagonal operator if $(Ae_n, e_n) = 0$ for some orthonormal basis $\{e_n\}$.


*A base $\{b_n\}$ is called a Riesz base if there is an invertible operator $T$, and an orthonormal basis $\{e_n\}$ such that $Te_n = b_n$ for all $n\ge 1$. Given a base $\mathcal B = \{b_n\}$ of $\mathcal H$, we can associate every operator $A: \mathcal H\to \mathcal H$ with a matrix $[A]_{\mathcal B}$ in a canonical way. Furthermore, we denote the base biorthogonal to $\mathcal B$ by $\mathcal B^* = \{b_n^*\}$, i.e., $(b_i, b_j^*) = \delta_{ij}$ for all $i,j\ge 1$. Then, the diagonal of $[A]_{\mathcal B}$ is $\{(Ab_n, b_n^*): n\ge 1\}$.
The first page of this paper says:

Thus, for an operator $A$, being similar to a zero diagonal operator can be described as having zeros everywhere in the diagonal of $[A]_{\mathcal B}$ (i.e., $(Ab_n, b_n^*) = 0$ for some Riesz base $\{b_n\}$).

In view of the statement above, I am trying to show that an operator $A$ is similar to a zero diagonal operator if and only if there exists a Riesz base $\{b_n\}$ such that $(Ab_n, b_n^*) = 0$ for all $n\ge 1$. I'd appreciate any help with the same.

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*Suppose $A$ is similar to a zero diagonal operator, i.e., $(S^{-1}AS e_n, e_n) = 0$ for some orthonormal basis $\{e_n\}$ and an invertible operator $S$. I must find a Riesz base $\{b_n\}$, and I tried to see if $\{Se_n\}$ or $\{S^{-1} e_n\}$ might work (they don't).


*Suppose there exists a Riesz base $\{b_n\}$ such that $(Ab_n, b_n^*) = 0$ for all $n\ge 1$. There is an invertible operator $T$ and an orthonormal basis $\{e_n\}$ such that $Te_n = b_n$. So, $(ATe_n, b_n^*) = 0$ for all $n\ge 1$.
Thank you!
 A: Both statements follow from the following facts:
$1.)$ Any $S: \mathcal{H}\to \mathcal{H}$ has an adjoint operator $S^*$ $\bigg($i.e $(Sx,y) = (x,S^*y)$ for all $x,y \in \mathcal{H} \bigg)$
$2.)$ If $\{e_n\}$ is an orthonormal basis, then the dual Riesz Basis of $Se_n$ is $(S^{-1})^*e_n$.
From here, all you need to see is that:
\begin{align}
(S^{-1}AS e_n , e_n) & = (ASe_n, (S^{-1})^*e_n) \\
& = (Ab_n, b_n^*)
\end{align}
The first fact is standard in intro Hilbert space theory.
The second fact is not too difficult to prove:
Suppose $e_n$ is an orthonormal basis and $S$ an operator. Then any $f \in \mathcal{H}$ can be expresses as
$$f = \sum_{n} (f,e_n)e_n $$ and this implies
\begin{align}
f = SS^{-1}(f) & = \sum_n (S^{-1}f,e_n)e_n \\
& = \sum_n(f,(S^{-1})^*e_n )S(e_n)
\end{align}
Letting $f = S(e_m) = b_m$  and $(S^{-1})^*(e_n) = b_n^*$, we obtain
$$b_m = \sum_n (b_m, b_n^*)b_n $$
By linear independence of $\{b_n \}$ (which follows from $S$ being an invertible operator), we must have $(b_n,b_n^*) = \delta_{mn}$.
