Does a bounded linear operator on a Hilbert space conjugate to its adjoint?

Question

Let $H$ be a Hilbert space over $\mathbb R$ or $\mathbb C$ and $f\in B(H)$. I wonder if there is a $g\in B(H)$ such that $fg=gf^*$, where $f^*$ is the adjoint of $f$.

We know a matrix over a field is always conjugate to its transpose. Is there a generalization of this fact to Hilbert spaces?

Answer 1

Presumably you mean $g$ to be invertible, otherwise the statement is trivially true (take $g=0$).

The answer is no. For example, if $H$ is infinite-dimensional take $f \in B(H)$ to be an isomorphism from $H$ to an infinite-dimensional closed subspace of $H$. $f$ is one-to-one, but $f^*$ is not, so $f$ and $f^*$ are not conjugate.

For an explicit example of this, on $\ell^2$ take $f((x_1,x_2,x_3,\ldots)) = (x_1, 0, x_2, 0, x_3, \ldots)$.

Answer 2

A matrix is conjugate to its transpose, not its adjoint (in general).

For example:

$$A = \begin{bmatrix} i & 0 \\ 0 & 1\end{bmatrix}, \quad A^* = \begin{bmatrix} -i & 0 \\ 0 & 1\end{bmatrix}$$ are clearly not conjugated since $\sigma(A) = \{i,1\}$ and $\sigma(A^*) = \{-i,1\}$.