Example of hyponormal operator which is not normal

Question

Let $F$ be a complex Hilbert space.

We recall that

• An operator $T\in \mathcal{B}(F)$ is normal if $A^*A=AA^*$ or equivalently $$\|Tx\|=\|T^*x\|,\;\forall\,x\in F.$$
• An operator $T\in \mathcal{B}(F)$ is hyponormal if $A^*A\geq AA^*$ (i.e. $\langle (A^*A-AA^*)x\;,\;x\rangle \geq 0$ for all $x\in F$) or equivalently $$\|Tx\|\geq\|T^*x\|,\;\forall\,x\in F.$$

Clearly every normal operator is hyponormal.

I want to find an example of an operator $A$ which is hyponormal but not normal.

Answer 1

A hyponormal operator on a finite-dimensional Hilbert space is necessarily normal.

Indeed, suppose that $A$ is hyponormal. Then $A^*A - AA^* \ge 0$ so $\sigma(A^*A - AA^*) \subseteq [0, +\infty\rangle$. But $\operatorname{Tr}(A^*A - AA^*) = 0$, which is the sum of eigevalues. Hence $\sigma(A^*A - AA^*) = \{0\}$ and because $A^*A - AA^*$ is normal, we have $A^*A - AA^* = 0$.

Therefore, an example of a hyponormal operator which is not normal must be on an infinite-dimensional Hilbert space.

Let $(e_n)_{n=1}^\infty$ be an orthonormal basis of a separable Hilbert space $H$ and let $(a_n)_{n=1}^\infty$ be a bounded sequence of scalars. Define a bounded linear map $T : H \to H$ with $Te_i = a_ie_{i+1}, \forall i \in \mathbb{N}$ or equivalently $$Tx = \sum_{n=1}^\infty \langle x, e_n\rangle a_ne_{n+1}$$

Show that the adjoint is given by $$T^*x = \sum_{n=1}^\infty \langle x, e_{n+1}\rangle \overline{a_n}e_{n}$$

Now show that $T$ is hyponormal if and only if the sequence $(|a_n|)_{n=1}^\infty$ is monotonically increasing.

On the other hand, $T$ is normal if and only if $a_n = 0, \forall n \in \mathbb{N}$, that is $T = 0$.

Answer 2

Over finite dimensional Hilbert spaces, every hypernormal operator is normal.

In particular, suppose that $A$ fails to be normal. Let $x_1,\dots,x_k$ be a list of all vectors satisfying $$ (A^*A)x_k = \lambda x_k = (AA^*)x_k $$ Let $\lambda$ be the largest eigenvalue of $A^*A$ that does not appear above. Because both $A^*A$ and $AA^*$ are Hermitian with the same eigenvalues and $A^*A \neq AA^*$, these vectors do not span $\Bbb C^n$. Let $S$ denote their span, and consider the restrictions $A^*A|_{S^\perp}$ and $AA^*|_{S^\perp}$.

Let $x \in S^\perp$ be the vector satisfying $$ x^*(A^*A)x = \max_{x \in S^\perp, \|x\| = 1} x^*(A^*A)x $$ By the Rayleigh-Ritz theorem, $x^*(A^*A)x = \lambda$. We also have $x^*(AA^*)x < \lambda$ (since otherwise we would have $x \in S$). Thus, $$ x^*[A^*A - AA^*]x > 0 $$ By a symmetrical argument, there exists a vector $y$ such that $$ y^*[A^*A - AA^*]y < 0 $$ Thus, $A^*A - AA^*$ fails to be positive or negative semidefinite.