Are self-inverse operators normal?

Question

Let $\mathcal{H}$ be an Hilbert space. Consider a bounded Operator $T:\mathcal{H}\to \mathcal{H}$. Suppose $TT=1$, does it hold, that $T^{*}T=TT^{*}$? If so, how does one show this? If not, what kind of counterexamples are there?

Answer 1

If you want to see whether a property like this holds for operators on a Hilbert space, it is often helpful to first check whether it holds for the special case of $2 \times 2$ matrices.

Multiplying $$ \left( \begin{array}{cc} a & b\\ c & d \end{array} \right) $$ by itself and setting it equal to the identity matrix imposes very strong conditions on the entries of the matrix. With this in mind, it is now straightforward to find a counterexample.

For example, consider $$ T= \left( \begin{array}{cc} 2 & \frac{3}{2} \\ -2 & -2 \end{array} \right). $$

This satisfies $TT=1$ but is not normal.

Answer 2

Thanks, Tom Cooney!

I also came up with: $$T = \left(\begin{array}[mc]{ccc} 1 &0 &0\\ 2 &1 &-2\\ 2 &0 &-1\\ \end{array}\right)$$

I somehow did not believe, it could work in dim = 2.

Unfortunate, that this be the case! Interestingly, though perhaps not surprisingly, since the spectral-theory result $Max_{\lambda\in\sigma(S)}|\lambda|=\|S\|$ is proven under the condition, that $S$ is normal, holds for this Matrix: $\|T\|=3,732\ldots > 1$.