Are self-inverse operators normal? 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?
 A: 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.
A: 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: $\|T\|=3,732\ldots > 1$.
