Spectrum and self-adjointness

Question

For bounded operator on a Hilbert space, if its spectrum is a subset of $\mathbb{R}$, then is this operator self adjoint? If not, what is a counterexample?

Answer 1

There are plenty of counterexamples. For example, consider any nilpotent operator $T: H \to H$ that is not self-adjoint. Then $T$ has spectrum $\{0\}$.

For example, consider $$M:=\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$$ Then $M^2 = 0$ but $M$ is not self-adjoint. If you really want this as an operator, it is $$M: \mathbb{C}^2 \to \mathbb{C}^2: \begin{pmatrix}x \\ y\end{pmatrix}\mapsto \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}= \begin{pmatrix}y \\ 0\end{pmatrix}$$