When a normal operator is also a self adjoint operator?

Question

Let $T$ be a normal operator on a complex inner product space. Then $T$ is a self adjoint operator if and only if

1) $T$ has distinct eigen values

2) $T$ has repeated eigen values

3) All eigen values of $T$ are real

4) $T$ has atleast one real eigen value

If $T$ is a self adjoint operator then all it's eigen values are real.

This rules out 1, 2 and 4.

But how to prove the converse? i.e. If all the eigen values of a normal operator $T$ are real then it is self adjoint.

Answer 1

Since $T$ is a normal operator on a complex inner product space $V$, the Spectral theorem implies that there is an orthonormal basis $\beta$ of $V$ such that $[T]_{\beta}$ is a diagonal matrix. Since all the eigenvalues are real, we have that $[T]_{\beta}$ is a real matrix. Now,

\begin{align} [T^*]_{\beta} &= \left( [T]_{\beta} \right)^* \tag{$\beta$ is orthonormal}\\ &= [T]_{\beta} \tag{the matrix is real and diagonal} \end{align} Hence, $T^* = T$.

Answer 2

If you assume that T acts on an $n$-dimensional complex inner product space, then $T$ is diagonalizable: There is a basis of orthogonal eigenvectors $v_j$ and in that basis $T$ is represented by a diagonal matrix with the eigenvalues $\lambda_j$ on the diagonal. The adjoint of T is represented in the same basis by a diagonal matrix with entries $\overline{\lambda_j}$, so $T$ is self-adjoint if f $\lambda_j=\overline{\lambda_j}$ for all $j$.