# When a normal operator is also a self adjoint operator?

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.

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, $$T^* = T$$.
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 $$j$$.