If $A$ is a normal matrix over $\mathbb{R}$ and all the eigenvalues of $A$ are real, is $A$ always symmetric?

Question

If $A$ is idoes $A$ always symmetric?

I know that if $A$ is normal it is diagonalizable, but does the fact it has only real eigenvalues make it symmetric?

Answer 1

Yes.

$A$ is normal so it diagonalizes in an orthonormal basis $\{e_1, \ldots, e_n\}$ with $Ae_i = \lambda_i e_i$. By assumption we have $\lambda_i \in \mathbb{R}$.

Then for $x = \sum_{i=1}^n \langle x, e_i\rangle e_i$ and $y = \sum_{i=1}^n \langle y, e_i\rangle e_i$ we have

\begin{align} \langle Ax,y\rangle &= \left\langle \sum_{i=1}^n \langle x, e_i\rangle \lambda_i e_i, \sum_{i=1}^n \langle y, e_i\rangle e_i\right\rangle \\ &= \sum_{i=1}^n \langle x, e_i\rangle \langle e_i, y\rangle \lambda_i \\ &= \left\langle \sum_{i=1}^n \langle x, e_i\rangle e_i, \sum_{i=1}^n \langle y, e_i\rangle \lambda_ie_i\right\rangle \\ &= \langle x, Ay\rangle \end{align}

so $A^* = A$.