Inverse of spectral theorem for symmetric matrices? I have another interesting problem. I intuitively think this is false.

If matrix $A \in M_n(\mathbb{R})$ has $n$ distinct eigenvalues and $n$ orthogonal eigenvectors $q_1, q_2, q_3,\dots, q_n$. Then $A$ is symmetric matrix!


This is kind of inverse of spectral theorem. Is it enough to use spectral theorem or does inverse doesn't follow? I can't make up any matrix that has there properties but it's not symmetric.
 A: It is true and relatively easy (compared to the converse). Let $P$ be the matrix whose columns are an orthonormal base of eigenvectors. Since they are orthogonal, $P^{-1}=P^t$ . Note that if $D$ is the diagonal matrix with the eigenvalues $q_i$ on the diagonal, then $P^{-1} A P = D$, hence $A = P D P^{-1} = P D P^{t}$, which is clearly symmetric.
A: A matrix $A $ has the properties as in "Problem "  if  and only if $A $ is normal. 
Example: $A= diag (i,-i)$ is normal,  but  not  symmetric. 
A: Symmetricity means that $\langle Ax,y\rangle = \langle x, Ay \rangle$ for all $x,y$. 
Since $\{q_i\}$ form a basis, write $x$ and $y$ in terms of this basis:
$$x = \sum x_i q_i$$
$$y = \sum y_i q_i$$
Now, recalling that $\{q_i\}$ are orthogonal, compute
$$\langle Ax, y \rangle = \langle \sum x_i \lambda_iq_i, \sum y_i q_i \rangle = \sum x_i y_i \lambda_i$$
$$\langle x, Ay \rangle = \langle \sum x_i q_i, \sum y_i \lambda_iq_i \rangle = \sum x_i y_i \lambda_i$$
Thus, $A$ is indeed symmetric. Note that the argument works even if $\lambda_i$ are not distinct.
