Inverse of spectral theorem for symmetric matrices?

Question

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!

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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.

Answer 1

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.

Answer 2

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.

Answer 3

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.