# Inverse of spectral theorem for symmetric matrices?

I have another interesting problem. I intuitively think this is false.

Problem: If matrix $$A\in M_n(\mathbb{R})$$ has $$n$$ distinct eiganvalues and $$n$$ orthogonal eigenvectors $$q_1,q_2,q_3,..,q_n$$. 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 folow? I can't make up any matrix that has there properties but it's not symetric.

Thanks!

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.

• Thank you. So to sum it up. I can say, aplying spectral theorem it's correct! – techno Jun 20 at 15:55
• You need to normalize your eigenvectors first. – Ted Shifrin Jun 20 at 15:55
• @TedShifrin, you are right. I corrected it. – mlainz Jun 20 at 15:58
• @techno Nothing to do with the spectral theorem. Just the change of basis formula. – Ted Shifrin Jun 20 at 16:02

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.

• Why the downvote? Is the complex case forbidden? – Fred Jun 20 at 15:53
• I didn't downvoted. Just edited, i forgot to put field mark R. Thank you for your reply. I will try your way. – techno Jun 20 at 15:54

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.

• Thank you for answering. Can i solve it using spectral theorem? It would be a lot easier for me because im new to linear algebra. – techno Jun 20 at 15:50