On powers of symmetric matrices.

Question

What is the best way to show if $A$ is symmetric then $A^2$ is as well using eigenvalues?

Answer 1

Just recall that for two matrices $B,C$ it holds that: $(BC)^T = C^TB^T$. Applying this with $B=A$, $C=A$, we get:

$$ (A^2)^T = (AA)^T = A^TA^T = AA = A^2$$

You can also do it with eigenvalues but it seems a lot more tedious to me (and requires much stronger results):

Let $A=VDV^T$ be the eigenvalue decomposition of $A$ (with $D$ diagonal, $V$ orthonormal, the decomposition exists by spectral theorem), then:

$$ A^2 = VD^2V^T $$

And it is easy to see that this is symmetric as well. In particular, apply the identity I cited at the beginning and note that $D^2$ is symmetric since it is diagonal.