Eigenvalues of $A$ where $A^2$ is real and symmetric? I want to prove that $\sigma(A^2) = \{\lambda ^2\: |\: \lambda \in \sigma(A) \}$ given that $A$ is a real symmetric matrix. It's easy to prove that $\lambda \in \sigma(A) \implies \lambda ^2 \in \sigma(A^2)$ however I am completely stuck trying to prove the converse. Any tips on how to proceed?
 A: This is true for any matrix, symmetric or not, so although Nightgap's suggestion to diagonalize using the spectral theorem is convenient, it is unnecessary. 
Suppose $\mu$ is an eigenvalue of $A^2$, so that $A^2 v = \mu v$ for some $v$, or equivalently $(A^2 - \mu) v = 0$. Factoring this expression gives
$$(A - \sqrt{\mu})(A + \sqrt{\mu}) v = 0$$
from which it follows that $A$ has eigenvalue either $\sqrt{\mu}$ or $- \sqrt{\mu}$, since either $u = (A + \sqrt{\mu}) v = 0$, in which case $v$ is an eigenvector of $A$ with eigenvalue $- \sqrt{\mu}$, or $u \neq 0$, in which case $(A - \sqrt{\mu}) u = 0$, so $u$ is an eigenvector of $A$ with eigenvalue $\sqrt{\mu}$. (We may need to pass to an algebraic closure of the ground field to get these square roots.) 
More generally, it's true that if $f(-)$ is any polynomial then $\sigma(f(A)) = f(\sigma(A))$. 
A: Someone should probably note that while it's true that $\sigma(A^2)=\{\lambda^2:\lambda\in\sigma(A)\}$ (proved in the other answer), it's not true that $\lambda^2\in\sigma(A^2)$ implies $\lambda\in\sigma(A)$ (for example $A=I$, $\lambda=-1$.)
I mention this because of the words "unable to prove the converse" in your question; the converse you say you can't prove is false, although the result you ask about in your first sentence is true. They're not the same thing at all.
