If $A$ is $n \times n$ matrix of complex numbers, then it's easy to show that if $c$ is an eigenvalue of $A$, then $c^2$ is an eigenvalue of $A^2$, like.. if $v$ is an eigenvector of $A$ corresponding to the eigenvalue $c$, then $Av = cv$, so $A(Av) = A(cv) = c^2 v$, we're done.
But I was not quite sure how to show the converse.. if $c$ is an eigenvalue of $A^2$, then either $-\sqrt{c}$ or $\sqrt{c}$ is an eigenvalue of $A$.
Any help would be greatly appreciated.
On
The forward direction and the converse can be proven together.
Let $A$ be your matrix and upper triangularize it to $T$. The eigenvalues of $T$, and hence the eigenvalues of $A$, appear precisely as the diagonal entries of $T$. Taking the $n$th power of $T$ has a simple effect on the diagonal where each diagonal entry is simply brought to its $n$th power. Therefore $\lambda$ is an eigenvalue of $A$ if and only if $\lambda^n$ is an eigenvalue of $A^n$.
Let $u$ be an eigenvector of $A^2$ for eigenvalue $c$, and let $v = A u$. Note that $A v = A^2 u = c u$. Then $A(\sqrt{c} u + v) = v + \sqrt{c} v + c u = \sqrt{c} (\sqrt{c} u + v)$ and similarly $A(-\sqrt{c} u + v) = -\sqrt{c} (-\sqrt{c} u + v)$. So at least one of $\sqrt{c} u + v$ and $-\sqrt{c} u + v$ is an eigenvector for $\pm \sqrt{c}$ (one of them could be $0$, but not both).