TITLE: Given a matrix $A$ find a matrix $C$ such that $C^3$=$A$
QUESTION [6 upvotes]: This is a question I had on a test, we were told not to use brute-force and figure out a smart way to solve the problem.
We have a matrix $A =$
$\displaystyle\begin{bmatrix}
2 & 3\\ 
3 & 2
\end{bmatrix}$. Find a matrix $C$ such that $C^{3}$=$A$.
What is the 'smart not brute-force' way to solve this, without picking numbers, looking for patterns and so on?
it was in eigenvalues section" in the end?

REPLY [2 votes]: In fact, the only 2 x 2 matrices that do not have cube roots (over the complex numbers) are those with Jordan canonical form $\left[ \matrix{0 & 1\cr 0 & 0\cr}\right]$.
The 3 x 3 matrices with no cube root are those with Jordan form 
$\left[ \matrix{0 & 1 & 0\cr 0 & 0 & 1\cr 0 & 0 & 0\cr}\right]$ or 
$\left[ \matrix{\lambda & 0 & 0\cr 0 & 0 & 1\cr 0 & 0 & 0\cr}\right]$.