Given a matrix $A$ find a matrix $C$ such that $C^3$=$A$ 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?

One thing that jumps out at me is that this matrix has two eigenvalues. Find them: det
$
(\begin{bmatrix}
2-\lambda & 3\\ 
3 & 2-\lambda
\end{bmatrix})$=
$(2-\lambda)^{2}-9$=$-5-4\lambda+\lambda^{2}$. Factoring this you get $(\lambda+1)(\lambda-5)=0$, so your eigenvalues are -1 and 5. Now, find bases for their respective eigenspaces.
Basis $\xi_{-1}$ = ker$(A+I)$=ker $(\begin{bmatrix}
3 & 3\\\ 
3 & 3
\end{bmatrix})$. Row-reducing, you get ker $(\begin{bmatrix}
1 & 1\\\ 
0 & 0
\end{bmatrix})$, which is equal to the span of $\begin{bmatrix}
-1\\ 
1
\end{bmatrix}$. So a basis of $\xi_{-1}$ = $\begin{bmatrix}
-1\\ 
1
\end{bmatrix}$.
Same for $\xi_{5}$ -- ker $(\begin{bmatrix}
-3 & 3\\\ 
3 & -3
\end{bmatrix})$, row-reducing we get ker $(\begin{bmatrix}
1 & -1\\\ 
0 & 0
\end{bmatrix})$, so ker = span $\begin{bmatrix}
1\\ 
1
\end{bmatrix}$. So a basis of $\xi_{5}$ = $\begin{bmatrix}
1\\ 
1
\end{bmatrix}$.
Now you know that $A$=$C^{3}$=$CCC$. So if $\vec{x}$ is an eigenvector of $A$, it is clearly an eigenvector of $C$, but perhaps with a different eigenvalue. You know that $A$$
\begin{bmatrix}
1\\ 
1
\end{bmatrix}$=$CCC$$
\begin{bmatrix}
1\\ 
1
\end{bmatrix}$=$\begin{bmatrix}
5\\ 
5
\end{bmatrix}$. So, $\sqrt[3]{5}$ must be an eigenvalue of $C$.
Same goes for $A$$\begin{bmatrix}
-1\\ 
1
\end{bmatrix}$=$CCC$$\begin{bmatrix}
-1\\ 
1
\end{bmatrix}$=$\begin{bmatrix}
1\\ 
-1
\end{bmatrix}$. This would be possible only if -1 was an eigenvalue of $C$, since $(-1)^{3}$=$-1$. 
So now you can construct a system of equations -- you know that $C$ is of some form $(\begin{bmatrix}
a & b\\\ 
c & d
\end{bmatrix})$, and based on the eigenvectors and eigenvalues of $C$ that we just found out, you can devise the following system:
$\left\{\begin{matrix}
a & + & b & = & \sqrt[3]{5}\\ 
c & + & d & = & \sqrt[3]{5}\\ 
-a & + & b & = & 1\\ 
-c & + & d & = & -1
\end{matrix}\right.$
Now solve for the variables -- $b=a+1$, substitute into the other $a, b$ equation to get $2a+1=\sqrt[3]{5}$, so $a=\frac{\sqrt[3]{5}-1}{2}$. $b$, then, is equal to $\frac{\sqrt[3]{5}-1}{2}+1$, as obvious from the third equation. Same for $c$ and $d$ -- $c=d+1$, so substituting, we get $2d+1=\sqrt[3]{5}$, so $d=\frac{\sqrt[3]{5}-1}{2}$, and hence $c=\frac{\sqrt[3]{5}-1}{2}+1$.
This results into $C$=$
\begin{bmatrix}
\frac{\sqrt[3]{5}-1}{2} & \frac{\sqrt[3]{5}-1}{2}+1 \\ 
\frac{\sqrt[3]{5}-1}{2}+1 & \frac{\sqrt[3]{5}-1}{2}
\end{bmatrix}$.
