<s>[INST] 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?
 [/INST] 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}$.
</s>