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?
 A: 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}$.
A: This would be easy for a diagonal matrix, because $\begin{bmatrix}a & 0\\ 0 & b\end{bmatrix}^3=\begin{bmatrix}a^3 & 0\\ 0 & b^3\end{bmatrix}$, which means that you could just take the cube root of each diagonal entry to solve the problem.  While $A$ is not diagonal, it is symmetric and therefore diagonalizable.  If you're comfortable with diagonalizing, find $S$ such that $SAS^{-1}=\begin{bmatrix}a & 0 \\ 0 & b \end{bmatrix}$.  You know how to find a matrix whose cube gives the right hand side.  Then notice how conjugation behaves with cubing: $(S^{-1}C'S)^3=S^{-1}C'^3S$.  Therefore, you can take $C=S^{-1}\begin{bmatrix}\sqrt[3]{a} & 0 \\ 0 & \sqrt[3]{b} \end{bmatrix}S$.  
(I was going to write more involving eigenvectors, but then other answers were posted covering this.)
An alternative approach using polynomial interpolation will work for all diagonalizable matrices having eigenvalues $-1$ and $5$, and does not require finding eigenvectors.  For more on this and generalizations, see Chapter 1 of Higham's Functions of matrices.
In this case, the Lagrange interpolating polynomial of the cube root function on the spectrum $\{5,-1\}$ of $A$ is $$p(t)=\sqrt[3]{5}\cdot\frac{t+1}{5+1}+\sqrt[3]{-1}\cdot\frac{t-5}{-1-5}=\frac{\sqrt[3]{5}+1}{6}\cdot t +\frac{\sqrt[3]{5}-5}{6},$$ so that a matrix cube root for $A$ can be obtained as $$p(A)=\frac{\sqrt[3]{5}+1}{6}\cdot A +\frac{\sqrt[3]{5}-5}{6}\cdot I=\frac{1}{2}\begin{bmatrix}\sqrt[3]5-1&\sqrt[3]5+1\\\sqrt[3]5+1&\sqrt[3]5-1\end{bmatrix}.$$
A: Remember that:

Matrices act on vectors.

Here the linearly independent vectors $u=\begin{pmatrix} 1\\ 1\end{pmatrix}$ and $v=\begin{pmatrix} 1\\ -1\end{pmatrix}$ are such that $Au=5u$ and $Av=-v$. 
Hence a suitable $C=\begin{pmatrix} a & b\\ c& d\end{pmatrix}$ could be defined by the conditions that $Cu=5^{1/3}u$ and $Cv=-v$. This gives you a linear system of two equations for $(a,b)$ and another linear system of two equations for $(c,d)$, and the matrix $C$ follows.
A: 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 
$\left[ \matrix{\lambda & 0 & 0\cr 0 & 0 & 1\cr 0 & 0 & 0\cr}\right]$.
A: Eigendecompose $\mathbf A$ (easily done since you have a symmetric matrix), take the cube root of the eigenvalues, and multiply back the matrix of eigenvectors appropriately.
I get
$$\frac12\begin{pmatrix}\sqrt[3]{5}-1&\sqrt[3]{5}+1\\\sqrt[3]{5}+1&\sqrt[3]{5}-1\end{pmatrix}$$
