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}$.

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 $$p(A)=\frac{\sqrt[3]{5}+1}{6}\cdot A +\frac{\sqrt[3]{5}-5}{6}\cdot I=\frac{1}{2}\begin{bmatrix}\sqrt[3]5-1&\sqrt[3]5+1\\\sqrt[3]5+1&\sqrt[3]5-1\end{bmatrix}.$$

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.

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 $\left[ \matrix{\lambda & 0 & 0\cr 0 & 0 & 1\cr 0 & 0 & 0\cr}\right]$.

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

$$\frac12\begin{pmatrix}\sqrt[3]{5}-1&\sqrt[3]{5}+1\\\sqrt[3]{5}+1&\sqrt[3]{5}-1\end{pmatrix}$$