Given matrix $A^2$, how to find matrix $A$? 
Let $$A^2 = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$$ Knowing that $A$ has positive eigenvalues, what is $A$?


What I did was the following:
$$A = \begin{pmatrix}
a & b \\
c & d 
\end{pmatrix}$$
so
$$A^2 = \begin{pmatrix}
a^2 + bc & ab+bd \\
ac+cd & bc+d^2 
\end{pmatrix}$$
I got stuck here after trying to solve the 4 equations. Can someone help, please?

Computing matrix powers can be done with diagonalization.
The eigenvalues of $A^2$ have sum $5$ (trace) and product $4$ (determinant), so they are $1$ and $4$.
The corresponding eigenvectors of $A^2$ are $\pmatrix{1\\-2}$ and $\pmatrix{1\\1}$, respectively.
Therefore, $A^2$ is diagonalized as follows:  $\pmatrix{1&1\\-2&1}\pmatrix{1&0\\0&4}\pmatrix{1&1\\-2&1}^{-1}=\pmatrix{1&1\\-2&1}\pmatrix{1&0\\0&4}\dfrac{\pmatrix{1&-1\\2&1}}3.$
Therefore, we can take $A=\pmatrix{1&1\\-2&1}\pmatrix{1&0\\0&2}\dfrac{\pmatrix{1&-1\\2&1}}3=\dfrac{\pmatrix{5&1\\2&4}}3$.
