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?
On
For a $2 \times 2$ matrix $B$ with distinct eigenvalues $\lambda_1$ and $\lambda_2$, a square root of $B$ would have eigenvalues $\pm \sqrt{\lambda_1}$ and $\pm \sqrt{\lambda_2}$. In your case you know you want the $+$ cases. Find $u$ and $v$ so that $\sqrt{\lambda_1} = u + \lambda_1 v$ and $\sqrt{\lambda_2} = u + \lambda_2 v$, and then take $A = u I + v B$.
On
Suppose we have $A=SJS^{-1}$ where $J$ is in Jordan normal form, then $A^2=SJ^2S^{-1}$.
By performing Jordan decomposition on $A^2$ given, we have $$
S=\begin{pmatrix}-1&1\\2&1
\end{pmatrix}
,\;J=\operatorname{diag}(1,4).$$
Then, obviously, $$S\cdot(\operatorname{diag}(1,2))\cdot S^{-1}=\dfrac13
\begin{pmatrix}5&1\\2&4
\end{pmatrix}$$ will be the only choice for $A$ if we prove that $A^2$ has the same $S$ for Jordan decomposition, as $A$ has.
EDIT
To prove the above, it suffices to realize that $A$ and $A^2$ share eigenvectors. Indeed, if $v$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda,$ then $$A^2v=A(Av)=A(\lambda v)=\lambda Av=\lambda^2v$$
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 $A=\pmatrix{1&1\\-2&1}\pmatrix{1&0\\0&2}\dfrac{\pmatrix{1&-1\\2&1}}3=\dfrac{\pmatrix{5&1\\2&4}}3$.