Given $A^2$ where A is matrix, how find A? Problem is simple.
Given $$A^2=\begin{bmatrix}13 & 9 &-9 \\ 0 & 4 & 0 \\ 12 & 12 & -8 \end{bmatrix}$$
How find $A$?
I think a method using eigenvalues and I find them.
But I can't find an actual $A$.
Is it right to use eigenvalues?
 A: I bet that when you say eigenvalues, you mean that you might diagonalize the matrix $A^2$ by expressing it as $A^2 = VDV^{-1}$ for some matrix $V$ and diagonal matrix $D$, where the entries in $D$ are precisely the eigenvalues of $A^2$.
Once you've done this, it is very easy to find a square root of $D$, since it is a diagonal matrix. For example, a square root of the matrix $\begin{pmatrix} 4 & 0 \\ 0 & 9 \end{pmatrix}$ might be $\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$.
Does this trail of breadcrumbs lead you to the answer?
A: Actually you don't need eigenvectors, just eigenvalues  (there is some complication in the case of eigenvalue $0$, but fortunately this example is nonsingular).  Find a polynomial $g(t)$ with the following property:  for each eigenvalue $\lambda$,  $g(\lambda) = \sqrt{\lambda}$ and, if $\lambda$ has multiplicity $k > 1$ in the minimal polynomial of $A^2$, the first $k-1$ derivatives of $g$ and $\sqrt{}$ agree at $\lambda$.  Then take $A = g(A^2)$.
This will work even when there does not exist a basis of eigenvectors, as long as $0$ is at most a simple zero of the minimal polynomial.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}$
Let's $B = A^{2}$. Let's assume you already found a base of $3$-D vectors
$\braces{v_{i}}$ with $v_{i}^{\sf T}v_{j} = \delta_{ij}$. Then
$A = \sum_{ij}A_{ij}v_{i}v_{j}^{\sf T}$ and
$B = A^{2} = \sum_{ij}B_{ij}v_{i}v_{j}^{\sf T}$. Also,
\begin{align}
A^{2}
&=
\sum_{ij}A_{ij}v_{i}v_{j}^{\sf T}\sum_{i'j'}A_{i'j'}v_{i'}v_{j'}^{\sf T}
=
\sum_{ij \atop i'j'}A_{ij}A_{i'j'}v_{i}\
\overbrace{v_{j}^{\sf T}v_{i'}}^{\delta_{ji'}}\
v_{j'}^{\sf T}
=
\sum_{ij \atop j'}A_{ij}A_{jj'}v_{i}v_{j'}^{\sf T}
=
\sum_{ij' \atop j}A_{ij'}A_{j'j}v_{i}v_{j}^{\sf T}
\\[3mm]&=
\sum_{ij}\pars{\sum_{j'}A_{ij'}A_{j'j}}\,v_{i}v_{j}^{\sf T}
\quad\imp\quad
B_{ij} = \pars{A^{2}}_{ij} = \sum_{j'}A_{ij'}A_{j'j}
\end{align}
$\braces{B_{ij}}$ is known. So, in principle you have 9 equations to determine
9 quantities e.g. $\braces{A_{ij}}$.
Around 1930, somebody call P. A. M. Dirac built the "matrix square root" ( roughly speaking ) to find a relativistic invariant equation in Quantum Mechanics.
A: Here is an answer obtained by simple inspection. It is not a generally applicable method, but every step below can be done very easily by hand. I have absolutely no intention to recommend this answer. I just want to see how far simple inspection can reach.


*

*Observe that the second row of $A$ has two off-diagonal zeroes. Therefore, if
$$
\pmatrix{4&0^\top\\ \color{blue}{v}&\color{red}{B}}
:=\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}A^2\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}
=\left(\begin{array}{c|cc}4&0&0\\ \hline\color{blue}{9}&\color{red}{13}&\color{red}{-9}\\ \color{blue}{12}&\color{red}{12}&\color{red}{-8}\end{array}\right)
=\pmatrix{2&0^\top\\ x&C}^2
\tag{1}
$$
for some $2$-vector $x$ and some $2\times2$ matrix $C$, one can set $A=\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}\pmatrix{2&0^\top\\ x&C}\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}$. 

*So, we try to solve
$$
\pmatrix{4&0^\top\\ v&B}
=\pmatrix{2&0^\top\\ x&C}^2
=\pmatrix{4&0^\top\\ (C+2I)x&C^2}.\tag{2}
$$
We need to find some $C$ such that $B=C^2$ and some $x$ such that $(C+2I)x=v$.

*Since $\det(B)=4$, we have $\det(C)=\pm2$. Let us assume that $\det(C)=2$ and see if $(2)$ is really solvable. By Cayley-Hamilton theorem, $C^2-\operatorname{tr}(C)C+\det(C)I=0$ and hence $\operatorname{tr}(C)C=B+2I$. Taking traces of both sides, we get $\operatorname{tr}(C)^2=9$. Set $\operatorname{tr}(C)=3$, we get $C=\operatorname{tr}(C)^{-1}(B+2I)=\pmatrix{5&-3\\ 4&-2}$.

*Solving $(C+2I)x=v$, we get $v=(3,4)^\top$ and hence we obtain the following solution:
$$
A=\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}
\pmatrix{2&0^\top\\ x&C}
\pmatrix{0&1&0\\ 1&0&0\\ 0&0&1}
=\pmatrix{5&3&-3\\ 0&2&0\\ 4&4&-2}.
$$

