<s>[INST] Prove that matrix can be square of matrix with real entries Prove that matrix 
\begin{bmatrix}1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}
can be square of matrix with all real entries.
I have found one such matrix to be 
\begin{bmatrix}1&0&0\\0&1&-1\\0&2&-1\end{bmatrix} but is there an elegant way to do it without any trial and error?
 [/INST] I will try to present  a solution "without any trial and error".
Here we have a matrix $F$ in the form 
$\begin{bmatrix}
1 &  0_{1 \times 2} \\ 0_{2 \times 1} & A_{2 \times 2} \end{bmatrix}$ so if $G^2=F$ then $G$ can be in the form $\begin{bmatrix}
\pm 1 &  0_{1 \times 2} \\ 0_{2 \times 1} & B_{2 \times 2} \end{bmatrix}$ where $B^2=A$.
Now let's concentrate on $B^2=\begin{bmatrix}
 -1 &  0 \\ 0  & -1 \end{bmatrix} $.
We have two matrix equations 
$B^2=-I$  i.e.
$B^2+I=0$   
and general equation from Cayley-Hamilton theorem  for $ 2 \times 2$ matrices
$B^2-\text{tr}(B)B+\det(B)I=0$.
Comparing both equations we obtain
$\text{tr(B)}=0$   , $ \det(B)=1$.
So if we denote $ B=\begin{bmatrix} a & b \\c  & d\end{bmatrix} $  then $d=-a$ and consequently $-a^2-bc=1$.
These conditions are sufficient  to obtain a solution not only with real numbers, but even with integer values.
For example
if $a=0$ then $b=1$, $c=-1$    
($b,c$  should have always opposite signs because -$a^2-bc$ has to be positive - possible other solution $-1,1$, also $a$ and $d$ can be switched)   
if $a=1$ then $b=2$, $c=-1$
if $a=2$ then $b=5$, $c=-1$
if $a=3$ then $b=5$, $c=-2$
...
if $a=8$ then $b=5$, $d=-13$   , etc... infinite number of solutions - for every integer $a$ we can find proper integer values of $b$ and $c$ from $-bc=a^2+1$..
Let's check the last listed solution.
Indeed
$  \begin{bmatrix} 8 & 5 \\-13  & -8\end{bmatrix}  \begin{bmatrix} 8 & 5 \\-13  & -8\end{bmatrix}    =   \begin{bmatrix} 8 \cdot {8} - 5\cdot 13  & 8\cdot 5 - 5 \cdot 8  \\ -13\cdot 8 + 8 \cdot 13 & -13\cdot 5 +8\cdot 8\end{bmatrix} =  \begin{bmatrix} -1 & 0 \\0  & -1\end{bmatrix} $ .
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