<s>[INST] Prove Why $B^2 = A$ exists? 
Define
  $$A =
\begin{pmatrix}
8 & −4 & 3/2 & 2 & −11/4 & −4 & −4 & 1 \\
2 & 2 & 1 & 0 & 1 & 0 & 0 & 0 \\
−9 & 8 & 1/2 & −4 & 31/4 & 8 & 8 & −2 \\
4 & −6 & 2 & 5 & −7 & −6 & −6 & 0 \\
−2 & 0 & −1 & 0 & 1/2 & 0 & 0 & 0 \\
−1 & 0 & −1/2 & 0 & −3/4 & 3 & 1 & 0 \\
1 & 0 & 1/2 & 0 & 3/4 & −1 & 1 & 0 \\
2 & 0 & 1 & 0 & 0 & 0 & 0 & 5 \\
\end{pmatrix} \in M_8(\mathbb R)$$ 
Justify why there is a matrix $B$ such that $B^2 = A$.

I think it has something to do with the Jordan normal form so I found it.
\begin{pmatrix}
2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 2 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 5 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 5 \\
\end{pmatrix}
(I know the ones are under the diagonal, but I think it doesn't make a difference.)
I have no idea what should I do next, a help will be appreciated!
Thanks!
 [/INST] I borrow (and slightly modify) an explanation that can be found in the excellent thread 
https://mathoverflow.net/questions/14106/finding-the-square-root-of-a-non-diagonalizable-positive-matrix
$f$ being any function, sufficient derivable, one can write, for example for a $4 \times 4$ Jordan block:
$$f\left(  \left[
\begin{array}
[c]{cccc}%
\lambda & 1 &  & \\
& \lambda & 1 & \\
&  & \lambda & 1\\
&  &  & \lambda
\end{array}
\right]  \right)  =\left[
\begin{array}
[c]{cccc}%
f\left(  \lambda\right)   & f^{\prime}\left(  \lambda\right)   & \frac{1}
{2!}f^{\prime\prime}\left(  \lambda\right)   & \frac{1}{3!}f^{\prime
\prime\prime}\left(  \lambda\right)  \\
& f\left(  \lambda\right)   & f^{\prime}\left(  \lambda\right)   & \frac
{1}{2!}f^{\prime\prime}\left(  \lambda\right)  \\
&  & f\left(  \lambda\right)   & f^{\prime}\left(  \lambda\right)  \\
&  &  & f\left(  \lambda\right)
\end{array}
\right]$$
(it is based on a certain Taylor expansion of $I+N$ where $N$ is a nilpotent matrix).
It suffices now to take $f(x)=\sqrt{x}$ to get for the square root of a Jordan block of similar form as above but $3 \times 3$:
$$\left[
\begin{array}
[c]{ccc}%
\sqrt{\lambda}   & \dfrac{1}{2\sqrt{\lambda}}  & -\dfrac{1}{8 (\lambda)^{3/2}}  \\
& \sqrt{\lambda}   & \dfrac{1}{2\sqrt{\lambda}}  \\
&  & \sqrt{\lambda}  
\end{array}
\right]$$ 
</s>