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!

By putting $A$ into Jordan normal form you have found a non-singular matrix $P$ such that 
$$
A = P^{-1}CP
$$
where $C$ is your matrix shown above.  It is easy to find a matrix sqare root of $C$, for example, in the upper left start with $\pmatrix{\sqrt{2} &0\\\frac14\sqrt{2}& \sqrt{2}}$
Then $$(P^{-1}DP)^2 = P^{-1}D(PP^{-1})DP = P^{-1}D^2P=P^{-1}CP=A
$$
