prove that for any nonsingular matrix $A$ there exist $X$ such that $X^2=A$ Prove that given any  matrix A, where $$\det(A)\neq0$$ $$A\in M_{n,n}(\mathbb C)$$
the following equation
$$X^2=A$$
always has a solution.
Should I do something with Jordan Normal form?
Any help will be appreciated
 A: Swear i have done this one fairly recently...
$$
\left(
\begin{array}{rr}
t & \frac{1}{2t} \\
0 & t \\
\end{array}
\right)^2 =
\left(
\begin{array}{rr}
t^2 & 1 \\
0 & t^2
\end{array}
\right)
$$
$$ $$
$$
\left(
\begin{array}{rrr}
t & \frac{1}{2t} & \frac{-1}{8 t^3} \\
0 & t & \frac{1}{2t} \\
 0 & 0 & t
\end{array}
\right)^2 =
\left(
\begin{array}{rrr}
t^2 & 1 & 0\\
0 & t^2 & 1 \\
0 & 0 & t^2
\end{array}
\right)
$$
$$ $$
$$
\left(
\begin{array}{rrrr}
t & \frac{1}{2t} & \frac{-1}{8 t^3} & \frac{1}{16 t^5} \\
0 & t & \frac{1}{2t}  & \frac{-1}{8 t^3}\\
 0 & 0 & t & \frac{1}{2t}\\
 0 & 0 & 0 & t
\end{array}
\right)^2 =
\left(
\begin{array}{rrrr}
t^2 & 1 & 0 & 0\\
0 & t^2 & 1 & 0 \\
0 & 0 & t^2 & 1 \\
0 & 0 & 0 & t^2
\end{array}
\right)
$$
$$ $$
$$
\left(
\begin{array}{rrrrr}
t & \frac{1}{2t} & \frac{-1}{8 t^3} & \frac{1}{16 t^5}& \frac{-5}{128 t^7} \\
0 & t & \frac{1}{2t}  & \frac{-1}{8 t^3}  & \frac{1}{16 t^5}\\
 0 & 0 & t & \frac{1}{2t}  & \frac{-1}{8 t^3}\\
 0 & 0 & 0 & t & \frac{1}{2t} \\
0 & 0 & 0 & 0 & t \\
\end{array}
\right)^2 =
\left(
\begin{array}{rrrrr}
t^2 & 1 & 0 & 0 & 0\\
0 & t^2 & 1 & 0 & 0 \\
0 & 0 & t^2 & 1 & 0\\
0 & 0 & 0 & t^2 & 1 \\
0 & 0 & 0 & 0 & t^2
\end{array}
\right)
$$
$$ $$
And
$$ \sqrt{t^2 + 1} \; \; = \; \; t \; \; \sqrt{1 + \frac{1}{t^2}} \; \; = \; \;  t +  \frac{1}{2t}  - \frac{1}{8 t^3}  + \frac{1}{16 t^5} -\frac{5}{128 t^7}  + \frac{7}{256 t^9} -\frac{21}{1024 t^{11}} \cdots $$
The resemblance is not cosmetic or accidental. We have, in a Jordan block of size $n,$ an identity matrix $I$ and a nilpotent matrix $N$ with $N^n=0.$ We are asking for $\sqrt{t^2 I + N}.$ As with other real analytic functions, we can use the facts that $IN=NI$ commute to give a power series for the resulting matrix, and the series is finite because $N$ is nilpotent. 
