Solutions for $A^2=B$ 
Show that there exists a neighborhood $U$ of $I$ in $\mbox{Mat}(2 \times 2)$ such that for all $B\in U$ there are at least two solutions in $\mbox{Mat}(2 \times 2)$ for the equation $A^2 = B$.


I have this question for homework and I would be happy to get a hint.  I think it uses the implicit function theorem.
 A: If $A^2 = B,$ then $(-A)^2 = B.$ This only fails when $A=-A,$ so $A=0,$ but $0$ is "far" from $I.$
Now it comes down to showing that there is such an $A.$ The soft solution is that the map $x\to x^2$ is an open map in a  neighborhood $V$ of identity, so its image is onto a neighborhood $U$ of identity.
The "hard" solution is to expand $(I+x)^{\frac12}$ in a power series (by the binomial theorem) and note that the series converges for $\rho(x)$ small. The "even harder" solution is to use the Jordan canonical form to note that $B$ is conjugate to either a diagonal matrix (in which case the solution is obvious) or to an upper triangular matrix, in which case a solution is easy to get.
A: In the (sufficiently small) neighborhood of $I$ matrices have two (not necessarily distinct) non-zero eigenvalues. It means that any such matrix, say $B$, has (so called Jordan Form) representation as $B = C*D*C^{-1}$ where $D$ is diagonal matrix. Now, $D$ has a root , it is just diagonal matrix with roots of $D$, let's denote it by $D^{\frac{1}{2}}$. Then if $A=C*D^{\frac{1}{2}}*C^{-1}$ one checks that $A^2=B$.
To get second solution change signs for the "root" matrix diagonals.
