Prove that for every n by n matrix M sufficiently close to the identity matrix there exists a square-root matrix (solution of $A^2 = M$) and the solution is unique if $A$ is required to be sufficiently close to the identity matrix?
Okay so call the inverse of that function $f$, its derivative $df = (A+H)^2-A^2 -A^2 = AH+HA$. We know that A is sufficiently close to the identity and so is AH+HA. Well now to apply the Inverse function theorem we need df to invertible. Well now let's say we prove that $df$ is invertible (because A is sufficiently close to $I$ we know that $df$ is basically $2H$ and I saw somewhere that that is invertible but why?). Now we can apply the inverse function theorem and say that indeed there is a unique solution to $A^2=M$ if both are required to be sufficiently close to the identity matrix?
Compute the derivative of the square-root function $M\text{->}A$?
Show that the binomial expansion can be used to compute A?