A problem that is well known in linear algebra is the existence of square root of a matrix, where square root of matrix $A \in M_n(\mathbb F)$ is defined to be $K \in M_n(\mathbb F)$ such that $K^2=A$
I looked in different places for a proof of this theorem and could not find any.
Let $A$ be a matrix over an algebraically closed field $\mathbb F$ and let $\lambda_1, \lambda_2,\dots,\lambda_m$ be distinct eigenvalues of $A$, if $0$ is not an eigenvalue of $A$ then exists a square matrix.
I tried finding a counter-example but couldn't, so I tried proving it, I could prove that for eigenspace, the matrix $A|_{V_{\lambda_i}}$ has a square root by looking at the Taylor polynomial $\sqrt{1+x}$ of the matrix $J= \lambda I+N$ and since $0$ is not an eigenvalue, the Taylor series converges since $N$ is nilpotent then the series trivially converges (as it becomes $0$ at some index $k\le n$)
I couldn't although prove that if the restriction of every eigenspace of $A$ has a square root then $A$ itself has a square root.