Can we reduce finding matrix roots to finding roots of Jordan blocks? I just found some interesting question about matrix square roots and I came to think of one way to find them, or at least reduce them to a set of simpler problems.
Assume we have a matrix $\bf A$ and it can be put on some Jordan form:
$${\bf A} = {\bf SJS}^{-1}$$
Where $\bf J$ is block-diagonal consisting of famous Jordan blocks of the shape:
$${\bf J_{k}} = \begin{bmatrix} \lambda_k&1&0&\cdots&0\\0&\ddots&\ddots& & 0\\\vdots&\ddots&\ddots&\ddots&\vdots\\0&\cdots&0&\lambda_k&1\\0&\cdots&\cdots&0&\lambda_k\end{bmatrix}$$
In other words main diagonal full with eigenvalue $\lambda_k$ and first off-diagonal filled with ones.
The problem of finding some n'th root to $\bf A$ can now be written $${\bf SJ}^{1/n}{\bf S}^{-1}$$ (Why?). 
So If I am correct so far.. we have reduced down to find some way of calculating square root of such Jordan blocks $\bf J_k$, and in simplest case blocks of dimensionality 1, finding some root over our scalar field (for the eigenvalues themselves).


*

*Firstly, is this reasoning correct so far?

*Secondly, how can we approach finding square root to matrices of the form $\bf J_k$. Is there some simplification or short-cut that can be done?
 A: Assume that the underlying field is algebraically closed and that we know the Jordan decomposition of $A$; thus we may assume that $A=diag(\lambda_1 I_{i_1}+J_1,\cdots,\lambda_k I_{i_k}+J_k)$ where $J_r$ is the nilpotent Jordan block of dimension $i_r$.
$\bullet$ The simple case is when $A$ is invertible and cyclic (that is, the eigenvalues $(\lambda_r)$ are distinct and non-zero); then $A$ admits exactly $2^k$ square roots: $diag(\pm L_1,\cdots,\pm L_k)$ where 
$L_r=\sqrt{\lambda_r}(I_r+(1/\lambda_r) J_r)^{1/2}$ and where the second factor is given by the Taylor's development, wrt $x$,  of $(1+(1/\lambda_r) x)^{1/2}$.
$\bullet$. Otherwise there are supplementary solutions -or eventually no solutions when $A$ is singular-. 
When $A$ is not invertible, cf. my post in
sufficient and necessary conditions for matrix to have pth roots
When $A$ is not cyclic, consider the case when $A=diag(I_2+J,I_2+J)$ (find in the first part $C(A)$).
EDIT. More precisely (for the above example) $C(A)$ is the vector space of dimension $8$ constituted by the matrices in the form $\begin{pmatrix}U_1&U_2\\U_3&U_4\end{pmatrix}$ where $U_j$ is in the form $a_j I_2+b_j J$. A particular square root of $A$ is 
$\begin{pmatrix}1/2&1&1&2/\sqrt{3}\\0&1/2&0&1\\3/4&-\sqrt{3}/2&-1/2&-1\\0&3/4&0&-1/2\end{pmatrix}$.
