Nth roots of square matrices Is there a general method (which can be implemented by hand) to finding the $n$-th roots of $2 \times 2$ matrices? Is there a similar method for a general $m \times m$ matrix? (for $n > 1$ and $n\in\mathbb{Z}$)
 A: By hand, you can do a Jordan decompositiotion, $A = S J S^{-1}$. If $J$ is diagonal, then it is easy to compute $J^{1/n}$, and so $A^{1/n} = S (J^{1/n}) S^{-1}$.
Note that the Jordan decomposition is not numerically stable, so it's generally not advised to use it.
This might still be usable for order $3$, because there are exact formulas for the eigenvalues and the eigenvectors, but for matrices of order higher than $3$, you should use computers.
For the algorithms (and the theory), I recommend Nick Higham's "Functions of Matrices: Theory and Computation".
A: For $2\times 2$ matrices, it's not so bad.
As Vedran mentions, compute the Jordan decomposition
$$M = P^{-1}JP.$$
The $n$th root is then just
$$M = P^{-1}J^{(1/n)}P.$$
There are a few cases.


*

*$J$ is diagonal. Then just take the $n$th roots of the diagonal entries.

*$J = \left[\begin{array}{cc}a & 1\\0 & a\end{array}\right]$, for $a\neq 0$. Then $J^{1/n} = \left[\begin{array}{cc} a^{1/n} & a^{(1-n)/n}/n\\ 0 & a^{1/n}\end{array}\right]$.

*Same as case 2, but $a=0$. Then no root exists.


Notice that the roots are not necessarily real matrices, unless $M$ is symmetric positive-semidefinite. 
For larger matrices, the possible configurations of the Jordan blocks become more complicated. Perhaps you can work out a general algorithm.
