Finding $A^k$ for non-diagonalizable $A$ Is there an easy way to find $A^k$ for a  square matrix $A$ that is NOT diagonalizable?
 A: Yes, it exists. You can do it by finding the Jordan normal form (which is done in a way similar to diagonalization) and finding the general form for the exponentiation of a block.
You will find an example of how it's done at the end of the Wikipedia page.
A: Even if the matrix $A$ is not diagonalisable, you can tridiagonalize over an algebraically closed field, i.e., $SAS^{-1}=U$ is upper-triangular. This is faster than computing the Jordan normal form, and helps to compute $A^k=S^{-1}U^kS$, because $U^k$ can be computed rather quickly.
A: There are two basic ways that I know of:
Over the complex field, any matrix $A$ can be written in Jordan normal form $A=P^{-1}BP$, where $B$ is a block diagonal matrix. Then in similar fashion to diagonalizable matrices, $A^k = P^{-1}B^k P$. It is easy to take powers of block diagonal matrices, so this is one possible general method.
However, it seems that Jordan normal form is not commonly used in numerical analysis due to "sensitivity to perturbations." (Wikipedia) One might then consider the Schur decomposition: any matrix $A$ can be written in Schur form $A=Q^{-1}CQ$, where $Q$ is unitary and $C$ is upper triangular. Moreover, there is a recursive algorithm to compute the Schur form; it is implemented in the Maple routine SchurForm in the LinearAlgebra package. This is apparently more stable, that is, more resistant to perturbations in the matrix $A$ than the Jordan normal form, so it seems to be preferred in numerical analysis.
A: Two friendly special cases should be mentioned where we even can use fractional powers $k$:   
In addition to the pointings to the Schur-form there might one more step possible: if the triangular form $U$ has all eigenvalues equal (that means, all elements on the diagonal are equal, say $c$) then it is an option to use the matrix-logarithm to even allow fractional powers because we can then write $$U^k = c^k \exp(k \cdot \log(U/c)) $$ where we use the mercator series of $U/c-I$ for the logarithm and the series will be finite (because $U/c-I$ is nilpotent) and thus converging.     
A further special case is if the matrix $U$ is the Pascal-matrix, then we need not even the matrix-logarithm but can achieve its arbitrary (complex) power by a similarity scaling with a diagonal-vector of powers of $k$.    
