I'm struggling to come up with a solution for the following problem.
Assume that a matrix A of size n by n has n distinct eigenvalues, all of which are $< \lvert 1\rvert$. Simplify the sum $$\sum_{k=2}^{\infty} A^k$$
So far, I've got that A can be expressed as $QDQ^{-1}$, where $D$ is a diagonal matrix with the eigenvalues along the diagonal and zeros everywhere else, and $Q$ is the matrix with columns containing the eigenvectors corresponding to the eigenvalues.
Then $A^k$ can be expressed as $QD^kQ^{-1}$.
However after this I'm not quite sure what I can do, or if I am even on the right path.
Since for each eigenvalue $\lambda$ of $A$ we have $|\lambda|<1$, the spectral radius of $A$ is $ <1.$ Hence $I-A$ is invertible and (Neumann series !)
$$(I-A)^{-1}= \sum_{k=0}^{\infty}A^k.$$
Thus
$$ \sum_{k=2}^{\infty}A^k=A^2\sum_{k=0}^{\infty}A^k= A^2(I-A)^{-1}.$$