# Jordan normal form powers

Let $$A$$ be a $$n\times n$$ such that $$A=PBP^{-1}$$ where $$B$$ is in Jordan normal form with $$\lambda_i(k)_j$$ Where $$i$$ is the size, $$k$$ is the eigenvalue and $$j$$ the order.

If $$A$$ was diagonal($$i=1$$) then $$A^n$$ in Jordan form has $$\lambda_1(k^n)_j$$.
If the Jordan form has Jordan blocks bigger then 1, how do we find $$A^n$$ In Jordan form?

A Jordon block of size $$i$$ and eigenvalue $$k$$ has the form $$(kI+t)$$ where $$t^i=0$$. Thus $$(kI+t)^n$$ will just be the truncated binomial expansion:$$\sum_{r=0}^{i-1} {n \choose r} k^{n-r}t^r$$
• Also, $t$ is a very specific matrix, is there a nice expression or pattern to its powers? – razivo Jul 19 at 9:25
• The non-zero entries of $t^i$ will be the diagonal row of $1$'s starting at $(1,1+i)$ and moving down and right. – tkf Jul 19 at 9:29