Higher order of Jordan form Let $\lambda =1$ is the eigenvalue corresponding to the single Jordan block $J$. Prove $J^m \sim J$ with an arbitrary positive integer $m$.
My try: Because $\lambda = 1$ is eigenvalue, $(J-I)^m =0$. After that $(J-I)^{m-1} J = (J-I)^{m-1}$. At this point, I do not know how to continue.
 A: I would do it this way: $J=I+N$ where $N$ is the $n\times n$ nilpotent upper triangular matrix. Taking the $m$-th power we obtain
$$
J^m=(I+N)^m=\sum_{k=1}^m\binom{m}{k}N^k=I+mN+(\text{super-super-diagonals}),
$$
from what we get $J^m-I=mN+(\text{super-super-diagonals})$ and, therefore,
$$
\text{rank}(J^m-I)=\text{rank}(N)=n-1.
$$
It gives us $\text{dim}\ker(J^m-I)=1$, which means only one Jordan block for $J^m$. The block is $J$.
A: First observe that any power $J^m$ of $J$ is an upper triangular matrix with all entries on the main diagonal equal to $1$. To prove that $J^m$ is similar to $J$, you can show that $J$ is the Jordan normal form of $J^m$. To do this, it suffices to show that $(J^m-I)^k$ is non-zero provided that $(J-I)^k$ is non-zero (since this implies that in the Jordan form of $J^m$ there must be one block of the same size as $J$). 
In the following computations, all matrices involved are polynomials in $J$ and hence commute. In particular, since $J$ and $I$ commute, you can use the standard formula to see that 
$$J^m-I=J^m-I^m=(J-I)(\sum_{i+j=m}J^iI^j)=(J-I)(\sum_{i=0}^mJ^i).$$
Now the second factor in this product is an upper triangular matrix with all entries on the main diagonal equal to $m+1$ and hence invertible. Forming powers (taking into account that all matrices we consider commute), you get $(J^m-I)^k=(J-I)^k(\sum_{i=0}^mJ^i)^k$. Since the second factor still is an invertible matrix, the claim follows. 
