# Non-zeroth power of a Jordan block for the eigenvalue $1$ is similar to itself

I'm trying to prove: If $J$ is a single Jordan block corresponding to an eigenvalue $\lambda = 1$, then $J^k$ is similar to $J$, where $k$ is a nonzero integer. Moreover, if $\lambda = 1$ is the only eigenvalue of a matrix $A$, then $A^k$ is similar to $A$.

Thanks

Let $J$ be an $n \times n$ matrix. We have that $J = I+N$, where $I$ is the identity matrix and $N$ is a nilpotent matrix. $J^2 = I+2N+N^2 = I + N^{(2)}$, where $N^{(2)}$ is, again, nilpotent with rank $n-1$. Since a nilpotent matrix has the eigenvalues zero, the Jordan form of $N^{(2)}$ is $N$, hence $N^{(2)} \sim N$. It follows that $J^2$ is similar to $J$. From here, it should be short work to prove that this holds for any positive power of $J$, for instance using induction. Assume that $J^{k-1} \sim J$. Then $J^k \sim J^2 \sim J$, so it follows that the similarity relation holds for any positive power of $J$.
If $\lambda = 1$ is the only eigenvalue to $A$, then $$A \sim J_{n_1} (1) \oplus ... \oplus J_{n_k}(1) =: J.$$ Applying the previous result blockwise, it follows that $A^k \sim J^k \sim J \sim A$, i.e. $A^k \sim A$, since matrix similarity is an equivalence relation.
Proposition. Let $\lambda\not=0,N\in M_n$, be a nilpotent matrix and $k$ be a positive integer. Then $(\lambda I_n+N)^k$ and $\lambda^k I_n+N$ are similar.
Proof. The two matrices have same spectrum. According to the Jordan's theory, it suffices to show that, for every positive integer $l$, $dim(\ker((\lambda I_n+N)^k-\lambda^k I_n)^l)=dim(\ker(N^l))$. In fact, one has $\ker((\lambda I_n+N)^k-\lambda^k I_n)^l=\ker(N^l)$. Indeed, one has $((\lambda I_n+N)^k-\lambda^k I_n)^l=(k\lambda^{k-1})^l N^l(I_n+\alpha N+\beta N^2+\cdots)$ where the matrix $(I_n+\alpha N+\beta N^2+\cdots)$ is invertible.