If the characteristic function of $A$ is $(\lambda - 1)^n$ then $A^k$ is similar to $A$ Let $A \in M_n(k)$ (where $k$ is a field of characteristic $0$) and suppose the characteristic function of $A$ is $(\lambda - 1)^n$
Prove that for any positive integer $k$, $A^k$ is similar to $A$.

I tried to prove that $A^k$ is similar to $A$ by Jordan form, but it seems to make it more complicated.
Any help will be appreciated.
 A: It suffices to show that $A$ and $A^k, k\ge 1$ have the same invariant factors. Each invariant factor of $A$ has the form $(x-1)^m$. Consider a cyclic subspace $W\subset F^n$ with cyclic vector $\alpha$ and minimal polynomial $(x-1)^m$. It is clear that $W$ is also $A^k$-invariant. It suffices to show that $\alpha$ is also a cyclic vector for $A^k$, namely, $W=\{f(A^k)\alpha: f\in F[x]\}$. If $f(A^k)\alpha=0$, then $(x-1)^m|f(x^k)$. Assume that $f=(x-1)^dg$ with $g(1)\ne 0$.We have $f(x^k)=(x^k-1)^dg(x^k)=(x-1)^dh^dg(x^k)$, where $h=(x^k-1)/(x-1)$. If $(h,(x-1))=1$, which is true $char(F)=0$, it should be easy to show that $d\ge m$ from $(x-1)^m|f(x^k)$. In particular, $(x-1)^m|f$. Thus the space $\{f(A^k)\alpha: f\in F[x]\}$ has dimension $m$, which implies that $W=\{f(A^k)\alpha: f\in F[x]\}$ and $A^k|_W$ also has invariant factor $(x-1)^m$. It is easy to realize the explicit similarity directly. The assumption shows that if we denote $\alpha_i=(A-I)^i\alpha$, then $\mathcal{B}=\{\alpha_1,\dots,\alpha_m\}$ is a basis of $W$. The above argument shows that if we denote $\beta_i=(A^k-I)^i\alpha$ and $\mathcal{B}'=\{\beta_1,\dots,\beta_m\}$. Then $\mathcal{B}'$ is also a basis of $W$. It is easy to see that $[A]_{\mathcal{B}}=[A^k]_{\mathcal{B'}}$, which equals the Jordan block.
