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Let $F$ be a field and $V$ a finite dimensional vector space over $F$. Let $T:V\to V$ be a linear transformation such that for every $v\in V$ there exists $n\in\Bbb{N}$ with $T^nv=v$.

  1. Show that it $F=\Bbb{C}$, then $T$ is diagonalizable.
  2. Show that if $\mathrm{char}(F)>0$, then there exists a non diagonalizable $T$ satisfying the above hypothesis.

Let $\{\alpha_1,\cdots,\alpha_k\}$ be a basis for $V$ over $F$. Let $n_i\in \Bbb{N}$ be an integer such that $T^{n_i}\alpha_i=\alpha_i$. Let $N=\max\{n_i:1\le i\le k\}$. Then $T^Nv=v$ for all $v\in V$. I could do this much. What next?

QED
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    Proofs that it is diagonalizable over $\mathbb{C}$ were done today here, probably there are previous questions as well. For the example in positive characteristic take $A=\begin{pmatrix}1&1\0&1\end{pmatrix}$ in characteristic $p$. Then $A(v_1,v_2)^T=(v_1+v_2,v_2)^T=r(v_1,v_2)^T$ only when $r=1$ and $v_2=0$. So, no basis of eigenvectors. And $A^p(v_1,v_2)^T=(v_1+pv_2,v_2)^T=(v_1,v_2)^T$. –  May 05 '18 at 05:55
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    Shouldn't you take the lcm instead of max? – Hagen von Eitzen May 05 '18 at 06:04

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