Suppose $T$ is a linear operator on a $\mathbb{C}$-vector space $V$. Further, assume $T^n$ is the identity operator, for some $n$. Then, $T$ is diagonalizable.

I think there is a proof using Jordan theory, but I wish to find one without using it.

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    $\begingroup$ would you count the fundamental theorem of finitely generated modules over a PID as "Jordan theory?" (probably you should since you can recover the Jordan form from it easily) $\endgroup$
    – hunter
    Feb 9 '15 at 10:04
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    $\begingroup$ An easy extension of this claim is "On a complex vector space, if a power of $T$ is diagonalizable, then so is $T$." $\endgroup$ Feb 9 '15 at 10:17
  • $\begingroup$ I must amend my above claim. You also need this power of $T$ to be not the zero operator. $\endgroup$ Feb 9 '15 at 10:23
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    $\begingroup$ @ChristopherA.Wong I think this is not enough. You need this power of $T$ to have trivial kernel, not just $T^n \neq 0$. $\endgroup$
    – sdcvvc
    Apr 5 '15 at 22:11
  • $\begingroup$ You are probably right. $\endgroup$ Apr 6 '15 at 0:01

$$T^n=I\implies (T-I)(T^{n-1}+\ldots+T+I)=0\implies $$

the minimal polynomial of $\;T\;$ divides $\;T^n-I\;$, all of which roots are different (why?). Since you're working on an algebraic closed field this means all the roots of that minimal polynomial are in the field and thus the min. pol. factors in different linear factors $\;\iff T\;$ is diagonalizable.

  • $\begingroup$ nice proof. Any good reference on the theorem you invoked? $\endgroup$
    – John
    Dec 20 '19 at 3:29

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