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Let $V$ a finite dimension space over $\mathbb{C}$ and $T:V\to V$, a linear transformation such that $T^k = Id$ for $k\ge 1$. Prove that $T$ is diagonalizable.

I'd be glad for an hint. How do I approach this question?

Thanks.

AlonAlon
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    Have you already studied Jordan's Canonical Forms? – Timbuc Dec 13 '14 at 13:28
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    Diagonalizable over $\mathbb C$? – Git Gud Dec 13 '14 at 13:29
  • Yes, although I'm not sure this assignment is related to this subject, but yes. – AlonAlon Dec 13 '14 at 13:29
  • @GitGud, Actually yes. How did you figure it out? Anyway, let me edit the question. – AlonAlon Dec 13 '14 at 13:29
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    @AlonAlon JNF helps immensely here. How did I figure it out? It would be false over the real numbers, take for instance $T=[i]$ and $k=4$. – Git Gud Dec 13 '14 at 13:31
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    Assuming that you are working over $\Bbb{C}$ then the duplicate from yesterday covers this (learn to look for dups, please). If not, the claim is false, so it is not too presumptious to think that you are working over $\Bbb{C}$ :-) – Jyrki Lahtonen Dec 13 '14 at 13:32

1 Answers1

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Hints:

1) The operator $\;T\;$ is regular (i.e., $\;\det T\neq 0\iff $ all its eigenvalues are non-zero)

2) Suppose $\;\lambda\neq 0\; $ , then over a field of characteristic $\;\neq 2\;$ (which contains all the eigenvalues of $\;T\;$ , say $\;\Bbb C\;$ )

$$\begin{pmatrix}\lambda&1&0\ldots&0\\ 0&\lambda&\ldots&\ldots\\ \ldots&\ldots&\ldots&\ldots\end{pmatrix}^2=\begin{pmatrix}\lambda^2&2\lambda&0\ldots&0\\ 0&\ldots&\ldots&\ldots\\ \ldots&\ldots&\ldots&\ldots\end{pmatrix}$$

3) Deduce that $\;T\;$ has to be diagonalizable

Timbuc
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