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let $V$ be a $\mathbb{R}$-vectorspace with $dim V < \infty$ and $F$ an endomorphism of V with $F^3 = F$.

Show: F is diagonalisable.

  • $F^3 = F$ is equivalent to $F^3 - F = 0$.

  • Now I know that $F$ is diagonalisable if the minimal-polynomial has linear-factors and every eigen-value is only a singular-null of the polynomial.

  • Now I should be able to conclude from $F^3 - F = 0$ that the minimal-polynomial... ?

Here I stuck :(

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    $\begingroup$ can you factorize $x^3-x$? $\endgroup$ – user87543 May 12 '14 at 9:26
  • $\begingroup$ Hint: The minimal polynomial of $F$ divides any polynomial $p$ with $p(F)=0$. $\endgroup$ – Sebastian Schoennenbeck May 12 '14 at 9:27
  • $\begingroup$ @ Sebastian Schoennenbeck - Thank you! Is this from definition? $\endgroup$ – Vazrael May 12 '14 at 10:07
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Let $p(x)=x^3-x=x(x-1)(x+1)$ and note that the minimal polynomial of $F$ divides $p(x)$. But $p(x)$ is a product of linear factors so the minimal polynomial itself must be a product of linear factors.

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