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 :(

  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.