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Given a matrix $A\in M_{n}(\mathbb{C})$ such that $A^8+A^2=I$, prove that $A$ is diagonalizable.

So let $p(x)=x^8+x^2-1$ and we know that $p(A)=0$.

The next step would be to show that the algebric and geometric multipliciteis of all the eigenvalues are equal.

But this polynomial is reducible in a very unpleasent way, so even checking for the minimal polynomial is not an option.

What can I do differently.

user1551
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Yinon Eliraz
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  • You only need to know there are no repeated roots to that polynomial... – Thomas Andrews May 26 '15 at 15:51
  • To be clear: you do not want of an argument saying that the minimal polynomial has distinct roots of multiplicity one, "because it divides $X^8+X^2-1$ which has distinct roots of multiplicity one"? – Clement C. May 26 '15 at 15:51

1 Answers1

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It suffices to show that all the eigenvalues are simple. If $\lambda$ is an eigenvalue with multiplicity $\ge2$ then we have $$\lambda^8+\lambda^2-1=0\tag 1$$ and $$8\lambda^7+2\lambda=0$$ but clearly $0$ isn't an eigenvalue so $$\lambda^6=-\frac14\tag2$$ so form $(1)$ we get $$-\frac14\lambda^2+\lambda^2-1=\frac34\lambda^2-1=0\iff \lambda=\pm \sqrt{\frac{4}{3}}$$ which contradicts $(2)$. Conclude.

  • You are assuming that $\lambda$ is an eigenvalue. How did you get from $\lambda ^8+\lambda ^2 -1$ to $8\lambda ^7+2\lambda=0$? – Yinon Eliraz May 26 '15 at 15:58
  • OR: A muiltiple root of $f(X)=X^8+X^2-1$ is a common root of $f$ and $f'$, but $f'(X)=8X^7+2X$ and $Xf'(X)-8f(X)=2X^2+8$, so a multiple root $\lambda$ would have $\lambda^2=-4$, $\lambda^8=256$, so $f(\lambda)=256-4+1\ne0$, i.e., there is no multiple root. – Hagen von Eitzen May 26 '15 at 15:58