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Let $A$ be an $n\times n$ matrix that satisfies the polynomial $p(x)=x^n-1$ (Update: $p$ is not necessarily the characteristic polynomial), that is $A^n-I=0$.

I am trying to prove that $A$ is diagonalizable over $\mathbb{C}$. (I am not sure if this is true yet.)


What I tried:

Let $\mu(x)$ be the minimal polynomial of $A$. Then $\mu(x)$ divides $p(x)$. Since $p(x)$ has no repeated roots, $\mu(x)$ has no repeated roots either.

Fact: The roots of $\mu(x)$ are precisely the eigenvalues of $A$.

Thus $A$ has distinct eigenvalues. If $A$ has $n$ distinct eigenvalues, then we are done.

However, $\mu(x)$ may be of degree less than $n$... I am stuck here.

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Well, the point is: regardless of $p$ being the actual characteristic polynomial, $\mu$ divides $p$ and $p$ is square-free. Since $$p(x)=(x-\lambda_1)\cdots (x-\lambda_n)$$

with $\lambda_i=\lambda_j\Rightarrow i=j$, the fact that $\mu\mid p$ yields that $\mu$ is square-free, i.e. it cannot have distinct roots. So the matrix is diagonalizable in $\Bbb C$.

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  • $\begingroup$ Ok I think I understand. $\endgroup$ – yoyostein Apr 27 '16 at 15:04
  • $\begingroup$ Wait... How do we know the eigenvalues are precisely the roots of $p$? ($p$ is not necessarily the characteristic polynomial here, sorry for the confusing notation) $\endgroup$ – yoyostein Apr 27 '16 at 15:10
  • $\begingroup$ @yoyostein we know that every eigenvalue is a root of $p$, but not that every root of $p$ is an eigenvalue. This is enough, however, for us to know that if $p$ has no repeated roots, there are no repeated eigenvalues in the minimal polynomial. $\endgroup$ – Omnomnomnom Apr 27 '16 at 15:16
  • $\begingroup$ @Omnomnomnom Shouldn't it be every eigenvalue is a root of the minimal polynomial and thus $p$, but not every root of $p$ is an eigenvalue? $\endgroup$ – yoyostein Apr 27 '16 at 15:18
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    $\begingroup$ @G.S please go ahead and copy them. I don't intend to put an answer together from my phone $\endgroup$ – Omnomnomnom Apr 27 '16 at 16:04
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All roots are different: use $(x^n-1,nx^{n-1})=1$, and there are $n$ of them. Therefore(/Moreover) the minimal polynomial is equal to the characteristic polynomial, since they have same roots and the latter has degree $n$.

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  • $\begingroup$ See Arturo's answer: math.stackexchange.com/questions/56745/… $\endgroup$ – student forever Apr 27 '16 at 14:52
  • $\begingroup$ "Equals"? We only have tat it divides it -- it may not be equals to the characteristic polynomial. $\endgroup$ – Clement C. Apr 27 '16 at 14:54
  • $\begingroup$ @ClementC. ... since all of roots are distinct and need to be root of minimal polynomial. $\endgroup$ – student forever Apr 27 '16 at 14:55
  • $\begingroup$ Yes, but that relies on the fact that you have $n$ distinct roots. Phrased as in your answer, this seems to read "if the minimal polynomial has all roots that are different, then it is equal to the characteristic polynomial" (which is false). Writing "Therefore, since there are $n$ of them (and the characteristic polynomial has degree $n$), [...]$ would remove the ambiguity. $\endgroup$ – Clement C. Apr 27 '16 at 14:57
  • $\begingroup$ I was supposing we are talking under this question and I can't see how someone understand "if the minimal polynomial has all roots that are different, then it is equal to the characteristic polynomial" because there is no "if the minimal polynomial has all roots that are ..." statement. $\endgroup$ – student forever Apr 27 '16 at 14:59

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