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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.

yoyostein
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2 Answers2

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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 $\Bbb C$.

  • Ok I think I understand. – yoyostein Apr 27 '16 at 15:04
  • 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) – yoyostein Apr 27 '16 at 15:10
  • @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. – Ben Grossmann Apr 27 '16 at 15:16
  • @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? – yoyostein Apr 27 '16 at 15:18
  • Yeah, I wrote it backwards the first time – Ben Grossmann Apr 27 '16 at 15:19
  • Oh, my! Sorry! All this time I thought $p$ was the characteristic polynomial! –  Apr 27 '16 at 15:57
  • @Omnomnomnom Since your comments solve the question, and correcting my post would amount to copy them, can you please write them as a full answer, so that I can remove this answer? –  Apr 27 '16 at 16:01
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    @G.S please go ahead and copy them. I don't intend to put an answer together from my phone – Ben Grossmann Apr 27 '16 at 16:04
  • I don't quite understand the meaning of "square free". In my consideration square free means that it is of the form $(x-a_{1})(x-a_{2})..(x-a_{k})$. With all roots distinct. But then why do you say that "it cannot have distinct roots"?. And how do you come to the conclusion that A is diagonalizable in $\mathbb{C}$? . (By the way almost every matrix is diagonalizable in $\mathbb{C}$). The criterion I know says the opposite. That if the minimal polynomial is of the form $(x-a_{1})(x-a_{2})..(x-a_{k})$ then the matrix is diagonalizable! –  Jul 02 '22 at 20:21
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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 $n$.

Clement C.
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student forever
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  • See Arturo's answer: http://math.stackexchange.com/questions/56745/minimal-polynomial-and-diagonalizable-matrix – student forever Apr 27 '16 at 14:52
  • "Equals"? We only have tat it divides it -- it may not be equals to the characteristic polynomial. – Clement C. Apr 27 '16 at 14:54
  • @ClementC. ... since all of roots are distinct and need to be root of minimal polynomial. – student forever Apr 27 '16 at 14:55
  • 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. – Clement C. Apr 27 '16 at 14:57
  • 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. – student forever Apr 27 '16 at 14:59
  • You literally wrote " All roots are different: [how to prove it]. Therefore (/Moreover) minimal polynomial equals to characteristic polynomial." (emphasis mine) – Clement C. Apr 27 '16 at 15:00
  • I really cant understand. Can you edit the answer? – student forever Apr 27 '16 at 15:02
  • OK -- I made a change, feel free to discard it. (Sorry for the extended conversation, and if it feels like nitpicking to you). – Clement C. Apr 27 '16 at 15:04
  • Update: $p$ isn't necessarily the characteristic polynomial – yoyostein Apr 27 '16 at 15:13
  • No, it is necessarily. Since degree of $x^n-1$ is $n$ and $A$ is an $n \times n$ matrix. – student forever Apr 27 '16 at 15:37
  • @studentforever I guess that here the OP is saying that $p(X)=X^n -1$ is not the characteristic polynomial, only that it is a polynomial such that $p(A)=0$. For instance, the characteristic polynomial could very well be $(X-1)^n$ – Clement C. Apr 27 '16 at 18:25