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Let $A\in \mathbb{C}^{n^2}$ such that $A^m=I_n$, for some $m,n\in \mathbb{N}$.

Please prove that $A$ is diagonalizable.

Now let $B\in \mathbb{C}^{n^2}$ such that $B^m=B$, for some $m\in \mathbb{N}$ such that $m>1$.

Please prove that $B$ is diagonalizable.

1 Answers 1

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Hints:
(1) A matrix $M$ is diagonalizable if and only if it's minimal polynomial $m_M(x)$ factors completely over the field (here $\mathbb{C}$) into distinct linear factors.The relevant theorem is here
(2) If $p(x)$ is a polynomial such that $p(M)=0$ then $m_M(x)|p(x)$.
(3) If we have two polynomials $f(x),g(x)$, $f(x)|g(x)$ and $g(x)$ factors completely into distinct linear factors, then so does $f(x)$.
(4) Observe that if $p(x)=x^m-1$ and $q(x)=x^m-x$ then $p(A)=0=q(B)$

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    The proof of (1) is ridiculously long and it's not on my notes, so I'm guessing I'm supposed to prove it in another way. My idea is to show that the matrices' Jordan Normal Form is a diagonal matrix. To do that it suffices to prove that $r(M-\lambda I)=r((M-\lambda I)^2)$ for every eigenvalue $\lambda$.2012-12-15
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    How do compute Jordan normal form? do you know that the size of the largest block corresponding to the eigenvalue $\lambda$ is equal to the power of $x-\lambda$ in $m_M(x)$?2012-12-15
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    I don't know that. In my notes the size of the largest block is the smallest natural $k$ such that $r((M-\lambda I)^k)=r((M-\lambda I)^{r+1})$, hence my comment above.2012-12-15
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    What is $r(\cdot)$? the rank?2012-12-15
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    Yes, it is the rank. Sorry.2012-12-15
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    I meant $r((M-\lambda I)^k)=r((M-\lambda I)^{k+1})$. Sorry again.2012-12-15
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    Just thought of another approach: Let $J_\lambda$ be a Jordan block corresponding to $\lambda$. Show that there exists $n$ such that $J_\lambda^n$ is diagonal if and only if $J_\lambda$ is $1\times 1$.2012-12-15
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    I don't think that's true. Consider $\lambda=0$, its Jordan block is nilpotent, therefore it is diagonal and it doesn't have to be a $1\times 1$ block. EDIT: I wanna thank you for your time before you get fed up with this problem.2012-12-15
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    Oh, nevermind, $\lambda$ can't be zero. Let me think about your idea.2012-12-15
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    I got another approach for the first part: Let $J_A$ be 'the' Jordan Normal Form of $A$. We have $P^{-1}AP=J_A$, for some $P$. It follows that $(J_A)^m=I_n$. From the last equality it will follow that $J_A$ is a diagonal matrix.2012-12-15
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    That's exactly what I said, formulated in reverse :-) You still need to prove that $J_A^m=I_n$ implies $J_A$ is diagonal.2012-12-15
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    I noticed what you said is basically the same. The fact that $J_A^m=I_n$ implies $J_A$ diagonal follows from this identity easily: http://en.wikipedia.org/wiki/Jordan_Normal_Form#Powers Thanks.2012-12-15