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Let $A\in\mathbb{C}^{n\times n}$ be a $n$ by $n$ matrix such that $A^k = I$ for some natural number $k$. Find a nonzero annihilating polynomial of A and prove that A is diagonalizable.

I will say beforehand that this is exam preparation, not homework. Now to my attempt:

Since $A^k = I$, the polynomial $X^k - 1$ annihilates A. Now we still must prove that A is diagonalizable. That is to say, A has $n$ different eigenvalues, or, the polynomial $X^k - 1$ has $n$ distinct zeros.

Is that true so far? How do I proceed from here? Are there better approaches to this?

Thank you.

user153012
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rehband
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    If $k < n$ then $X^k-1$ certainly won't have $n$ roots, but also it is not necessary to have all distinct eigenvalues to be diagonalizable. Just consider the identity matrix! – KCd Aug 27 '14 at 07:13
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    One can prove the contrapositive quickly: If your matrix $A$ is not diagonalizable, its Jordan normal form $PJP^{-1}$ has a Jordan block of size at least $2$; what happens when you raise $A = PJP^{-1}$ to the $k$th power? – Travis Willse Aug 27 '14 at 07:20

2 Answers2

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Hint: The minimal polynomial of $A$ will divide $X^k-1$. Since there are $k$ distinct $k$th roots of unity in $\mathbb{C}$, $X^k-1$ will be a product of distinct linear factors, and hence the minimal polynomial of $A$ will be a product of distinct linear factors. Now, a matrix is diagonalizable if and only if its minimal polynomial is a product of distinct linear factors. Using this result, the matrix $A$ will be diagonalizable.

Branimir Ćaćić
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Ripan Saha
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$A=QJQ^{-1}$ where $J$ is its Jordan canonical form. Therefore: $$ A^k=(QJQ^{-1})^k=QJ^kQ^{-1}=I\Rightarrow J^k=I $$ This will force off-diagonal elements of J to be zero and therefore $A$ is diagonalizable.

Troy Woo
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