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Consider the matrix:

$M = \begin{pmatrix} 8&-3\\ 3&-1 \end{pmatrix}$

What are ways of showing $M$ has infinite order? I guess one is to find a closed form for the powers of $M$, but is there a more elegant way?

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    Do you mean, how do you show that $M$ does not have finite multiplicative order? The characteristic polynomial is $x^2 +7x +1$, so not all eigenvalues are roots of unity (the sum of the two roots has complex norm 7, but two roots of unity cannot have a sum with complex norm greater than 2); hence the minimal polynomial cannot divide $x^n-1$ for any $n\gt 0$, so we never have $M^n = I$ for any $n\gt 0$.2011-11-29
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    yes, that no power of $M$ equals the identity matrix.2011-11-29
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    @Arturo I didn't notice your comment when I wrote the answer. Apologies...2011-11-29
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    @Srivatsan: No problem. Happens to all of us.2011-11-29

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