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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 $n\gt 0$. –  Arturo Magidin Nov 29 '11 at 4:02
    
yes, that no power of $M$ equals the identity matrix. –  Learner Nov 29 '11 at 4:03
    
@Arturo I didn't notice your comment when I wrote the answer. Apologies... –  Srivatsan Nov 29 '11 at 4:13
    
@Srivatsan: No problem. Happens to all of us. –  Arturo Magidin Nov 29 '11 at 4:14
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2 Answers

One approach is to compute the eigenvalues. Writing the characteristic equation, $\lambda^2 - 7 \lambda + 1 = 0$, we find the eigenvalues $$ \lambda_1 = \frac{7 + 3\sqrt{5}}{2}, \quad \lambda_2 = \frac{7 - 3\sqrt{5}}{2} . $$ Check that $0 < \lambda_2 < 1 < \lambda_1$.

The matrix $A^n$ has eigenvalues $\lambda_1^n$ and $\lambda_2^n$. For any $n \geqslant 1$, we have $0 < \lambda_2^n < 1 < \lambda_1^n$; in particular, $A^n$ cannot be the identity matrix.

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My first attempt was this: if the determinant of a real matrix is different from $\pm 1$, then $A$ has infinite order. But in this case, the question is framed in such a way to make $\det A = 1$. –  Srivatsan Nov 29 '11 at 4:19
    
+1 Indeed, the determinant seems to have been tailored to not make this exercise too easy. Consequently your eigenvalues must be units of the ring of integers of $\mathbf{Q}(\sqrt5)$. A closer look shows that they are the fourth power of the golden ratio and its reciprocal. –  Jyrki Lahtonen Nov 29 '11 at 8:32
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The trace of $M$ (which equals the sum of the eigenvalues) is $7$, which means that its two eigenvalues cannot both be roots of unity (since a complex root of unity has complex norm $1$, the sum of two complex roots of unity has norm less than or equal to $2$).

But if $M^k = I$, then $M$ satisfies the polynomial $x^k-1$, which means that the minimal polynomial of $M$ divides $x^k-1$. Since the roots of the minimal polynomial are the (complex) eigenvalues of $M$, a necessary condition for $M$ to have finite order is that its eigenvalues be complex roots of unity. As this is not the case, $M$ cannot have finite order.

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+1 Nice way to conclude that the roots can't both be roots of unity. –  Srivatsan Nov 29 '11 at 4:16
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