Prove this matrix has infinite order 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?
 A: 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. 
A: 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.
