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What techniques are there for ensuring nonnegativity of various entries of matrix powers?

Specific Question: Consider a matrix $A\in SL_2(\mathbb R)$. Let $(A^n)_{i,j}$ denote the $(i,j)$ entry of the matrix power $A^n$. Under what conditions on $A$ does the following hold: $$ \text{sgn}\ (A^n)_{i,j}=(-1)^{j+1} $$

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    $\begingroup$ How is this "non-negativity"? $\endgroup$
    – Igor Rivin
    Commented Jul 7, 2015 at 19:58
  • $\begingroup$ More precisely, the question is concerned with control of the signs of the matrix entries. $\endgroup$
    – pre-kidney
    Commented Jul 7, 2015 at 21:53
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    $\begingroup$ If one diagonalises $A$ using a basis of (possibly complex) eigenvectors, then the coefficients of $A^n$ can be computed quite explicitly. (The parabolic case when $A$ has a repeated eigenvalue can be treated separately, and is actually rather easier than the general case as $A^n$ basically depends linearly on $n$ in that case.) $\endgroup$
    – Terry Tao
    Commented Jul 7, 2015 at 23:04

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To follow up on Terry's comment (actually, I did the computation before seeing it, but whatever): in the parabolic case, where the matrix has the form

$$A = \begin{pmatrix} d & -b \\ -c & a\end{pmatrix} \begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix} \begin{pmatrix} a & b \\ c & d\end{pmatrix},$$ Then

$$A^n = \begin{pmatrix}1 + c d n x & d^2 n x\\ - c^2 n x & 1 - c d n x\end{pmatrix}.$$ So, if $c \in \mathbb{R},$ then $x < 0$ (from the bottom left), and so $c < 0$ from the bottom right, and we are fine (that is, if $|c d x| > 1,$ then any $a, b$ works). Since $x$ is real, it is clear from the bottom left that $c$ is also, so this finishes the parabolic case.

In the hyperbolic case, $$A = \begin{pmatrix} d & -b \\ -c & a\end{pmatrix} \begin{pmatrix} x & 0 \\ 0 & 1/x\end{pmatrix} \begin{pmatrix} a & b \\ c & d\end{pmatrix},$$ where $x>1.$

and $$A^n = \begin{pmatrix}=b c x^{-n} + a d x^n & b d x^{-n}(-1 + x^{2 n})\\ a c(x^{-n}+ x^n) & ad x^{-n} - b c x^n\end{pmatrix}.$$ For $n \gg 1,$ we have $$A^n \sim x^n \begin{pmatrix} ad & b d\\ -ac & -bc\end{pmatrix},$$ which implies that either $a < 0, b>0, c> 0, d<0,$ or $a > 0, b< 0, c<0, d> 0.$ Neither of which is compatible with the base case $n=1$ satisfying your condition.

In the elliptic case, the eigenvalues are either roots of unity, so $A^n = I,$ infinitely often, or not, in which case $A^n$ is very close to $I$ infinitely often, so either way, your condition fails.

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