Exponential growth of shortest vector norm for successive lattices corresponding to powers of a matrix

Question

Let $A\in M_{2\times 2}(\mathbb{Z}) $ be a two by two integer matrix such that $0,\pm 1$ are not eigenvalues of $A$ and $\left|\det(A)\right|>1$. I am interested in the growth of the norm shortest vector of the lattice $A^n \mathbb{Z}^2$ when $n$ is big. More precicely, let $$ \delta_n:=\min\{ \Vert x \Vert_2 : x \in A^n\mathbb{Z}^2 \setminus \{0\} \}. $$ I would like to understand for which matrices $A$ there exists $\alpha >1 $ such that,

\begin{equation} \tag{$*$} \label{exp_growth} \alpha^n = \mathcal{O} ( \delta_n), \end{equation}

i.e. $\delta_n$ has at least exponential growth. In particular is it true for all such $A$ (such that $0,\pm 1$ are not eigenvalues of $A$ and $\left|\det(A)\right|>1$) ?

For example if $A$ is expanding, i.e. there exists $ \alpha >1 $ such that $ \Vert A x \Vert_2 \geq \alpha \Vert x \Vert_2, \forall x \in \mathbb{R}^2 $, $\delta_n$ grows exponentially.

With some more thought one can construct examples of matrices which are not expanding but still \eqref{exp_growth} holds. Consider for example the matrix $$ B = \begin{bmatrix} 3 & 1 \\ 1 & 1 \end{bmatrix}. $$ Then $B^2 = 2 U, \,\, U \in GL(2,\mathbb{Z})$, hence $B^{2n} \mathbb{Z}^2 = 2^n \mathbb{Z} \times 2^n \mathbb{Z}$, therefore $ \delta_{2n} = 2^{n} $, hence $\delta_n$ grows like $2^{\frac{n}{2}}$.

Finally Notice that Minkowski's bound for the shortest vector problem of a lattice gives that $$ \delta_n \leq \sqrt{2}\det(A)^{\frac{n}{2}} $$ which means that for the matrix $B$ the growth of $\delta_n$ is the maximum possible (up to a factor of $\sqrt{2}$).

Answer 1

Since $|\det A|>1$, $A$ has at least one eigenvalue $\lambda$ with $|\lambda|>1$.

If $A$ has another eigenvalue $\mu$ with $1<|\mu|<|\lambda|$, then let $u,v$ be eigenvectors for $\mu,\lambda$, respectively, and let $x\ne0$ be in ${\bf Z}^2$. Then $x=\alpha u+\beta v$ with $\alpha,\beta$ not both zero, so $A^nx=\alpha\mu^nu+\beta\lambda^nv$, and $\|A^nx\|_2$ grows at least as fast as $C|\mu|^n$ for some constant $C>0$.

If $A$ has an eigenvalue $\mu$ with $|\mu|<1$, then $\mu$ is an algebraic integer with $0<|\mu|<1$, so $\mu$ is irrational, and so is $\lambda$. Then the eigenvectors $u,v$ corresponding to $\mu,\lambda$ must have irrational entries. Then for $x\ne0$ in ${\bf Z}^2$, $x=\alpha u+\beta v$, we must have $\beta\ne0$ (else, $x$ would be an eigenvector for $\mu$). $A^nx=\alpha\mu^nu+\beta\lambda^nv,$ so $\|A^nx\|_2$ grows at least as fast as $C|\lambda|^n$ for some constant $C>0$.

Now suppose $|\mu|=|\lambda|$ but $\mu\ne\lambda$. Then (with notation as previously) $A^nx=\lambda^n(\alpha(\mu/\lambda)^nu+\beta v)$, aand $\alpha(\mu/\lambda)^nu+\beta v\ne0$ since $u,v$ are linearly independent, so $\|A^nx\|_2$ grows at least as fast as $C|\lambda|^n$ for some constant $C>0$.

If $A$ has a repeated eigenvalue, and is diagonalizable, then $A=\lambda I$, and $A^nx=\lambda^nx$, and $\|A^nx\|=|\lambda|^n\|x\|$.

Finally, if $A$ has a repeated eigenvalue, and is not diagonalizable, then $A^nx=\lambda^n(\alpha v+\beta w)+\beta n\lambda^{n-1}v$ where $w$ is a generalized eigenvector. Again, $\alpha v+\beta w\ne0$, so $\|A^nx\|$ grows as $C|\lambda|^n$ for some $C>0$.