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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, \forall x \in \mathbb{R}^3 $, $\delta_n$ 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 $\sqrt{2}$).

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 $\sqrt{2}$).

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 $|\det(A)|>1$. 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 $|\det(A)|>1$) ?

For example if $A$ is expanding, i.e. there exists $ \alpha >1 $ such that $ \Vert A x \Vert_2 \geq \alpha \Vert x \Vert, \forall x \in \mathbb{R}^3 $, $\delta_n$ 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 $\sqrt{2}$).

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, \forall x \in \mathbb{R}^3 $, $\delta_n$ 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 $\sqrt{2}$).

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 $A$. 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 $A$) ?

For example if $A$ is expanding, i.e. there exists $ \alpha >1 $ such that $ \Vert A x \Vert_2 \geq \alpha \Vert x \Vert, \forall x \in \mathbb{R}^3 $, $\delta_n$ 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 $\sqrt{2}$).

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 $|\det(A)|>1$. 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 $|\det(A)|>1$) ?

For example if $A$ is expanding, i.e. there exists $ \alpha >1 $ such that $ \Vert A x \Vert_2 \geq \alpha \Vert x \Vert, \forall x \in \mathbb{R}^3 $, $\delta_n$ 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 $\sqrt{2}$).