Proposition to prove or give counterexample regarding matrix and its inverse

Question

Suppose that $\mathbb H \in M_n(\mathbb R)$ is a set of real $n \times n$ matrices and $A \in M_n(\mathbb R)$ is invertible such that $AH \in \mathbb H$ for every $H \in \mathbb H$. Does it follow that $A^{-1}H \in \mathbb H$ for every $H \in \mathbb H$? Prove or give a counterexample.

Now, I tried for a counterexample and what is better than to find one in $2 \times 2$ matrices? After hours I decided that it cannot be done. Taking the general form for a matrix $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and for its inverse, I took a general $H$ matrix, made the multiplications $AH,A^{-1}H$ and tried to make some elements $0$ or $1$ both in $A,AH$ while forcing them to have different values in $A^{-1}H$. I couldn't get it done without having $det(A)=0$, which is not possible since $A$ is invertible. So, I need to constrict a proof but I don't know where to begin. Any help?

Answer 1

Let $n=1$ so our $1\times 1$ matrices are just numbers and let $\mathbb{H}$ be real numbers greater than or equal to 1, and $A$ be the number 2. Then $A$ is invertible and it carries elements of $\mathbb{H}$ to $\mathbb{H}$, but $A^{-1}$ carries $1$ to $1/2$ which is not in $\mathbb{H}$.

Answer 2

Fix $A=\pmatrix{ 1 & 1 \\0&1}$, and set $$ \mathbb H = \{ A^n, n \in \mathbb N\} = \left\{ \pmatrix{1 & n \\ 0 & 1}, n\in \mathbb N\right\} $$ Then $A\mathbb H \subset \mathbb H$. It holds $A^{-1}=\pmatrix{1 & -1 \\ 0 & 1}$ and $A^{-1}A\ne \mathbb H$.