As you can embedd the complex numbers or the quaternions in $M_m(\mathbb{R})$ I was wondering if their is a nice criteria for a given subset $\left\{A_1,...,A_n \right\} \subset M_m(\mathbb{R})$ such that it satisfies the following criteria.
$\sum_{i = 1}^{n}x_i A_i$ is invertible if and only if $0 \neq (x_1,...,x_n) \in \mathbb{R}^n$.
Your criterion is true if and only if your set $\{A_1, \ldots, A_n\}$ consists of a single invertible matrix.
Namely, if some $A_j$ is not invertible then the nontrivial linear combination $$0\cdot A_1 + \cdots + 0 \cdot A_{j-1} + 1 \cdot A_j + 0\cdot A_{j+1} + \cdots + 0\cdot A_n = A_j$$ is not invertible.
On the other hand, if there are two different invertible matrices $A_i$ and $A_j$ then
$$\alpha A_i + \beta A_j \text{ is invertible } \iff A_{i}^{-1}(\alpha A_i + \beta A_j) \text{ is invertible } \iff \alpha I + \beta A_{i}^{-1}A_j \text{ is invertible } $$ which is true if and only if $\frac{\alpha}{\beta} \notin \sigma( A_{i}^{-1}A_j)$, which is nonempty. Hence, if you pick $\alpha \in \sigma( A_{i}^{-1}A_j)$ and $\beta = 1$ then $\alpha A_i + \beta A_j$ is not invertible.