Linearly independent vectors generated from matrices

Question

I came across this question recently:

Do there exist $n$ real $n\times n$ matrices $A_1, A_2, \cdots, A_n$, such that for any $n$-dimensional non-zero vector $v$, $A_1 v, A_2 v, \cdots, A_n v$ are always linearly independent?

It seems like a usual linear algebra question, but I can't think of any idea to approach it, and I can't come up with a counterexample either. Is there anything I've missed?

Any help would be appreciated.

Edit: So for $n=2$, this is obvious as we can just take two rotation matrices with different angles. The resulting two vectors will always be on different lines. For $n>2$ however, rotation won't work as there'll be an axis of rotation, on which the vectors are eigenvectors, i.e., on the same line after rotation.

Answer 1

Note that $(1)\iff(2)\iff(3)$ in the following:

1. $A_1v,A_2v,\ldots,A_nv$ are linearly independent whenever $v\ne0$.
2. $\sum_{i=1}^nc_iA_iv\ne0$ whenever $v\ne0$ and $(c_1,c_2,\ldots,c_n)\in\mathbb R^n\setminus0$.
3. $\sum_{i=1}^nc_iA_i$ is invertible whenever $(c_1,c_2,\ldots,c_n)\in\mathbb R^n\setminus0$.

So, you are essentially asking whether there exists an $n$-dimensional subspace of $n\times n$ real matrices in which all nonzero members are invertible. This is a non-trivial but solved problem: the answer is “yes” if and only if $n=1,2,4,8$. See here for some references.