Invertible combinations of linear maps on infinite-dimensional vector spaces

Question

Let $V$ be a real infinite-dimensional vector space of cardinality $\kappa$. Does there exist a set $\Omega$ of cardinality $\kappa$ of linear maps from $V$ to $V$ such that for every $n\geq 1$, every nonzero vector $(x_1,\ldots, x_n) \in \mathbb{R}^n$, and distinct $A_1,\ldots, A_n \in \Omega$, the map $$x_1A_1+\cdots+x_nA_n$$ is an isomorphism from $V$ to $V$?

When $V$ is an infinite-dimensional separable Hilbert space, there exists a countable set $\Omega$ such that $$U^2=-I,~UV+VU=0,~~~~~~~~~~~~~~~(1)$$ for all distinct $U,V \in \Omega$. Then every nontrivial linear combination of elements in $\Omega$ is invertible. The question is that if generally one can forego the conditions (1) but increase the cardinality of the set and still have invertible nontrivial linear combinations.

Answer 1

I think this is true for any infinite $\kappa$, which is presumably the case you are interested in. (For finite $\kappa$, the situation is more interesting but well understood: see for instance here.)

If $\kappa$ is infinite, you can construct such a family essentially by using the Clifford algebra. Namely, take a real vector space $W$ of dimension $\kappa$. Equip it with a positive definite quadratic form, and generate the Clifford algebra $V:=Cliff(W)$, which will be the union of Clifford algebras corresponding to finite-dimensional subspaces of $W$. It is easy to see that $\dim V=\dim W$, and that (say, left) action of $W$ on $V$ is invertible. Now let your family of operators be the action of some basis vectors of $W$.

Remark. I find it interesting that if you ask the same question over the field $\mathbb{C}$, you must assume that $\kappa\ge c$ instead of $\kappa$ being infinite (and then you can use fields of rational functions instead of Clifford algebras).