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. 



Invertible combinations of linear maps on infinite-dimensional vector spaces
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](https://en.wikipedia.org/wiki/Vector_fields_on_spheres).) 


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).


