If I have (finite) $k$ vectors, $u_1,...,u_k\in\mathbb{R}^N$ that are in general linearly dependent is it possible to take positive affine transformations of the form: $$u'_i=\alpha_i u_i +\beta_i\vec{1}_N$$ $$~\text{with}~~\alpha\in\mathbb{R}_{++} \text{ and } \beta\in\mathbb{R}$$
(The idea is each $u$ is an expected utility function and these transformations preserve the preference ordering on lotteries)
So that the collection of the transformed vectors are linearly independent?
I'm working on a project and this would be a useful condition to have. Any help is appreciated!
We can assume that there are 2 dimensions (call them $b$ and $w$) such $\forall u_i$ the $b^{th}$ entry of vector is the largest and the $w^{th}$ entry is the smallest.
It seems like I should be able to rescale such that for every $u$ has $u'(w)=-1$, which let's me focus on convex combinations rather than linear combinations. This seems useful though not quite enough.
This can be only done if the following two conditions are satisfied: $$ \dim \mathrm{span}(v_1\dots v_{k}) = k-1 $$ and $$ e\not\in \mathrm{span}(v_1\dots v_{k}). $$ where $e=(1,\dots,1)^T$.
If $\dim \mathrm{span}(v_1\dots v_{k}) < k-1$, then $$ \dim \mathrm{span}(v_1\dots v_{k}, e) < k, $$ which implies that there are not $k$ linear independent vectors in that set.
The same can be said, if $e\in \mathrm{span}(v_1\dots v_{k})$.
Now let me argue that the conditions above are also sufficient: Let $i$ be an index such that $v_i \in \mathrm{span}( \{v_1\dots v_k\}\setminus v_i)$. Then $v_i+e$ is not in that span. Moreover, $$ \dim \mathrm{span}( \{v_1\dots v_{k}\}\setminus \{v_i\} \cap \{v_i+e\})=k, $$ hence these vectors are linearly independent. Thus the vectors $$ u'_j = u_j \quad j\ne i, \ u'_i = u_i + e $$ satisfy the requirements.