Sum of bases of a vector space for an arbitrary permutation. I have two lists of $n$ distinct vectors $L_1$ and $L_2$. Both lists are bases for a vector space $\mathbf{V}$ over a field $\mathbb{F}$ with dimension $d>1$.
Let $L_1=(a_1,\ldots,a_n)$ and $L_2=(b_1,\ldots,b_n)$, where $\{a_i;b_j\}\in \mathbf{V}$ are basis vectors. Let
$$L_1+\sigma(L_2)=(a_1+b_{\sigma(1)},\ldots,a_n+b_{\sigma(n)})$$I need to prove or disprove wether if every $L_1+\sigma(L_2)$ is also a basis for $\mathbf{V}$ for some $\sigma\in\mathbf{S}_n$. I made a similar question for the case in wich $\sigma$ is the identity, so i’ll post the link here. This one would be a sort of generalization of the previous question, wich showed that, at least for the identity in $\mathbf{S}_n$, the sum of bases might not be a base.
link to original question: Sum of bases for a vector space
 A: Actually, from my comments, I think I've talked myself into thinking this is not true, and cannot be obviously salvaged.
Take $L_1 = (a_1, \ldots, a_n)$ to be any basis, and form $L_2$ by $b_i = -a_i$. then, for any $\sigma \in S_n$. Then $L_1 + \sigma(L_2)$ is not a basis.
Why? If $\sigma$ is not a derangement, then $L_1 + \sigma(L_2)$ contains a $0$ vector, which precludes it from being a basis. Otherwise, map this basis under the coordinate vector map with respect to $L_1$. Then, every coordinate vector contains exactly one $1$, exactly one $-1$, and $0$s elsewhere. If $L_1 + \sigma(L_2)$ were a basis, we would expect these coordinate vectors to span $\Bbb{F}^n$. But, this is not the case, as all these coordinate vectors have their entries sum to $0$. Either way, $L_2 + \sigma(L_1)$ is not a basis.
EDIT: Or, slightly more generally, one could also just take any bases $L_1$ and $L_2$ such that the sum of all vectors from both bases is $0$. Then, the sum of $L_1 + \sigma(L_2)$ will be $0$, which would contradict linear independence.
