To add the proof for my claim in Todd's answer, which essentially repeats Läuchli's original [1] arguments with minor modifications (and the addition that the resulted model satisfies $DC_\kappa$).
We will show that it is consistent to have a model in which $DC_\kappa$ holds, and there is a vector space over $\mathbb F_2$ which has no linear functionals.
Assume that $M$ is a model of $ZFA+AC$ and that $A$, the set of atoms has $\lambda>\kappa$ many atoms, where $\lambda$ is a regular cardinal. Endow $A$ with a structure of a vector space over $\mathbb F=\mathbb F_2$. Now consider the permutation model $\frak M$ defined by the group of linear permutations of $A$, and by ideal of supports generated by subsets of dimension $\le\kappa$.
Denote by $\operatorname{fix}(X)$ the permutations which fix every element of $X$, by $\operatorname{sym}(X)$ the permutations that fix $X$ as a set, and by $[E]$ the span of $E$ as a subset of $A$. We say that $E\subseteq A$ is a support of $X$ if $\pi\in\operatorname{fix}(E)\Rightarrow\pi\in\operatorname{sym}(X)$.
Final word of terminology, since $A$ will play both the role of set of atoms as well the vector space, given $U\subseteq A$ the complement will always denote a set complement, whereas the direct complement will be used to refer to a linear subspace which acts as a direct summand with $U$ in a decomposition of $A$.
Claim 1: If $E$ is a subset of $A$ then $\operatorname{fix}(E)$ is the same as $\operatorname{fix}([E])$.
Proof: This is obvious since all the permutations considered are linear. $\square$
From this we can identify $E$ with its span, and since (in $M$) the $[E]$ has the same cardinality of $E$ we can conclude that without loss of generality supports are subspaces.
Claim 2: $\frak M$$\models DC_\kappa$.
Proof: Let $X$ be some nonempty set, and $\lt$ a binary relation on $X$, both in $\frak M$. In $M$ we can find a function $f\colon\kappa\to X$ which witness $DC_\kappa$ in $V$.
Since $\frak M$ is transitive, we have that $\alpha,f(\alpha)\in\frak M$ and thus $\langle\alpha,f(\alpha)\rangle\in\frak M$. Let $E_\alpha$ be a support for $\lbrace\langle\alpha,f(\alpha)\rangle\rbrace$ then $\bigcup_{\alpha<\kappa} E_\alpha$ is a set of cardinality $<\kappa^+$ and thus in our ideal of suports. It is simple to verify that this is a support of $f$, therefore $f\in\frak M$ as wanted. $\square$
Claim 3: If $x,y\in A$ are nonzero (with respect to the vector space) then in $M$ there is a linear permutation $\pi$ such that $\pi x=y$ and $\pi y=x$.
Proof: Since $x\neq y$ we have that they are linearly independent over $\mathbb F$. Since we have choice in $M$ we can extend this to a basis of $A$, and take a permutation of this basis which only switches $x$ and $y$. This permutation extends uniquely to our $\pi$.
Claim 4: If $U\subseteq A$ and $U\in\frak M$ then either $U$ is a subset of a linear subspace of dimension at most $\kappa$, or a subset of the complement of such space.
Proof: Let $E$ be a support of $U$, then every linear automorphism of $A$ which fixes $E$ preserves $U$. If $U\subseteq [E]$ then we are done, otherwise let $u\in U\setminus [E]$ and $v\in A\setminus [E]$, we can define (in $M$ where choice exists) a linear permutation $\pi$ which fixes $E$ and switches $u$ with $v$. By that we have that $\pi(U)=U$ therefore $v\in U$, and so $U=A\setminus[E]$ as wanted. $\square$
Claim 5: If $U\subseteq A$ is a linear proper subspace and $U\in\frak M$ then its dimension is at most $\kappa$.
Proof: Suppose that $U$ is a subspace of $A$ and every linearly independent subset of $U$ of cardinality $\le\kappa$ does not span $U$, we will show $A=U$. By the previous claim we have that $U$ is the complement of some "small" $[E]$.
Now let $v\in A$ and $u\in U$ both nonzero vectors. If $u+v\in U$ then $v\in U$. If $u+v\in [E]$ then $v\in U$ since otherwise $u=u+v+v\in[E]$. Therefore $v\in U$ and so $A\subseteq U$, and thus $A=U$ as wanted.$\square$
Claim 6: If $\varphi\colon A\to\mathbb F$ a linear functional then $\varphi = 0$.
Proof: Suppose not, for some $u\in A$ we have $\varphi(u)=1$, then $\varphi$ has a kernel which is of co-dimension $1$, that is a proper linear subspace and $A=\ker\varphi\oplus\lbrace 0,u\rbrace$. However by the previous claim we have that $\ker\varphi$ has dimension $\kappa$ at most, and without the axiom of choice $\kappa+1=\kappa$, thus deriving contradiction to the fact that $A$ is not spanned by $\kappa$ many vectors.
Aftermath: There was indeed some trouble in my original proof, after some extensive work in the past two days I came to a very similar idea. However with the very generous help of Theo Buehler which helped me find the original paper and translate parts, I studied Läuchli's original proof and concluded his arguments are sleek and nicer than mine.
While this cannot be transferred to $ZF$ using the Jech-Sochor embedding theorem (since $DC_\kappa$ is not a bounded statement), I am not sure that Pincus' transfer theorem won't work, or how difficult a straightforward forcing argument would be.
Lastly, the original Läuchli model is where $\lambda=\aleph_0$ and he goes on to prove that there are no non-scalar endomorphisms. In the case where we use $\mathbb F=\mathbb F_2$ and $\lambda=\aleph_0$ we have that this vector space is indeed amorphous which in turn implies that very little choice is in such universe.
Bibliography:
- Läuchli, H. Auswahlaxiom in der Algebra. Commentarii Mathematici Helvetici, vol 37, pp. 1-19.
Best Answer
I'll echo Ben Webster's comment from the other thread, but in the other direction: find some natural commuting operators whose simultaneous eigenspaces are all one-dimensional. This is how, for example, one gets the root space decomposition in Lie theory and the decomposition of spaces of modular forms into eigenforms. (Admittedly, the former is only unique up to a choice of Cartan subalgebra.)
I guess that to get an actual basis instead of a basis-up-to-scalar-multiplication one needs a little more. For cusp forms one usually takes the normalization where the coefficient of $q$ in the Fourier expansion is $1$, whereas for the Lie algebra I guess we take a normalization where the structure constants are as nice as possible.