Let V be a finite dimensional vector space over a field F. Let $V^*$ be the dual space of $V$. Consider the dual basis $f_i$ given by
$$
f_i(v_j)=1\;\text{if}\;i=j\;\text{and}\;0\;\text{if}\;i\neq j
$$
Consider the following expression $a_1f_1+….+a_nf_n=0$
Then evaluating the maps at each basis $v_i$ we get $a_i=0$
The question is, are the scalars assumed to be constant?
Edit:
Read the following edit, please:
what I'm essentially asking is what if the scalars depended on the input vector, would the proof still hold?
Best Answer
I will interpret your question as asking what happens when instead of taking the $a_i\in F$, we take $a_i \in V^*$, and whether then $a_1f_1 + \cdots + a_nf_n = 0$ implies $a_i = 0$ for all $i$. Consider $a_1 = f_2$, $a_2 = -f_1$, $a_i = 0$ otherwise. Clearly $a_1f_1 + \ldots + a_nf_n = f_2f_1 - f_1f_2 = 0$, but $a_1, a_2 \neq 0$. Note that once you do this though, the $a_if_i$ are not necessarily elements in the dual space anymore, as for example for any nonzero $f \in V^*$, we have that $f^2$ is not linear (so in particular, I don't know if there is any reason to consider this)