Is every vector a function on the basis of the vector space?

Question

Let $B= \left\{ a_ 1 , \dots a_n \right\}$ a basis of a vector space $V$ on a field $k$. Let $v \in V$ a vector such that $v = \sum_{i=1}^n \alpha_i a_i$. Is it true that $v$ corresponds to a well defined function $f_v : B \to k$ such that $f_v(a_i) = \alpha_i$? If this is reasonable, let $\circ$ be the pointwise product of vectors, i.e. such that $\sum_{i} \alpha_i a_i \circ \sum_{i} \beta_i a_i = \sum_{i} \alpha_i \beta_i a_i$. What does the operation $\circ$ correspond to when we view vectors as functions on $B$? In other words, is there a way to endow the functions $f_v$ with a notion of product, such that they form an algebra?

Answer 1

If $k^B$ is the set of all function from $B$ to $k$, then the map $$\phi : k^B \to V, \ g \mapsto \sum_{i=1}^n g(a_i)a_i$$ is a bijection (the inverse sends $v \in V$ to $f_v$, as you can check). Thus, a natural way of defining the pointwise product $\bullet$ is by $$(\forall g_1,g_2 \in k^B) \quad g_1 \bullet g_2 := \phi^{-1}(\phi(g_1) \circ \phi(g_2)).$$ In particular, $$(\forall v_1,v_2 \in V) \quad f_{v_1} \bullet f_{v_2} = f_{v_1 \circ v_2}.$$ Observe that, with $\bullet$ defined that way, $\phi$ is some kind of homomorphism: $\phi(g_1 \bullet g_2) = \phi(g_1) \circ \phi(g_2)$ for all $g_1,g_2 \in k^B$.

Answer 2

Is it true that $v$ corresponds to a well defined function $f_v : B \to k$ such that $f_v(a_i) = \alpha_i$?

This is true because $V$ is finite dimensional ($B$ is a finite set) and the choice of one of its bases fixes a unique isomorphism of vector spaces $V \cong V^*=\hom_k(V,k)$, precisely in the way that you write. That is to say, since $\langle B\rangle=V$, a linear map $f : V \to k$ is completely determined by the image of the basis $B$, i.e. by a function $B\to k$, i.e. by a $n$-tuple of scalars $(\alpha_1,\dots,\alpha_n)$.