I have a vector space $V$ over the field $K$ with a basis $(b_j)_{j\in \mathbb N}$ and the bilinear scalar product $\iota:V\times V \rightarrow K$ given by: $b_i * b_j = \delta_{ij}$.
What I want to find is a basis for the subspace of the dual vector space, which contains all linear forms $a^*:V\rightarrow K$ induced by my scalar product, i.e I am looking for a basis for the set: $$\{a^*\in V|\exists x\in V: \iota(x, .)= a^*\} $$
whereas $\iota(x,.) $ stands for a linear form $V\rightarrow K : y \mapsto \iota(x,y)$
So my thoughts were, that the basis simply must be the basis for the whole dual vector space, since $\iota(x,y) = \sum_{i \in \mathbb N} x_i y_i$. Since it says, I am dealing with a subspace of the dual vector space, I am not sure however whether this is true. Any help would be greatly appreciated!
First of all: you're looking for a basis, so let's actually describe that basis. Let $S \subset V^*$ denote the subspace in question. Then the functionals $a_i$ defined by $\alpha_i = \iota(b_i, \cdot)$ form a basis. To see that set forms a basis, note that these functionals are linearly independent and that since any $x$ can be written as a finite sum $\sum c_j b_j$, we may also write $$ \iota(x,v) = \iota \left(\sum_j c_jb_j, v\right) = \sum_{j}c_j\iota(b_j,v) = \left[\sum_j c_j \alpha_j \right](v) $$ However, these vectors will not generally form a basis for the entirety of $V^*$. As an example: take $V$ to be the space of all real sequences with finitely many non-zero entries. We can take $b_j: \Bbb N \to \Bbb R$ to be defined by $b_j(k) = \delta_{jk}$.
Now, define $f \in V^*$ by $f(x) = \sum_{k \in \Bbb N} x(k)$. This is a linear map, and it's defined on all sequences in $V$. However, $f$ is not a linear combination of the dual vectors $\alpha_j$.