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**Drexel28** The basic idea is this. Suppose that $\displaystyle V$ is a $\displaystyle F$-space. You can completely specify a linear transformation from $\displaystyle V$ to any other $\displaystyle F$-space $\displaystyle W$ by demanding that $\displaystyle v_i\mapsto w_i$ where $\displaystyle \{v_1,\cdots,v_n\}$ is a basis for $\displaystyle V$ and the $\displaystyle w_i$ are just any vectors in $\displaystyle W$ (not necessarily different). How? Well, say you have made a choice about what the $\displaystyle v_i$ go to, you still haven't defined a transformation on $\displaystyle V$ itself. That said, if the map which takes $\displaystyle v_i\to w_i$ is to be a linear transformation you must take each $\displaystyle v=\alpha_1v_1+\cdots+\alpha_n v_n$ to $\displaystyle \alpha_1w_1+\cdots+\alpha_nw_n$. Thus, if $\displaystyle T:V\to W$ is linear and satisfies $\displaystyle T(v_i)=w_i$ then the function must be defined by the rule $\displaystyle T(v)=\alpha_1w_1+\cdots+\alpha_nw_n$ where $\displaystyle \alpha_1v_1+\cdots+\alpha_nv_n$ is the UNIQUE representation of $\displaystyle v$ as a linear combination of the basis $\displaystyle \{v_1,\cdots,v_n\}$. Conversely, you can check that the map defined that way does, in fact, satisfy the condition of being a linear transformation $\displaystyle V\to W$ with $\displaystyle v_i\mapsto w_i$. So, in your case you have that $\displaystyle W=F$ and $\displaystyle w_j=\delta_{i,j}\in F$

For the second part, what would happen if $\displaystyle \displaystyle \sum_{i=1}^{n}v^\ast_i=0(v)$ (where I put $\displaystyle 0(v)$ to emphasize that it's the zero function)? What happens if you plug in $\displaystyle v_j$ for $\displaystyle j=1,\cdots,n$?