Why is this proof sufficient for all vectors in the dual space

I'm reviewing this proof given in Friedberg (page 120) on dual spaces. I'm specifically curious about why showing that $$g(x_j) = f(x_j)$$ is sufficient to claim that any $$f$$ can be rewritten as $$\sum^n_{i=1}f(x_i)f_i$$.

I could be misunderstanding, but it looks to me like we just fed in a basis vector, and didn't think about the cases where we're constructing other vectors by linear combinations of the basis.

Thank you!

• A linear map is uniquely determined by where it maps a (any) basis. That has been proved long before this. Sep 30 at 18:33
• Friedberg mentions a linear transformation is completely determined by its action on a basis, but the theorem following this statement was hard for me to understand how to use. He states, "Let $V$ and $W$ be vector spaces over $F$, and suppose that $\{v_1,v_2,...,v_n\}$ is a basis for $V$. For $w_1,w_2,...,w_n$ in $W$, there exists exactly one linear transformation $T:V\rightarrow W$ such that $T(v_i)=w_i$ for $i=1,2,...,n$." (Theorem 2.6, 73) So he explicitly states some set of size $n$ in $W$, I suppose this is the same as saying "for every element of $W$? Sep 30 at 18:42
• In other words, I'm not sure how to finish this statement - "If I can state where T takes a every element of a basis, then I know - " would I finish it as "that T holds for every other element in the range?" or "where T goes for every other element in the range?" Further, would citing the theorem above be sufficient (and how?) Sep 30 at 18:46
• In this case you should let $v_i = x_i$ and $w_i = f(x_i)$. Since we proved that $g(x_i) = f(x_i)$ and it's obvious that $f(x_i) = f(x_i)$, both $f$ and $g$ satisfy the conditions of the statement you give. But the theorem says there's a unique such map satisfying those conditions - hence $f$ and $g$ are the same map. I think it's fine to say "since linear maps are determined by their action on a basis" as justification, so long as you make sure you actually believe it's true! Well done for being so thorough with checking your understanding. Sep 30 at 18:53
• It sounds like you're confusing domain and range. $n$ is the dimension of the domain. Once you specify the values of $T(v_1),\dots,T(v_n)$, then the linear map $T\colon V\to W$ is completely determined. No, we're never saying "for every element of $W$." We're saying that we know $T(v)$ for every element $v\in V$. Sep 30 at 18:56

Proof: Suppose this equation were true; then because $$\beta^{*}$$ is right dimension and spans $$V^{*}$$, it would be a basis. Now, it is sufficient to see if this equation holds for the basis vectors in the source. Since a linear transformation is completely determined by its action on the basis vectors, if that is true, then we are done. Set $$g=\sum^n_{i=1}f(x_i)f_i$$, and inspect its behavior for each element in the basis. We iterate through and bookkeep which element we are at with $$j$$. Then for $$1\leq j \leq n$$, we have: \begin{align*} g(x_j) &= \sum^n_{i=1}f(x_i)f_i(x_j) \\ LHS &= \sum^n_{i=1}f(x_i)\delta_{ij} && \text{But note } f_i(x_j) \text{is just } \delta_{ij} \\ LHS &= f(x_1)\delta_{1j}+...+f(x_n)\delta_{nj} && \text{Expand} \\ LHS &= f(x_j) && \delta_{ij} \text{ sends every element to 0 except } f(x_j) \end{align*} Since $$g(x_j) = f(x_j)$$, LHS=RHS as claimed and we are done.