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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 $\sum^n_{i=1}f(x_i)f_i$.

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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!

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    $\begingroup$ A linear map is uniquely determined by where it maps a (any) basis. That has been proved long before this. $\endgroup$
    – Ted Shifrin
    Sep 30 at 18:33
  • $\begingroup$ 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 $W$? $\endgroup$
    – Flying Spaghetti
    Sep 30 at 18:42
  • $\begingroup$ 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?) $\endgroup$
    – Flying Spaghetti
    Sep 30 at 18:46
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    $\begingroup$ 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. $\endgroup$
    – Izaak van Dongen
    Sep 30 at 18:53
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    $\begingroup$ 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 $v\in V$. $\endgroup$
    – Ted Shifrin
    Sep 30 at 18:56

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Thank you, all, I was able to rewrite it in my own words and hopefully I understand the proof now. Any more elucidation is welcome.

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.

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