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I've been doing a bit of self study into the formalism of dual spaces. In the book that I've been reading the author introduces the notion of a dual space, $V^{\ast}$ to a given vector space $V$ as the set of linear functionals $f:V \rightarrow \mathbb{F}$ which map $V$ to its underlying field $\mathbb{F}$ such that $$f\left(\mathbf{v} +\mathbf{w}\right) = f\left(\mathbf{v}\right) +f\left(\mathbf{w} \right)\quad\forall\;\mathbf{v},\mathbf{w}\in V$$ and $$f\left(c\mathbf{v}\right) = cf\left(\mathbf{v}\right)$$ The author then goes on to introduce an ordered basis $\mathcal{B} = \lbrace\mathbf{e}_{i}\rbrace _{i=1, \ldots , n}$ for the vector space $V$ such that for a given $\mathbf{v} \in V$ $$\mathbf{v}= \sum_{i=1}^{n}v^{i}\mathbf{e}_{i}= v^{1}\mathbf{e}_{1}+ \cdots + v^{n}\mathbf{e}_{n}$$ and then defines the $i^{th}$ dual basis vector $\mathfrak{e}^{i}$ as a linear functional that satisfies $$\mathfrak{e}^{i}\left(\mathbf{v}\right) = f\left( v^{1}\mathbf{e}_{1}+ \cdots + v^{n}\mathbf{e}_{n}\right) = v^{i}$$ such that the linear functional $\mathfrak{e}^{i}$ "picks off" the $i^{th}$ component of the vector $\mathbf{v} \in V$.

Is there any particular motivation behind this? Or do we just choose the functional this way such that the basis vectors $\mathbf{e}_{i}$ and their corresponding dual basis vectors $\mathfrak{e}^{j}$ satisfy $$\mathfrak{e}^{i}\left(\mathbf{e}_{j}\right) = \delta^{i}_{j}$$

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Do we just choose the functional this way such that the basis vectors ei and their corresponding dual basis vectors $\mathfrak e^j$ satisfy $$ \mathfrak e^i(e_j)= \delta _{ij} $$

Yes, that's exactly the reason. This, in particular, makes computation extremely easy.

If $f = \sum_{i}a_i \mathfrak e^i$ and $v = \sum_j b_j e_j$, then we have $$ f(v) = \sum_{i} a_i b_i $$ Which you may think of as being tantamount to a "dot product". What we're doing here is ostensibly the same as selecting an "orthonormal basis" in a situation where we've forgone the notion of an inner product.

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  • $\begingroup$ Thanks for your help. Do we just define it this way to guarantee a mapping that preserves linear independence such that we can form a basis for the dual space? $\endgroup$
    – Will
    Sep 9 '14 at 0:01
  • $\begingroup$ What do you mean by "a mapping that preserves linear independence"? Why would we want to/ have to guarantee such a thing? $\endgroup$ Sep 9 '14 at 2:24
  • $\begingroup$ Sorry, this was a bit of a faux-pas on my part, I wrote it late last night and didn't really think it through properly! I guess I was trying to justify in my mind a bit more why we have such a definition for a dual vector?! $\endgroup$
    – Will
    Sep 9 '14 at 13:58
  • $\begingroup$ Why wouldn't we have such a definition for a dual vector? I'm trying to understand what it is that you feel the need to justify. Perhaps the following will help: for any functional $f$ on a finite dimensional inner product space, there must exist a vector $v$ such that for any $v$, $f(v) = \langle u, v\rangle$. $\endgroup$ Sep 9 '14 at 14:46
  • $\begingroup$ I think I was just expecting there to be a bit more motivation behind the definition to be given in the text, but as far as I understand (from the text and what you've said) we choose a set of linear functionals which satisfy this property and call this the dual vector space to a given vector space, $V$. I guess the crux of my confusion really is, what is a dual space to a given vector space and what is the motivation behind it? $\endgroup$
    – Will
    Sep 9 '14 at 15:18
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One of the reasons for doing this is to mimic the inner product

Often one writes $\langle v,f \rangle$ for $f(v)$ where $v \in V$ and $f \in V^*$. Note that for a general Hilbert space $\mathcal{H}$, all liner functionals are given by inner products, i.e. we have a correspondence of a linear functional $\phi$ with a vector $x \in \mathcal{H} $(Riesz representation theorem). So, in this case, an orthonormal basis of the Hilbert space will become an orthonormal basis of the dual space under the identification $\mathcal{H^*}$ with $\mathcal{H}$.

In case of a general vectors space (with or without the structure of norm, inner product etc) we would still like to think of $f(v)$ as "$\langle v,f \rangle$", i.e. some sort of inner product pairing. This intuitive notion makes a lot of sense in calculation- and makes computations really easy- and one can use the intuitions from the Hilbert space setting in this case to "guess" and prove certain formula.

I can write a few more examples if you want- so let me know in comments- so that I can come back and edit the answer.

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  • $\begingroup$ Cheers, that's very useful. Is there any way to prove the relation $\mathfrak{e}^{i}\left(\mathbf{e}_{j}\right)= \delta^{i}_{j}$, or is this just a definition that we use so we can guarantee a mapping that preserves linear independence of a set of vectors so that we have a basis for the dual space? $\endgroup$
    – Will
    Sep 8 '14 at 23:57
  • $\begingroup$ @Will: It's a definition :). $\endgroup$
    – voldemort
    Sep 9 '14 at 0:03
  • $\begingroup$ Cheers, thanks for the help :) $\endgroup$
    – Will
    Sep 9 '14 at 0:45
  • $\begingroup$ Forgot to say, would appreciate a few more examples if you have the time. Thanks :) $\endgroup$
    – Will
    Sep 9 '14 at 1:22

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