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I was hoping I could clarify my understanding on some basic notions of dual spaces.

Suppose I have a vector space [itex]V[/itex] along with a basis [itex]\lbrace\mathbf{e}_{i}\rbrace[/itex], then there is a unique linear map [itex]\tilde{e}^{i}: V\rightarrow \mathbb{F}[/itex] defined by [tex]\tilde{e}^{i}(\mathbf{v})=v^{i} \Rightarrow \tilde{e}^{i}(\mathbf{e}_{j})= \delta^{i}_{j}[/tex]

which maps each vector [itex]\mathbf{v}\in V[/itex] to its [itex]i^{th}[/itex] component [itex]v^{i}\in \mathbb{F}[/itex] with respect to the basis vector [itex]\mathbf{e}_{i}[/itex]. The set of linear maps [itex]\lbrace\tilde{e}^{i}\rbrace[/itex] form a basis for the dual space, [itex]V^{\ast}[/itex], of [itex]V[/itex].

To prove that this map is unique suppose that we have some other linear map [itex]\tilde{f}^{i}[/itex] which also satisfies [itex] \tilde{f}^{i}(\mathbf{e}_{j})= \delta^{i}_{j}[/itex], then as [itex]\lbrace\mathbf{e}_{i}\rbrace[/itex] is a basis for [itex]V[/itex] we can express a vector [itex]\mathbf{v}\in V[/itex] as a unique linear combination [itex]\sum_{j}v^{j}\mathbf{e}_{j}[/itex], and so

[itex] \tilde{e}^{i}\left(\mathbf{v}\right)= \tilde{e}^{i}(\sum_{j}v^{j}\mathbf{e}_{j})= \sum_{j}v^{j} \tilde{e}^{i}(\mathbf{e}_{j})= \sum_{j}v^{j}\delta^{i}_{j} = \sum_{j}v^{j}\tilde{f}^{i}(\mathbf{e}_{j}) = \tilde{f}^{i}\left(\sum_{j}v^{j}\mathbf{e}_{j}\right) = \tilde{f}^{i}(\mathbf{v}) [/itex]

Hence, as [itex]\mathbf{v}[/itex] was chosen arbitrarily, we conclude that [itex]\tilde{e}^{i}=\tilde{f}^{i}[/itex] and as such the mapping [itex]\tilde{e}^{i}: V\rightarrow V^{\ast}[/itex] defined by [tex]\tilde{e}^{i}(\mathbf{v})=v^{i}[/tex] is unique.

To prove that such a mapping exists, let [itex]\mathbf{v}[/itex], and as above, express it in terms of the basis [itex]\lbrace\mathbf{e}_{i}\rbrace[/itex], [itex]\sum_{j}v^{j}\mathbf{e}_{j}[/itex]. Now, as the scalars [itex]v^{i}[/itex] are uniquely determined by [itex]\mathbf{v}[/itex] and therefore, [itex]v^{i}[/itex] is a uniquely determined element of [itex]\mathbb{F}[/itex]. This gives us a well defined rule for obtaining an element of [itex]\mathbb{F}[/itex] from [itex]V[/itex], i.e. a function from [itex]V[/itex] to [itex]\mathbb{F}[/itex]. Thus, there is a function [itex]\tilde{e}^{i}:V \rightarrow \mathbb{F}[/itex] satisfying [tex] \tilde{e}^{i}\left(\mathbf{v}\right)= v^{i}.[/tex]

Now we show that this is a linear map. Let [itex]\mathbf{u}, \mathbf{v}\in V[/itex] and [itex]\alpha, \beta \in\mathbb{F}[/itex]. Let [itex]\mathbf{u}=\sum_{i}a^{i}\mathbf{e}_{i}[/itex] and [itex]\mathbf{v}=\sum_{i}b^{i}\mathbf{e}_{i}[/itex], with respect to the basis [itex]\lbrace\mathbf{e}_{i}\rbrace[/itex]. Then

[tex]\alpha\mathbf{u} + \beta\mathbf{v} = \sum_{i}\left(\alpha a^{i} + \beta b^{i}\right)\mathbf{e}_{i},[/tex]

and the definition of [itex]\tilde{e}^{i}[/itex] gives

[tex]\tilde{e}^{i}\left(\alpha\mathbf{u} + \beta\mathbf{v}\right) = \left(\alpha a^{i} + \beta b^{i}\right) = \alpha a^{i} + \beta b^{i}= \alpha\tilde{e}^{i}\left(\mathbf{u}\right) + \beta\tilde{e}^{i}\left(\mathbf{v}\right)[/tex]

and so [itex]\tilde{e}^{i}[/itex] is linear.

Sorry for the long-windedness of this post, but I just wanted to check whether my understanding is correct?

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# Dual vector spaces and linear maps

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