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I am trying to develop a feeling for duality and stumbled upon the following question:

Say we are in $\mathbb R^2$ and let $U=\operatorname{span}([2,1]')$ be our subspace. My question is: what is $U^*$ (the dual space of $U$)?

I assume that since $\dim V = \dim V^*$ for any finite-dimensional vector space (and $U$ is one) $U^*$ should be a line as well, a line in $(\mathbb R^2)^*$ with row vectors so to speak. Yet it looks like anything from $(\mathbb R^2)^*$ would go: $$ \varphi: U \to \mathbb R, [x,y]' \mapsto [3,4]*[x,y]'$$ seems to be linear as well and so by definition $ \varphi \in U^* $. So whatever we have instead of $[3,4]$ the function is still linear.

And how should we construct the dual basis for $[2,1]'$? I understand the $\delta_{ij}$ way when we are dealing the the dual space of a "true" vector space (meaning it's not a subspace of any other vector space except itself).

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up vote 3 down vote accepted

A member of the dual space is a linear functional on $U$. Such a linear functional is determined by its value on the vector $[2,1]'$, i.e. $f(t [2,1]') = t f([2,1]')$. Yes, the restrictions of members of $({\mathbb R}^2)^*$ to $U$ are members of $U^*$, but different members of $({\mathbb R}^2)^*$ can have the same restriction to $U$, e.g. $u \mapsto [3,4] \cdot u$ is the same as $u \mapsto [10,0] \cdot u$. So it's still a one-dimensional set. As such, any nonzero member of it constitutes a basis.

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Or you could apply the Riesz representation theorem that tells you that if $H$ is a Hilbert space then the map $h \mapsto \langle \cdot , h \rangle$ defines an isometric isomorphism.

$U$ itself is a Hilbert space hence using Riesz you "see" that $U^\ast$ looks like $U$ itself.

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