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The recent topics I studied were linear functionals and dual spaces. I like to think about a linear functional as a stack of hyperplanes like it is described here.

In "Finite dimensional vector spaces" by Paul Halmos I read that every vector space is isomorphic to its double dual. I wonder if there is an intuitive way to see that they are isomorphic? Also I am not sure how I could graphically or geometrically think about the double dual space (in the sense I think about the dual space as a stack of hyperplanes in every direction). Is there a way to visualize the double dual space? Maybe then the isomorphism would become clearer. I guess there is some intuition behind it since someone had to think about it first before he or she invented the concept (double dual space). I hope my question makes sense?

Thanks for any responses!

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    $\begingroup$ Every finite dimensional vector space is isomorphic to its dual, and hence to its double dual, triple dual, etc., but that is just a matter of computing the dimension. The interesting thing is that the space is naturally isomorphic to its double dual - and that carries much more information than a mere dimensionality computation. $\endgroup$ Commented Aug 6, 2016 at 21:01

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Let $x \in V$ and let $z \in V^*$ (where $V$ is a vector space). You can think of $z$ as doing something to $x$ (in other words, $z$ takes $x$ as input and returns $z(x)$ as output). But, you can equally well think of $x$ as doing something to $z$! In other words, you can imagine that $x$ itself takes $z$ as input and returns $z(x)$ as output. From this viewpoint, $x$ is a linear functional on $V^*$.

I like the notation \begin{equation} z(x) = \langle z, x \rangle, \end{equation} because it treats $z$ and $x$ symmetrically, and emphasizes that both viewpoints are equally valid.

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  • $\begingroup$ That makes much more sense now to me thanks! $\endgroup$
    – mannequin
    Commented Aug 12, 2016 at 19:35
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Let $V$ be a vector space and $V'$ its dual, you have $i:V\rightarrow V"$ defined by $i(v)(f)=f(v)$. Remark here that $f$ is an element of $V'$, that is a linear function defined on $V$. Show that $i$ is injective: $i(v)=0$ implies that for every $f\in V'$, $f(v)=0$. Suppose $v\neq 0$, you can find a basis $(e_1=v,...,e_n)$ of $V$ and define $f_v(e_1)=1, f_v(e_i)=0, i>1$, you have $i(v)(f_v)=f_v(v)=1$. Contradiction. since $dim V=dim V"$ $i$ is a linear isomorphism.

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Every (finite dimensional) vector space has the same dimension as its dual.

Hence, $dimV=dimV^*=dimV^{**}$. As the dimensions are equal, $V\cong V^{**}$. However, it is worthwhile to note that there exists a natural isomorphism between $V$ and $V^{**}$. That is, an isomorphism that does not depend on the basis of $V$. In which case, one may identify $V$ with $V^{**}$.

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