# Dual space of a dual space = What?

Suppose $V$ is a real vector space.

Then $V^*$ --- its dual space --- is the vector space of linear maps $V\to \mathbb R$

How then do I interpret $(V^*)^*$?? The dual space of the dual space??

Thanks!

• Yes, it is the dual space of the dual space. Why is that a problem? – wj32 Jun 11 '13 at 10:53
• It's the space of linear maps $V^* \to \mathbb R$. – Najib Idrissi Jun 11 '13 at 10:58
• @nik: Thanks, nik, So they are linear maps of linear maps? I just find it a bit difficult to get my head around... – Dr Strangelove Jun 11 '13 at 11:00
• @DrStrangelove They are linear maps having a linear map as an input and giving a real number as an output. It is quite difficult to get ones head around at first. It was the same for me. You'll get used to it. – Julian Kuelshammer Jun 11 '13 at 11:06
• Luckily, if $V$ is finite dimensional, $(V^*)^*$ is naturally isomorphic to $V$; you can think of $v\in V$ as a map $V^*\to\mathbb{R}$ by $v:f\mapsto f(v)$, and it turns out all linear maps $V^*\to\mathbb{R}$ are of this form. – mdp Jun 11 '13 at 11:07

The space of linear maps $\ell : V \rightarrow \mathbb{R}$ is itself a vector space, with pointwise addition and scalar multiplication of functions. Thus, $(V^*)^*$ is the dual of this vector space.
There is a canonical linear transformation $\xi : V \rightarrow (V^*)^*$ defined by $\xi(v) = \xi_v$, where $\xi_v : V^* \rightarrow \mathbb{R}$ is the linear map given by $\xi_v(\ell) = \ell(v)$. The map $\xi$ is injective, so when $V$ is a finite dimensional vector space, the map $\xi$ is a (canonical) isomorphism $V \cong (V^*)^*$. However, $\xi$ is not necessarily an isomorphism if $V$ is infinite dimensional.