# How to interpret $(V^*)^*$, the dual space of the dual space?

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?

• 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 $V^* \to \mathbb R$. 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... 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. 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.