Math 2.0

statement theorem

Canonical isomorphism with the double dual

Statement

Let FF be a field and let VV be a vector space over FF. Let V∗∗=(V∗)∗V^{**} = (V^*)^* be the dual of the dual space. The evaluation map ev⁡ ⁣:V→V∗∗\operatorname{ev} \colon V \to V^{**} is defined by ev⁡(v)(f)=f(v)\operatorname{ev}(v)(f) = f(v) for v∈Vv \in V and f∈V∗f \in V^*. This map is linear and injective. If VV is finite-dimensional, then ev⁡\operatorname{ev} is an isomorphism.

Remarks

The evaluation map does not depend on a choice of basis, so it is called canonical or natural. In finite dimensions, it is an isomorphism because it is injective and dim⁡V∗∗=dim⁡V∗=dim⁡V\dim V^{**} = \dim V^* = \dim V. For infinite-dimensional VV, the evaluation map is not surjective.

Explanations

L1 human

A vector vv can act on linear forms: it sends each form ff to the number f(v)f(v). This turns each vector into an element of the double dual. In finite dimensions, every element of the double dual comes from exactly one vector in this way.

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  1. Vector space definition
  2. Linear combination definition
  3. Linear map definition
  4. Linear subspace definition
  5. Linear independence definition
  6. Linear form definition
  7. Linear span definition
  8. Space of linear maps definition
  9. Dual space definition
  10. Spanning set definition
  11. Basis definition
  12. Dual basis definition
  13. Dimension of the dual space statement
  14. Canonical isomorphism with the double dual statement

Sources

  1. Web https://en.wikipedia.org/wiki/Dual_space, Injection into the double-dual