### Dual

##### 4. Double dual

Let `V` be a vector space. Its dual `V`^{*} is also a vector space. The double dual of `V` is the dual `V`^{**} of the dual space `V`^{*}. There is a natural relation between `V` and `V`^{**}. In contrast, there is no natural relation between `V` and `V`^{*}. For example, the relation such as `ι`: (**R**^{n})^{*} ↔ **R**^{n} *depends on the choice* of the standard basis.

A vector `v` ∈ `V` induces a linear function `v`^{**}(`f`) = `f`(`v`) on `V`^{*}. The vector `v`^{**} ∈ `V`^{**} is the double dual of **v**. The transformation `v` → `v`^{**}: `V` → `V`^{**} is the double dual transformation.

Of course we need to justify the definition by showing that `v`^{**} is indeed a linear function on `V`^{*} and `v` → `v`^{**} is linear. The following verifies `v`^{**}(`f` + `g`) = `v`^{**}(`f`) + `v`^{**}(`g`).

`v`^{**}(`f` + `g`)

= (`f` + `g`)(`v`) (definition of `v`^{**})

= `f`(`v`) + `g`(`v`) (definition of `f` + `g`)

= `v`^{**}(`f`) + `v`^{**}(`g`). (definition of `v`^{**})

The equality `v`^{**}(`cf`) = `c `**v**^{**}(`f`) can be similarly verified. As for the linearity of `v` → `v`^{**}, the following verifies (`u` + `v`)^{**} = `u`^{**} + `v`^{**}.

(`u` + `v`)^{**}(`f`)

= `f`(`u` + `v`) (definition of (`u` + `v`)^{**})

= `f`(`u`) + `f`(`v`) (`f` is linear)

= `u`^{**}(`f`) + `v`^{**}(`f`) (definition of `u`^{**} and `v`^{**})

= (`u`^{**} + `v`^{**})(`f`). (definition of the addition `u`^{**} + `v`^{**} of linear functions)

The equality (`c`**v**)^{**} = `c`(**v**^{**}) can be similarly verified.

The double dual transformation takes a basis to a basis.

Let {**b**_{1}, **b**_{2}, ..., **b**_{n}} be a basis of `V`. Let {`β`_{1}, `β`_{2}, ... , `β`_{n}} be its dual basis. Then {**b**_{1}^{**}, **b**_{2}^{**}, ..., **b**_{n}^{**}} is the dual basis of {`β`_{1}, `β`_{2}, ... , `β`_{n}}.

Proof We simply need to verify the defining property of the dual basis for {**b**_{1}^{**}, **b**_{2}^{**}, ..., **b**_{n}^{**}}:

**b**_{i}^{**}(`β`_{j}) = `β`_{j}(**b**_{i}) = { |
1, |
if `i` = `j` |

0, |
if `i` ≠ `j` |

where the first = is due to the definition of **v**^{**}, and the second = is due to the assumption that {`β`_{1}, `β`_{2}, ... , `β`_{n}} is the dual of the basis {**b**_{1}, **b**_{2}, ..., **b**_{n}}.

Given the basis and its double dual, the linearity of `v` → `v`^{**} then implies that the double dual transformation is the same as

**v** = `c`_{1}**b**_{1} + `c`_{2}**b**_{2} + ... + `c`_{n}**b**_{n} → **v**^{**} = `c`_{1}**b**_{1}^{**} + `c`_{2}**b**_{2}^{**} + ... + `c`_{n}**b**_{n}^{**}.

This is the composition of two linear transformations

**v** = `c`_{1}**b**_{1} + `c`_{2}**b**_{2} + ... + `c`_{n}**b**_{n} → (`c`_{1}, `c`_{2}, ..., `c`_{n}) → **v**^{**} = `c`_{1}**b**_{1}^{**} + `c`_{2}**b**_{2}^{**} + ... + `c`_{n}**b**_{n}^{**}.

where the first → is the coordinates with respect to {**b**_{1}, **b**_{2}, ..., **b**_{n}} and the second → is the inverse to the coordinates with respect to {**b**_{1}^{**}, **b**_{2}^{**}, ..., **b**_{n}^{**}}. Since both are isomorphisms, the compisition is an isomorphism.

For finite dimensional vector spaces, the double dual transformation `v` → `v`^{**}: `V` → `V`^{**} is an isomorphism.

Finally, for a linear transformation `T`: `V` → `W`, we may the dual twice and get the double dual `T`^{**}: `V`^{**} → `W`^{**} of **T**. Prove that `T`^{**}(`v`^{**}) = `T`(`v`)^{**}.

`T`^{**}(`v`^{**}) = `T`(`v`)^{**}.

Proof We only need to show that test `T`^{**}(`v`^{**})(`f`) = `T`(`v`)^{**}(`f`) for any linear function `f` ∈ `V`^{*}.

`T`^{**}(`v`^{**})(`f`)

= `v`^{**}(`T`^{*}(`f`)) (definition of `T`^{**}, the dual of `T`^{*})

= `T`^{*}(`f`)(`v`) (definition of `v`^{**})

= `f`(`T`(`v`)) (definition of `T`^{*}, the dual of `T`)

= `T`(`v`)^{**}(`f`). (definition of `T`(`v`)^{**}, the double dual of `T`(`v`))

The equality `T`^{**}(`v`^{**}) = `T`(`v`)^{**} means that, after `V`^{**} and `W`^{**} are naturally identified with `V` and `W` via the double dual transformation, `T`^{**}: `V`^{**} → `W`^{**} is identified with the original linear transformation `T`: `V` → `W`. The situation is illustrated by the following diagram.

In the special case **V** = **R**^{m} and **W** = **R**^{n}, it is easy to verify that `v` → `v`^{**}: **R**^{m} → (**R**^{m})^{**} is the same as combining the identification `ι` and its dual together: `ι`^{*}`ι`^{-1}: **R**^{m} → (**R**^{m})^{*} → (**R**^{m})^{**}. Then the natural identification between `T`^{**} and `T` explains the involutive property of the transpose of matrices.