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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: (Rn)*Rn depends on the choice of the standard basis.

A vector vV induces a linear function v**(f) = f(v) on V*. The vector v**V** is the double dual of v. The transformation vv**: VV** 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 vv** 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 vv**, 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 (cv)** = c(v**) can be similarly verified.

The double dual transformation takes a basis to a basis.

Let {b1, b2, ..., bn} be a basis of V. Let {β1, β2, ... , βn} be its dual basis. Then {b1**, b2**, ..., bn**} is the dual basis of {β1, β2, ... , βn}.

Proof We simply need to verify the defining property of the dual basis for {b1**, b2**, ..., bn**}:

bi**(βj) = βj(bi) = { 1, if i = j
0, if ij

where the first = is due to the definition of v**, and the second = is due to the assumption that {β1, β2, ... , βn} is the dual of the basis {b1, b2, ..., bn}.

Given the basis and its double dual, the linearity of vv** then implies that the double dual transformation is the same as

v = c1b1 + c2b2 + ... + cnbnv** = c1b1** + c2b2** + ... + cnbn**.

This is the composition of two linear transformations

v = c1b1 + c2b2 + ... + cnbn → (c1, c2, ..., cn) → v** = c1b1** + c2b2** + ... + cnbn**.

where the first → is the coordinates with respect to {b1, b2, ..., bn} and the second → is the inverse to the coordinates with respect to {b1**, b2**, ..., bn**}. Since both are isomorphisms, the compisition is an isomorphism.

For finite dimensional vector spaces, the double dual transformation vv**: VV** is an isomorphism.

Finally, for a linear transformation T: VW, 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 fV*.

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: VW. The situation is illustrated by the following diagram.

In the special case V = Rm and W = Rn, it is easy to verify that vv**: Rm → (Rm)** is the same as combining the identification ι and its dual together: ι*ι-1: Rm → (Rm)* → (Rm)**. Then the natural identification between T** and T explains the involutive property of the transpose of matrices.


[exercise]
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