dual of a vector space

Warning Although nothing below is really very deep it is highly abstract (technically this is abstract nonsense) so take it slowly and have a mug of strong coffee handy.

Fix V a finite dimensional vector space over a field k. (One can consider these ideas for infinite dimensional vector spaces but I make this restriction for simplicity.) Define V^* to be the set of all linear transformations f:V-->k. The important thing to realise is that V^* is itself a vector space called the dual vector space, or simply dual space, of V. Here, the operations are given as follows

• If a in k and f in V^* then a.f in V^* is defined by a.f(v)=af(v), for v in V.
• If f,g are in V^* then f+g in V^* is defined by (f+g)(v)=f(v)+g(v), for v in V.

Even better not only can we apply duality * to a vector space we can also apply it to a linear transformation. For if h:V-->W is a linear transformation then we can associate h^*:W^*--> V^* defined by h^*(f)(v)=f(h(v)), for f in W^*, and v in V. The duality operation * is an example of a functor. It is a functor from the category of vector spaces to itself.

If the abstraction hasn't killed you thus far (and there is plenty of time for that later... :-) then you might want to think about this a little more concretely in terms of bases. So suppose that e₁,...,e_n is a basis of V. How might we fashion a basis for V^*?

Well a typical vector in v in V can be written uniquely as

v=a₁e₁ + ... + a_ne_n (*)

f(v)=a₁f(e₁) + ... + a_nf(e_n)

Consider then e₁^*,...,e_n^*. We will show that these elements are a basis of V^* called the dual basis.

• These linear maps are linearly independent. For if

  b₁e₁^* + ... + b_ne_n^*=0

  for some scalars b₁,...,b_n then apply both sides to a vector v as in (*) then we obtain

  b₁e₁^*(v) + ... + b_ne_n^*(v)=0

  and so

  b₁a₁ + ... + b_na_n=0

  Now we can take the a_i to be any values we like. For example if we take a₁=1 and the others to be zero then the above equation tells us that b₁=0. Similarly all the other b_i=0 and we have linear independence.
• These maps span. For, take an arbitrary f in V^* and suppose that f(e_i)=a_i. Then it is easy to see that

  f= a₁e₁^* + ... + a_ne_n^*

This shows that V and V^* both have the same dimension and hence are isomorphic as vector spaces. But it turns out that there is a deeper and more natural isomorphism between V and V^**. This latter object is the dual vector space of the dual vector space of V.

To explain this we associate to an arbitary vector v in V^* an element of V^**. Define f_v in V^** by f_v(g)=g(v), for g in V^*. Then we have a linear tranformation F_V:V-->V^** defined by mapping v to f_v.

Theorem

1. F_V is an isomorphism
2. Further, whenever h:V-->W is a linear transformation then

   h^**F_V=F_Wh

This theorem means that not only is V isomorphic to V^** but that this isomorphism works well with linear transformation between vector spaces.

Proof: F_V is injective for suppose that F_V(v)=0. This means that g(v)=0 for all linear transformations. Applying this to e_i^* we deduce that when we write v as in (*) then a_i=0. Surjectivity is now clear because we know that e₁^**,...,e₁^** is a basis of V^**. It is obvious that F_V(e_i)=e_i^**.

It remains to prove the naturality property. Well h^**F_V is a mapping from V to W^** so we have to see what this map does to a vector v in V. So consider h^**F_V(v) Apply this to g in W^* to get h^**(F_V(v))(g) This is the same as F_V(v)(h^*(g)) which in turn equals h^*(g)(v). This is simply g(h(v)) which equals F_W(h(v))(g), as needed.

What about infinite dimensional vector spaces? Well the isomorphism above is only an injection in the more general case. To have any hope of an isomorphism one has to have some sort of topology and make restrictions on the linear transformations under consideration.