The dual of a vector space is one of those fairly simple things that wasn't understood properly until relatively recently (this century). It is used extensively in modern mathematics.

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.
It is not difficult to check that these operations make V* into a k-vector space.

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,...,en 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 + ... + anen (*)
so if f is in V* then
f(v)=a1f(e1) + ... + anf(en)
so that we can write down what f does to a general element of V once we know where it sends the basis vectors e1,...,en. Conversely, given n values for the images of these basis vectors there is a uniquely determined element of V*. Let ei* in V* be the unique linear transformation that vanishes on ej, for j not i and takes the value 1 on ei.

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

• These linear maps are linearly independent. For if
b1e1* + ... + bnen*=0
for some scalars b1,...,bn then apply both sides to a vector v as in (*) then we obtain
b1e1*(v) + ... + bnen*(v)=0
and so
b1a1 + ... + bnan=0
Now we can take the ai 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 bi=0 and we have linear independence.
• These maps span. For, take an arbitrary f in V* and suppose that f(ei)=ai. Then it is easy to see that
f= a1e1* + ... + anen*

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 fv in V** by fv(g)=g(v), for g in V*. Then we have a linear tranformation FV:V-->V** defined by mapping v to fv.

Theorem

1. FV is an isomorphism
2. Further, whenever h:V-->W is a linear transformation then
h**FV=FWh

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

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

It remains to prove the naturality property. Well h**FV 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**FV(v) Apply this to g in W* to get h**(FV(v))(g) This is the same as FV(v)(h*(g)) which in turn equals h*(g)(v). This is simply g(h(v)) which equals FW(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.

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