There is one glaring omission in our Linear Algebra curriculum – it avoids talking about the dual space of a vector space. This makes talking about relationship between subspaces and equations that define them exceedingly difficult. Better late than never, so here it comes.

Let be a vector space over a field Denote by the set of linear functions

**Examples**

Let the space of continuous functions on Then the function given by is linear on

Let be the vector space of polynomials with real coefficients.

Then the function given by is linear on

Note that as is linear, one has for any Thus we have defined by so that To simplify notation, we will write instead As well, we can define for any and more generally, And there is the zero function for any Thus we have all the ingredients of a vector space, as can be easily checked.

is a vector space over It is called the

dual spaceof

So far, we haven’t used the linearity of our functions at all (we actually did not need the fact that ). Indeed, any closed under addition and multiplication set of functions would form a vector space.

What makes the dual space so special is that to define it suffices to define on a basis of Indeed, so we can compute for any once we know the ‘s.

Thus for a finite-dimensional vector space one sees a (dependent upon the choice of a basis in ) bijection between and This bijection, that is even an isomorphism of vector spaces, is defined by the *dual basis* of given by * coordinate functions* where ‘s are the coefficients of is the decomposition of in the basis

Finite-dimensionality is crucial here. E.g. let us consider the vector space of polynomials It is a countable space: one can view it as the set of infinite 0-1 strings, with only finitely many 1’s occurring in each string. On the other hand, can be viewed as the set of all the infinite 0-1 strings, which is uncountable, so there cannot be a bijection between and

Given one can define a function as follows: It is linear, as Here we do not see any dependence on the choice of a basis in and we have

The vector space of linear functions on is (canonically) isomorphic to via the mapping

Indeed, we see immediately that and so we need only to check that this mapping is bijective. Let be a basis in and its dual basis in Then if and 0 otherwise. Thus is the basis of which is dual to the basis of and the mapping sends the vector with coordinates to the vector with the same coordinates in

the basis of Hence the latter is bijective.

In view of the latter, we can identify with and write instead The set of such that is a subspace, called *annihilator of * of dimension More generally, the following holds.

Let be a subspace of and Then the annihilator of is a subspace of of dimension

Indeed, we can choose a basis in so that is a basis of where Then we have the dual basis of and is the subspace with the basis

In view of this, each can be obtained as the set of solutions of a system of homogeneous linear equations for of rank

** Dual spaces and annihilators under a basis change**

Let be a linear transformation of and a subspace of Then is a subspace. How can one look at By writing out in a basis for any in the dual basis we get equation Thus, considering as a matrix, we get where denotes the action of on It follows that i.e. We have, considering that acts on by right multiplication, and not by left ones, to take the transpose, too.

acts on as

**An example.**

Let We work in the standard basis of Then the dual basis of is , so that

Let be the group of matrices It fixes, in its left action on by multiplication, the vector . Let be the 1-dimensional subspace of generated by Then is generated by and The group preserves in its action on As is 2-dimensional, there should be a nontrivial kernel in this action, and indeed, it consists of the elements of the form

A particularly simple case is Then is isomorphic to the symmetric group on 4 letters, as can be seen in its action on the 4 elements of outside On the other hand, it acts on the 3 nonzero elements of as

4 May, 2009 at 21:15 |

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