Let be a vector space over , then the space dual to is , the space of all linear mappings into the field called also linear functionals or linear forms.
Notice that for any subspace of and , there is a fuctional such that and. To see this, complete a basis of to a basis , is such a way that vector is in the completed basis and define the values of the fuctional in this basis accordingly.
E.g., let and and . Then functional , to satisfy the above conditions has to be the following and . Thus, .
Therefore, given a basis of a finite-dimensional space : , , we can find functionals , such that , but for , . Notice, that those functionals are a basis of space and this basis is called the dual basis to . Ususaly we denote by
E.g., for basis , the dual basis is , and .
Notice also that given a functional , its coordinates in the basis are the values of this functional on vectors . Indeed, if we get that , because the left and right side of this equation are equal when calculated on argument equal to for any .
The existence of the dual basis shows that if , then , and for given basis we have an isomorphism defined as for all .
This is not true in the case of infinitely dimensional spaces. Notice that, for example, the dual space to the space of all sequences which are zeroes from some point on is isomorphic to , so it is not isomorphic to . Indeed, the isomorphism given as , where , (so ) defined as (the series is convergent because sequence equals zero from some point on).
Notice that every vector is related to a functional (so ) given by the formula . Moreover, the mapping defined as is linear (indeed and ), and actually is a monomorphism. Thus, if , and so , then is an isomorphism!
If is a linear mapping, then given as is called its dual mapping.
It is easy to notice, that , , and that if is an isomorphism, then is an isomorphism as well.
Moreover, one can check that if is a monomorphism, then is an epimorphism. On the other hand, if is an epimorphism, then is a monomorphism.
Some things get easier under the assumption that the dimensions are finite. It becomes clear that then , and if is a basis of , and is a basis of , and and are respective dual bases then , where is the transposition, meaning that the rows are swapped with columns. Indeed, if is the entry in the -th row and -th column of , we get , and this is the entry in -th row and -th column of matrix , where are vectors from , and are vectors from .
It is also worth mentioning that in the finite dimensional case, after identifying we get .
Dual space to an unitary space
Notice that if is an unitary space, we get an easy identification between and . Consider a mapping defined as . One can prove that is a bijection and , so is an isomorphism , where is the space with modified operation of multiplication . From this point on we will be identifying with .
It is easy to prove that assuming this identification for any endomorphisms , if and only if for any . Therefore, . The mapping is self-adjoint if and only if . Moreover, a mapping is self-adjoint if and only if its matrix in an orthonormal basis is hermitian.
There is a common property of self-adjoint endomorphisms and unitary automorphism. For both of them we get . The mappings satisfying this condition are called normal endomorphims. One can prove that an endomorphism has an orthonormal basis consisting of eigenvectors if and only if it is normal.