Studying Dual Spaces can be confusing. I know it was for me. I’m going to try to break down the arguments into a more coherent whole. I am following Serge Lang’s “Linear Algebra” (the Master is my favourite author). However, I am not strictly following the order he follows to develop the theory.
Say we have a vector space over field
satisfying
. Then this vector space is isomorphic to the vector space of
-tuples over
, provided the operations defined are component-wise addition and scalar multiplication across all components.
Convince yourself of the fact above. It really is simple, and the -tuple representation is most clearly suggestive of the steps of the proof.
Now in such an -tuple vector space, every term can multiply with every vector in
to give a map into
. Note that it is not a map into
. It is a map into
. This
-tuples vector space is called the dual space of
, as it mimics the properties of
amazingly well. Notationally, the dual space of
is represented by Lang as
.
I feel studying dual spaces in this order gives the main motivation behind the nomenclature- it is isomorphic to , just like any other
-tuple vector space over
.
Now onto a major property. Say has a basis
. Take
to be a vector subspace of
. Then
for obvious reasons. Let
be generated by
, where
.
Is a vector space? No. It does not contain
. However, it is generated (almost) by
basis vectors of
. Hence, if it were a vector space, we could write
! But we can’t, as
is not a vector space. However, if we could somehow embed
into a vector space which is of
dimension, we’d be done.
There may be multiple ways of doing this. We’re going to look at one in particular- the set of functionals which maps to
. Note that it can map vectors in
to
too. But it definitely maps all vectors in
to
, regardless of what it does with the vectors in
.
Let us call this set , and pick an arbitrary functional
. Then
for all
. Check this for yourself. It is also easy to check that
is a vector space over
,
. If the basis of
is written as
, then each element of
can be visualised as
, where
are the elements in
that
are mapped to respectively.
Hence, we have , as
. It is strange that we are adding dimensions of subspaces that belong to different vector spaces (
belongs to
while
belongs to
). However, we are only adding natural numbers, and nothing else. Hence there is no contradiction.
Note: Another possibility that we could have looked into of embedding into a suitable vector space would have been the vector subspace of elements with coefficients of
as
. This is obviously isomorphic to
. The point here is that we didn’t necessarily need to traverse to a different vector space to find a suitable subspace of dimension
. We had one right at home.