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.