Dual spaces- An exposition

by ayushkhaitan3437

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 V over field \Bbb{K} satisfying \dim V=n. Then this vector space is isomorphic to the vector space of n-tuples over \Bbb{K}, 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 n-tuple representation is most clearly suggestive of the steps of the proof.

Now in such an n-tuple vector space, every term can multiply with every vector in V to give a map into \Bbb{K}. Note that it is not a map into V. It is a map into \Bbb{K}. This n-tuples vector space is called the dual space of V, as it mimics the properties of V amazingly well. Notationally, the dual space of V is represented by Lang as V^*.

I feel studying dual spaces in this order gives the main motivation behind the nomenclature- it is isomorphic to V, just like any other n-tuple vector space over \Bbb{K}.

Now onto a major property. Say V has a basis \{v_1,v_2,\dots, v_n\}. Take W to be a vector subspace of V. Then 0\in W for obvious reasons. Let W be generated by \{a_1,a_2,\dots,a_i\}, where \{a_1,a_2,\dots,a_i\}\subseteq \{v_1,v_2,\dots, v_n\}.

Is V\setminus W a vector space? No. It does not contain 0. However, it is generated (almost) by n-i basis vectors of V. Hence, if it were a vector space, we could write \dim W+\dim V\setminus W=\dim V=n! But we can’t, as V\setminus W is not a vector space. However, if we could somehow embed V\setminus W into a vector space which is of n-i 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 W to 0. Note that it can map vectors in V\setminus W to 0 too. But it definitely maps all vectors in W to 0, regardless of what it does with the vectors in V\setminus W.

Let us call this set \beta, and pick an arbitrary functional T\in \beta. Then T(a_k)=0 for all k\in \{1,2,3,\dots,i\}. Check this for yourself. It is also easy to check that \beta is a vector space over \Bbb{K}, \dim \beta=n-i. If the basis of V is written as \{a_1,a_2,\dots,a_i,b_1,b_2\dots,b_{n-i}\}, then each element of \beta can be visualised as \{0,0,\dots,0,c_1,c_2,\dots,c_{n-i}\}, where c_1,c_2,\dots,c_{n-i} are the elements in \Bbb{K} that b_1,b_2,\dots,b_{n-i} are mapped to respectively.

Hence, we have \dim W+\dim\beta=n, as i+(n-i)=n. It is strange that we are adding dimensions of subspaces that belong to different vector spaces (W belongs to V while \beta belongs to V^*). 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 V\setminus W into a suitable vector space would have been the vector subspace of elements with coefficients of a_1,a_2\dots,a_i as 0. This is obviously isomorphic to \beta. The point here is that we didn’t necessarily need to traverse to a different vector space to find a suitable subspace of dimension n-i. We had one right at home.

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