# Dual spaces- An exposition

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.