The dual space of \(V\), named \(V’\), is the vector space formed by linear functionals on \(V\) (because recall set of linear maps between two vector spaces form a vector space).

constituents

A vector space \(V\)

requirements

\(V’ = \mathcal{L}(V, \mathbb{F})\) , and its a vector space.

additional information

dimension of dual space is equivalent to the original space

\begin{equation} \dim V’ = \dim V \end{equation}

Proof:

Because \(\dim \mathcal{L}(V,W) = (\dim V)(\dim W)\), and \(V’ = \mathcal{L}(V, \mathbb{F})\). Now, \(\dim V’ = \dim \mathcal{L}(V,\mathbb{F}) = (\dim V)(\dim \mathbb{F}) = \dim V \cdot 1 = \dim V\).

dual basis

Let \(v_1, …, v_{n}\) be a basis of \(V\), then, we can construct a basis of \(V’\) with linear functionals \(\varphi_{1}, …, \varphi_{n}\):

\begin{equation} \varphi_{j}(v_{k}) = \begin{cases} 1, if\ k=j \\ 0, if\ k \neq j \end{cases} \end{equation}

Now, we can show that \(\varphi_{j}\) are indeed linear functionals by basis of domain: we defined its behavior of each \(\varphi_{j}\) based on where it sends each \(v_{j}\) (i.e. the basis of \(V\), the domain of elements in \(V’\)) into values in \(\mathbb{F}\) (i.e. \(1\) or \(0\)).

We can now show that these \(\varphi_{j}\) is indeed a basis of \(V’\) by only showing that it is linearly independent because we have already a list of \(n\) \(\varphi_{j}\) elements (i.e. \(\dim V’=\dim V = n\) number of \(\varphi_{j}\)), and linearly independent list of length dim V are a basis of V.