Houjun Liu

dual space

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).


A vector space \(V\)


\(V’ = \mathcal{L}(V, \mathbb{F})\) , 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}


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.