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).
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.