Dual vector spaces

We have a n \in \mathbb{N} finite dimensional vector space V such that \dim V = n, with a basis \{\mathbf{e}_i\}_{i=1}^n and a field F. In the following sections we make use of the Einstein summation convention introduced.

Definition 1: let \mathbf{\hat f}: V \to F be a covector or linear functional on V if for all \mathbf{v}_{1,2} \in V and \lambda, \mu \in F we have

\mathbf{\hat f}(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2) = \lambda \mathbf{\hat f}(\mathbf{v}_1) + \mu \mathbf{\hat f}(\mathbf{v}_2).

Throughout this section covectors will be denoted by hats to increase clarity.

Definition 2: let the the dual space V^* \overset{\text{def}} = \mathscr{L}(V, F) denote the vector space of covectors on the vector space V.

Each basis \{\mathbf{e}_i\} of V therefore induces a basis \{\mathbf{\hat e}^i\} of V^* by

\mathbf{\hat e}^i(\mathbf{v}) = v^i,

for all \mathbf{v} = v^i \mathbf{e}_i \in V.

Theorem 1: the dual basis \{\mathbf{\hat e}^i\} of V^* is uniquely determined by

\mathbf{\hat e}^i(\mathbf{e}_j) = \delta_j^i,

for each basis \{\mathbf{e}_i\} of V.

??? note "Proof:"

Let $\mathbf{\hat f} = f_i \mathbf{\hat e}^i \in V^*$ and let $\mathbf{v} = v^i \mathbf{e}_i \in V$, then we have

$$
    \mathbf{\hat f}(\mathbf{v}) = \mathbf{\hat f}(v^i \mathbf{e}_i) = \mathbf{\hat f}(\mathbf{e}_i) v^i = \mathbf{\hat f}(\mathbf{e}_i) \mathbf{\hat e}^i(\mathbf{v}) = f_i \mathbf{\hat e}^i (\mathbf{v}), 
$$

therefore $\{\mathbf{\hat e}^i\}$ spans $V^*$. 

Suppose $\mathbf{\hat e}^i(\mathbf{e}_j) = \delta_j^i$ and $\lambda_i \mathbf{\hat e}^i = \mathbf{0} \in V^*$, then

$$
    \lambda_i = \lambda_j \delta_i^j = \lambda_j \mathbf{\hat e}^j(\mathbf{e}_i) = (\lambda_j \mathbf{\hat e}^j)(\mathbf{e}_i) = \mathbf{0},
$$

for all $i \in \mathbb{N}[i \leq n]$. Showing that $\{\mathbf{\hat e}^i\}$ is a linearly independent set.

Obtaining a vector and consequent covector space having the same dimension n.

From theorem 1 it follows that for each covector basis \{\mathbf{\hat e}^i\} of V^* and each \mathbf{\hat f} \in V^* there exists a unique collection of numbers \{f_i\} such that \mathbf{\hat f} = f_i \mathbf{\hat e}^i.

Theorem 2: the dual of the covector space (V^*)^* \overset{\text{def}} = V^{**} is isomorphic to V.

??? note "Proof:"

Will be added later.

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Dual vector spaces

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