The Dual Base for E* of a Finite-Dimensional Vector Space E

Let $E$ be a finite-dimensional vector space, say $\mathrm{dim} (E) = n$ and let $\{ e_1, e_2, ..., e_n \}$ be a basis for $E$. Then every $x \in E$ can be written uniquely in the form:

(1)

\begin{align} \quad x = \lambda_1 e_1 + \lambda_2 e_2 + ... + \lambda_n e_n \end{align}

For each $1 \leq i \leq n$, let $e_i^*$ be the linear form on $E$ defined for all $x \in E$ by:

(2)

\begin{align} \quad \langle x, e_i^* \rangle := \lambda_i \end{align}

Then $\{ e_1^*, e_2^*, ..., e_n^* \}$ is a base for $E^*$. Indeed, $\{ e_1^*, e_2^*, ..., e_n^* \}$ is linearly independent, for if:

(3)

\begin{align} \quad \sum_{i=1}^{n} \mu_i e_i^* = 0 \end{align}

for some $\mu_1, \mu_2, ..., \mu_n \in \mathbf{F}$, then for each $1 \leq j \leq n$:

(4)

\begin{align} 0 = \left \langle e_j, \sum_{i=1}^{n} \mu_i e_i^* \right \rangle = \mu_j \langle e_j, e_j^* \rangle = \mu_j \end{align}

so that $\mu_1 = \mu_2 = ... = \mu_n = 0$.

Furthermore, $\{ e_1^*, e_2^*, ..., e_n^* \}$ spans $E^*$, since for each $f \in E^*$ we have that:

(5)

\begin{align} \quad \langle x, f \rangle = \left \langle \sum_{i=1}^{n} \lambda_i e_i, f \right \rangle = \sum_{i=1}^{n} \lambda_i \langle e_i, f \rangle \end{align}

So for each $1 \leq i \leq n$, let $\mu_i := \langle e_i, f \rangle$. Then from above we have that:

(6)

\begin{align} \quad \langle x, f \rangle = \sum_{i=1}^{n} \lambda_i \mu_i = \sum_{i=1}^{n} \mu_i \langle x, e_i^* \rangle \end{align}

Thus, $f = \sum_{i=1}^{n} \mu_i e_i^*$.