Duality and Inner Products

Dual space

• Let $V$ be a vector space over a field $k$.
• A linear functional $\lambda$ on $V$ is a linear transformation $\lambda: V \to k$.
• Linear functionals form a vector space, called the dual space $V^*$ of $V$.

Duality in Euclidean spaces

• Assume now that $V$ is a finite-dimensional vector space over $\mathbb{R}$ equipped with an inner product $\langle\cdot,\cdot\rangle$.
• Let $w\in V$ be a fixed vector. Then we can define a linear functional $\lambda_w: V \to\mathbb{R}$ by $\lambda_w(v) = \langle v, w \rangle$ for an arbitrary vector $v\in V$.
• We thus obtain a linear map $ev: V \to V^*$ given by $ev: w \mapsto \lambda_w$.
• This is injective: if $w \neq 0$, $\lambda_w = \langle w, w \rangle >0$ and so $\lambda_w \neq 0$.

In fact, the linear transformations $\lambda_w$ are all the linear transformations, i.e., the map $ev: V \to V^*$ is bijective, as we see from the following theorem.

Theorem: Let $\alpha: V \to \mathbb{R}$ be any linear functional. Then there exists a vector $w \in V$ such that $\alpha = \lambda_w$, i.e., for all $v\in V$, $\alpha(v) = \lambda_w(v)$.

Proof:

• Fix a linear functional $\alpha: V \to \mathbb{R}$.
• We construct below a vector $w$ with $\alpha = \lambda_w$.
• Recall that $V$ has an orthonormal basis $v_1$, $v_2$, $\dots$, $v_n$.
• Let $w = \sum\limits_{i=1}^n \alpha(v_i) v_i$.
• For $1\leq j \leq n$, we see that
• $\lambda_w(v_j) = \langle v_j, w \rangle$ by definition.
• $\langle v_j, w \rangle = \sum\limits_{i=1}^n \alpha(v_i) \langle v_j, v_i \rangle = \alpha(v_j)$, as $w = \sum\limits_{i=1}^n \alpha(v_i) v_i$ and $\langle v_j, v_i \rangle = \delta_{ji}$.
• Thus, $\lambda_w(v_j) = \alpha(v_j)$.
• As linear transformations are determined by their values on basis elements, it follows that $\alpha = \lambda_w$, completing the proof of the theorem.
• As $\alpha$ was arbitrary and $\alpha = \lambda_w = ev(w)$ (the second equality is by definition of $ev$), it follows that $ev$ is surjective, hence bijective.