Linear Forms on a Vector Space and its Algebraic Dual

# Linear Forms on a Vector Space and its Algebraic Dual

Definition: Let $E$ be a vector space over $\mathbf{F}$. A Linear Form (or Linear Functional) on $E$ is a linear operator from $E$ to $\mathbf{F}$. |

Definition: Let $E$ be a vector space. The Algebraic Dual of $E$ denoted by $E^*$, is the set of all linear forms on $E$. |

## The Existence of Nonzero Linear Forms

Proposition 1: Let $E$ be a vector space. Then, for each $a \in E$ with $a \neq 0$ there exists an $f \in E^*$ such that $f(a) \neq 0$. |

**Proof:**Let $a \in E$ with $a \neq 0$. Since every vector space has a base (as proven on the Every Vector Space has a Base page) and since $\{ a \}$ is a linearly independent set, there exists a subset $A \subset E$ containing $a$ such that $A$ is a base of $E$.

- Let $f : E \to \mathbb{R}$ be defined for each $a \in A$ by $f(a) = 1$, and extend $f$ linearly. That is, for an arbitrary $x \in E$, write $x = \lambda_1 a_1 + \lambda_2 a_2 + ... + \lambda_n a_n$ where $\lambda_1, \lambda_2, ..., \lambda_n \in \mathbf{F}$ and where $a_1, a_2, ..., a_n \in A$. Then define:

\begin{align} \quad f(x) = f(\lambda_1 a_1 + \lambda_2 a_2 + ... + \lambda_n a_n) := \lambda_1 f(a_1) + \lambda_2 f(a_2) + ... + \lambda_n f(a_n) = \lambda_1 + \lambda_2 + ... + \lambda_n \end{align}

- Then $f \in E^*$ and is such that $f(a) = 1 \neq 0$. $\blacksquare$

## Nonzero Linear Forms on a Topological Vector Space are Open Maps

Proposition 1: Let $E$ be a topological vector space. Then every nonzero linear form $f$ on $E$ is an open map, i.e., every nonzero linear form $f$ on $E$ maps open sets in $E$ to open sets of scalars. |

**Proof:**Let $A \subseteq E$ be open and take any point $x \in A$. Then $A - x$ is a neighbourhood of the origin, since $o = x - x \in A - x$. By the proposition on the Bases of Neighbourhoods for a Point in a Topological Vector Space, $A - x$ is absorbent.

- Since $f$ is a nonzero linear form, there exists a point $a \in E$ such that $f(a) = 1$.

- Since $A - x$ is absorbent and since $a \in E$, there exists a $\lambda > 0$ such that:

\begin{align} a \in \delta (A - x) \end{align}

- whenever $\delta \in \mathbf{F}$ is such that $|\delta| \geq \lambda$, or equivalently, $\frac{1}{\delta} a \in A - x$ whenever $\delta \in \mathbf{F}$ and $|\delta| \geq \lambda$. Set $\mu := \frac{1}{\delta}$. Then:

\begin{align} \mu a \in A - x \end{align}

- whenever $\mu \in \mathbf{F}$ and $\left | \frac{1}{\mu} \right | \geq \lambda$ or equivalently when $\mu \in \mathbf{F}$ and $|\mu| \leq \lambda$. So $x + \mu a \in A$ whenever $\mu \in \mathbf{F}$ and $|\mu| \leq \lambda$, but then:

\begin{align} \quad f(x + \mu a) = f(x) + \mu f(a) = f(x) + \mu \in A \end{align}

- for all $\mu \in \mathbf{F}$ with $|\mu| \leq \lambda$. If $B(f(x), \lambda)$ denotes the open ball centered at $f(x)$ with radius $\lambda > 0$, then $B(f(x), \lambda) \subseteq f(A)$. So every point of $f(A)$ is an interior point of $f(A)$, and thus $f(A)$ is open.

- So $f$ is an open map. $\blacksquare$