# Linear functionals

• Jan 1st 2010, 03:03 AM
math.dj
Linear functionals
Let V be a finite dimensional vector space over field F and let v belong to V,v not equal to 0.Then there exists some f in V* such that f(v) is not equal to 0.
(where V* is dual space of V)
• Jan 1st 2010, 04:15 AM
Opalg
Quote:

Originally Posted by math.dj
Let V be a finite dimensional vector space over field F and let v belong to V,v not equal to 0.Then there exists some f in V* such that f(v) is not equal to 0.
(where V* is dual space of V)

There is a basis $\displaystyle \{v, e_2,\ldots,e_n\}$ of V with v ($\displaystyle =e_1$) as its first element. Every element x of V has a unique expression $\displaystyle x = \sum_{i=1}^n\alpha_ie_i$ as a linear combination of the basis vectors. Define $\displaystyle f(x) = \alpha_1$.
• Jan 1st 2010, 11:09 AM
Roam
Quote:

Originally Posted by math.dj
Let V be a finite dimensional vector space over field F and let v belong to V,v not equal to 0.Then there exists some f in V* such that f(v) is not equal to 0.
(where V* is dual space of V)

f is a mapping $\displaystyle f: V \rightarrow F$, isn't it? It is termed a linear functional if it's a linear mapping from V into F.

If $\displaystyle \{v_1,...,v_n \}$ is a basis of Vover F, let $\displaystyle f_1, ...,f_n \in V^*$ be the linear functionals defined by Kronecker delta notation:

$\displaystyle f_i(v_j) = \delta_{ij} =\left \{\begin{matrix} 1, & \mbox{if } i=j \\ 0, & \mbox{if } i \ne j \end{matrix}\right.$

Then set $\displaystyle {f_1,...,f_n}$ is a basis for $\displaystyle V^*$. Anf those linear functionals f_i are unique since they're defined on a basis of V.