Linear functionals

**math.dj** · January 1st 2010, 03:03 AM

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)

**Opalg** · January 1st 2010, 04:15 AM

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 $\{v, e_2,\ldots,e_n\}$ of V with v ( $=e_1$ ) as its first element. Every element x of V has a unique expression $x = \sum_{i=1}^n\alpha_ie_i$ as a linear combination of the basis vectors. Define $f(x) = \alpha_1$ .

**Roam** · January 1st 2010, 11:09 AM

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

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

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

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

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