Chapter 1: Problem 3
Ist \(X\) eine nichtleere Menge, \(V\) ein \(K\)-Vektorraum und \(\mathrm{Abb}(X, V)\) die Menge aller Abbildungen von \(X\) nach \(V\), so ist auf Abb \((X, V)\) durch $$ (f+g)(x):=f(x)+g(x), \quad(\lambda \cdot f)(x):=\lambda f(x) $$eine Addition und eine skalare Multiplikation erklärt. Zeige, daß \(\mathrm{Abb}(X, V)\) mit diesen Verknüpfungen zu einem \(K\)-Vektorraum wird.
Short Answer
Step by step solution
Understand the Given Operations
Vector Space Axioms
Prove Addition Properties
Prove Scalar Multiplication Properties
Prove Distributive Properties
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function space
Vector space axioms
- Addition Axioms: Associativity, Commutativity, and the existence of an Additive Identity. Functions \( f \) and \( g \) within this space should add such that the operation remains closed within the same space. Also, \( f+g \) must equal \( g+f \) (commutativity), and grouping doesn't affect the sum, i.e., \( (f+g)+h = f+(g+h) \) (associativity). Finally, an element like a zero function must exist so that \( f+0 = f \).
- Scalar Multiplication Axioms: There needs to be a scalar multiplication identity, \( 1 \cdot f = f \). Another important rule is associativity with scalars: \( \lambda(\mu f) = (\lambda \mu) f \).
- Distributive Properties: For any function \( f \) and \( g \), and scalars \( \lambda \) and \( \mu \), distributive laws such as \( \lambda(f+g) = \lambda f + \lambda g \) and \((\lambda + \mu) f = \lambda f + \mu f \) must hold.
Scalar multiplication
- Identity Property: If you scale a function by \( 1 \), it stays unchanged, so \( 1 \cdot f(x) = f(x) \).
- Associative Property: For scalar multiplication, \( \lambda (\mu f) = (\lambda \mu) f \), ensuring that combining scalars is consistent, irrespective of how the order of multiplication is structured.
Function addition
- Commutative Property: Order doesn’t matter in addition, so \( f+g = g+f \).
- Associative Property: Regardless of grouping, the result is the same, such as \((f+g)+h = f+(g+h)\).
- Existence of Identity Element: Just like zero in numbers, the zero function \( 0(x) = 0_V \) satisfies that \( f + 0 = f \).