Is B(X,Y) a Vector Space of Bounded Linear Operators over the Same Scalar Field?

In summary, to prove that a set is a vector space, the 4 axioms of a vector space must be satisfied. These include the associative and commutative properties of addition, the existence of a zero vector and additive inverse, and the distributive property of scalar multiplication. To show that B(X,Y) is a vector space, we must show that it satisfies these axioms. This can be done by defining operations such as addition and scalar multiplication on linear transformations F and G, as well as showing that these operations satisfy the axioms. Additionally, since X and Y are vector spaces over the same scalar field, L(X,Y) is also a vector space.
  • #1
gtfitzpatrick
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0

Homework Statement



Show L(X,Y) is a vector space. Then if X,Y are n.l.s. over the same scalar field define B(X,Y) = set of all bounded linear operators for X and Y
Show B(X,Y) is a vector space(actually a subspace of L(X,Y)

Homework Equations



The Attempt at a Solution


im not sure if i have taken this question down properly.
To prove some set is a vector space you have to show the 4 axiom of a vector space hold. namely-for u, v, w be arbitrary vectors in V, and a, b be scalars in F
1. u + (v + w) = (u + v) + w.
2. v + w = w + v.
3. There exists an element 0 ∈ V, called the zero vector, such that v + 0 = v for all v ∈ V.
Inverse elements of addition For all v ∈ V, there exists an element w ∈ V, called the additive inverse of v, such that v + w = 0.
4. a(v + w) = av + aw
Im not sure how to progress with this
 
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  • #2
Well, how about doing what you just said!

If F and G are linear transformations from vector space X to vector space Y, F: u-> F(v) and G:u-> G(u), how would you define "F+ G"? F+ G would map a vector u in X to what vector in Y? Does that satisfy F+(G+ H)= (F+ G)+ H for any linear transformations, from X to Y, F, G, and H? Do that satsfy F+ G= G+ F? What would the "0" linear transformation be? What would the adidtive inverse of linear transformation F be?

If F is a linear transformation from X to Y, F:u->F(u), and a is a number how would you define aF? aF would map vector v in X to what vector in Y?
Does that satisfy a(F+ G)= aF+ aG for linear transformations F and G?
 
  • #3
Im not sure how to show these axioms hold. When it says the are over the same scalar field it is saying the they are more or less the same? so if X is a vector space and Y is a vector space, then because they are over the same scalar field L(X,Y) is also a vector space?
 

Related to Is B(X,Y) a Vector Space of Bounded Linear Operators over the Same Scalar Field?

1. What is a bounded linear operator?

A bounded linear operator is a mathematical function that maps one vector space to another, while preserving certain properties such as linearity and boundedness. It is an important concept in functional analysis and is often used in the study of linear transformations.

2. How is a bounded linear operator different from a general linear operator?

A bounded linear operator is different from a general linear operator because it satisfies the additional condition of being bounded. This means that the operator's output is always within a certain range, regardless of the input. In contrast, a general linear operator may produce unbounded outputs for certain inputs.

3. What is the importance of bounded linear operators in mathematics?

Bounded linear operators play a crucial role in many areas of mathematics, including functional analysis, differential equations, and operator theory. They provide a powerful framework for studying linear transformations between vector spaces and have numerous applications in various branches of mathematics and physics.

4. How do you determine if a linear operator is bounded?

A linear operator is bounded if its norm, which is a measure of its size, is finite. The norm of a bounded linear operator is the smallest number that bounds the operator's output for all possible inputs. In other words, the operator's output cannot exceed its norm regardless of the input.

5. Can a bounded linear operator be unbounded in some cases?

No, a bounded linear operator cannot be unbounded in any cases. By definition, a bounded linear operator must have a finite norm, which means its output is always within a certain range. If the operator were unbounded, it would not satisfy this condition and would not be considered a bounded linear operator.

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