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Bounded linear operators

  • #1

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
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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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?
 

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