Set of all linear maps between spaces - [ilmath]L(U,V)[/ilmath]

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Definition

Let [ilmath]\mathbb{K} [/ilmath] be a field and let [ilmath](U,\mathbb{K})[/ilmath] and [ilmath](V,\mathbb{K})[/ilmath] be vector spaces over [ilmath]\mathbb{K} [/ilmath]. We define:

• [ilmath]L(U,V):\eq\{f:U\rightarrow V\in\mathcal{F}(U,V)\ \vert\ f\ \text{is a } [/ilmath][ilmath]\text{linear map} [/ilmath][ilmath]\} [/ilmath], here [ilmath]\mathcal{F}(U,V)[/ilmath] denotes the set of all functions from [ilmath]U[/ilmath] to [ilmath]V[/ilmath] -
  TODO: this notation isn't fixed yet

That is to say [ilmath]L(U,V)[/ilmath] denotes the set of all linear maps from [ilmath]U[/ilmath] to [ilmath]V[/ilmath].

• Claim 1: [ilmath]L(U,V)[/ilmath] is a vector space over [ilmath]\mathbb{K} [/ilmath] in its own right

Proof of claims

Claim 1: [ilmath]L(U,V)[/ilmath] is a vector space over [ilmath]\mathbb{K} [/ilmath]

1. Addition operation:
   • Let [ilmath]f,g\in L(U,V)[/ilmath] then we define:
     • [ilmath](f+g):U\rightarrow V[/ilmath] by [ilmath](f+g):u\mapsto f(u)+g(u)[/ilmath]
   • Explicitly, the operation [ilmath]+:L(U,V)\times L(U,V)\rightarrow L(U,V)[/ilmath] is [ilmath]+:(f,g)\mapsto(f+g)[/ilmath] as defined above.
2. Scalar multiplication operation:
   • Let [ilmath]\alpha\in\mathbb{K} [/ilmath] and let [ilmath]f\in L(U,V)[/ilmath] then define:
     • [ilmath](\alpha f):U\rightarrow V[/ilmath] by [ilmath](\alpha f):u\mapsto \alpha f(u)[/ilmath]
   • Explicitly, the operation [ilmath]*:\mathbb{K}\times L(U,V)\rightarrow L(U,V)[/ilmath] is [ilmath]*:(\alpha,f)\mapsto (\alpha f)[/ilmath] as defined above.

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It remains to be shown that with these operations that [ilmath]L(U,V)[/ilmath] is actually a vector space, however the remainder of the proof is easy and routine

This proof has been marked as an page requiring an easy proof

See also

• The set of all continuous linear maps between spaces - [ilmath]\mathcal{L}(U,V)[/ilmath]
• The set of all continuous maps between spaces - [ilmath]C(X,Y)[/ilmath]

References

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