# The vector space of all linear maps between two 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:

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]

• 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

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## Notes

1. You may have seen this before as [ilmath]V^U[/ilmath] - the set of all maps from [ilmath]U[/ilmath] into [ilmath]V[/ilmath]