Set of all linear maps between spaces - $L(U,V)$

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Definition

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

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

That is to say $L(U,V)$ denotes the set of all linear maps from $U$ to $V$.

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

Proof of claims

Claim 1: $L(U,V)$ is a vector space over $\mathbb{K}$

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

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

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See also

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

References

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