# How is $\mathcal{L}(V,W) \cong W\mathcal{^B}$ where $\mathcal{B}$ is a basis for $V$?

How is there a bijective map between the sets $$\mathcal{L}(V,W) = \{T ~|~ T:V \to W\}$$ and $$W^\mathcal{B} = \{T ~|~ T: \mathcal{B} \to W\}$$ where $$T$$ is a linear function and $$\mathcal{B}$$ is a basis for $$V$$.

Denote $$\mathcal{B} = \{b_1,b_2, \cdots, b_n\}$$. It makes sense that for any $$v \in V$$, $$v = \sum \alpha_i b_i$$ and so each $$v$$ is composed of a unique tuple $$(a_1,a_2, \cdots, a_n)$$ to ensure the injectivity criteria. For surjectivity, we have the spanning property of the basis that does the work.

But this doesn't make much sense to me because $$T(v) \in W$$ where $$v\in V$$ and $$T(\sum \alpha_i b_i) \in W$$ where $$\sum a_i b_i \notin \mathcal{B}$$.

• $\mathcal{L}(V,W)$ is not the set of all functions from $V$ to $W$ but the set of all the linear functions from $V$ to $W$. – Robin Carlier Nov 6 '19 at 18:17
• In $\mathcal{L}(V,W)$ you take all linear maps. In $W^\mathcal{B}$ you take all maps. the symbol $\cong$ denotes a bijection. This is what happens when we choose a basis $\mathcal{B}$ for the vector space $V$. – GEdgar Nov 6 '19 at 18:19
• So is the notation $\{T ~| ~T : V \to W\} \cong \{T ~|~ T : \mathcal{B} \to W\}$ correct? I'm sorry I still don't get it – charlesh Nov 6 '19 at 18:28
• @charlesh No, it is not. $\{T\, |\, T: V \to W\}$ denotes the set of all maps from $V$ to $W$ while $\mathcal{L}(V, W)$ denotes only the set of linear maps. – Robin Carlier Nov 6 '19 at 18:32
• I forgot to mention that $T$ is a linear function. Will it satisfy then? That, $\mathcal{L}(V,W) = \{T | T:V \to W\}$ and $W^\mathcal{B}= \{T | T : \mathcal{B} \to W \}$. I have edited the original post as well – charlesh Nov 6 '19 at 18:40

Say $$\mathcal{B}$$ is a basis of $$V$$. Then, the set $$W^{\mathcal{B}}$$ is the set $$\\{T|T: \mathcal{B} \to W\ \text{is a map} \\}$$. This is a set of all maps, without restriction, from the finite set $$\mathcal{B}$$ to $$W$$.
The set $$\mathcal{L}(V, W)$$ can be written as $$\\{T | T: V \to W, T\ \text{is linear}\\}$$.
Now why are those sets in bijection? Let $$T$$ be a linear map, and $$v \in W$$, as you noted, $$v = \sum a_i b_i$$ for some unique $$n$$-uple $$(a_1,\ldots,a_n)$$. Since $$T$$ is linear, $$T(v) = \sum a_i T(b_i)$$. So the information about the values of $$T$$ on $$b_i$$ is enough to reconstruct, by linearity, the value of $$T$$ for any element of $$V$$. This is why the sets are in bijection. In fact, both sets are vector spaces and you can even prove they are isomorphic as such.