If the set of all linear maps is itself a vector space, then what is its basis and dimension?

Question

In linear algebra, we learn about linear mappings and it is an easy proof that the set of all linear maps is itself a vector space over the same field. But if it's a vector space how would one go about figuring out what a basis for the set of linear maps are? To be more specific we can take the set of all linear map from R^3 to R^2 over the field R.

Answer 1

A bit more basic: Let $V,W$ be vector spaces over a field $F$. Let $B_V$ and $B_W$ be bases of these vector spaces. Then each linear map $L:V\longrightarrow W$ can be uniquely written as $$\begin{align}L\left(\sum_{x\in B_V} a_x x\right)&=\sum_{x\in B_V}a_x L(x)\\ &=\sum_{x\in B_V}a_x\sum_{y\in B_W}b_y y\\ &=\sum_{x\in B_V}\sum_{y\in B_W}a_xb_y y\end{align}$$ The coefficients $c_{xy}:=a_x b_y$ uniquely determine the function. That's the idea of the matrix representation of a linear map. Those coefficients are the matrix entries in the finite dimensional case. Now let's define the linear maps $L_{xy}:V\longrightarrow W$ which map $x\in B_V$ to $y\in B_W$ and map every other element of $B_V$ to $0$. We can see that using the coefficients from above, we can write $L=\sum_{x\in B_V}\sum_{y\in B_w} c_{xy}L_{xy}$., and there is only one way to do so. That means, the maps $L_{xy}$ are a basis of the space of linear maps $V\longrightarrow W$. There is exactly one such map for each combination of vectors in $B_V$ and $B_W$, that is, one for each element of $B_V\times B_W$. So the dimension is the cardinality of this product. In the finite dimensional case, that's the product of the cardinalities of $B_V$ and $B_W$, so the product of the dimensions of $V$ and $W$.

So that's the takeaway: if $\dim V=n,\dim W=m$, then $\dim\operatorname{Hom}(V,W)=nm$.