Proof that $\dim\big(\mathcal{L}(V,W)\big)=\dim V *\dim W$, when $V, W$ are finite dimensional. I've been looking a few proofs of this, and I have a problem with the standard one, which uses the concept of isomorphism. Suppose $\dim V=n$ and $\dim W=M$. The proof uses the function $M$ from $\mathcal{L}(V,W)$ (the linear maps from $V$ to $W$), to $\text{Mat}_{m\times n}(F)$ (matrices of $m\times n$), which assigns to each linear map $T$ it's matrix $M(T)$. To do this, they fix first some basis $(v_1,....v_n)$ of $V$ and $(w_1,...w_m)$ of $W$. My problem is this fixing of the basis. Because then the function $M$ is not a function, at least not from $L(V,W)$, because each linear map has many matrices depending of the basis. How can we then state that $\dim\big(\mathcal{L}(V,W)\big)=(\dim V)(\dim W)$ ?
 A: A more conceptual proof:
Le $\mathcal B$ a basis of $V$. By definition of a basis, we have an isomorphism $L(V,W)\simeq W^\mathcal B$, so $$\dim(L(V,W))=\dim(W^\mathcal B)=\lvert\mathcal B\rvert\cdot \dim W =\dim V\cdot \dim W.$$
A: Let $V$ and $W$ be finite dimensional vector spaces and $\left\{v_1,\dots , v_n\right\}$ and $\left\{w_1,\dots , w_m\right\}$ bases of $V$ and $W$ respectively. Define $L_{ij}:V\rightarrow W$ by $L_{ij}(v_k)=\delta_{ik}w_j$. Then $L_{ij}\in L(V,W)$ for all $1\leq i\leq n, 1\leq j\leq m$. Show that any linear map $T\in L(V,W)$ can be uniquely written as a linear combination of the $L_{ij}$'s. It follows that the $L_{ij}$'s are a basis of $L(V,W)$.
Now this proof indeed fixes bases at the beginning, however this is not a problem since any linear map corresponds uniquely to a matrix after choice of basis. When we have two matrices corresponding to the same linear map but w.r.t. different bases, they are linked by some matrix of base change.
A: Suppose, m = dim($V$) and n = dim($W$). Since $L(V,W)$ and $\mathbb{F}^{m,n}$ are isomorphic, dim $(L(V,W))$ = dim $\mathbb{F}^{m,n}$ where dim $\mathbb{F}^{m,n} = mn$ = dim($V$) dim($W$).
A: Fix a basis $\beta$ of $V$ and fix a basis $\gamma$ of $W$. Now $M$ takes $T$ to its matrix $[T]_{\beta}^{\gamma}$ with respect to these bases. This is a function.
