Finite vector space V,W so is L(V,W)

Question

Preceding example:

If V,W are vector spaces then $Map\left ( V,W \right )$ is an infinite dimensional vector space with operation defined by $\left ( f+g \right )\left ( v \right ):=f\left ( v \right )+g\left ( v \right )$ and $ \left ( \alpha f \right )\left ( v \right ):=\alpha f\left ( v \right )$. The linear map $L\left ( V,W \right )$ form a subspace of Map$\left ( V,W \right )$

I believe the question below is preceded by this example:

Prove that if V,W are finite dimensional then so is $L\left ( V,W \right )$.

Here's my attempt:

Let V,W be finite dimension n. Hence, V,W contains a set of n linearly independent vectors but no n+1 linearly independent vectors.

Let $L:V\rightarrow W$

with $ \left ( f+q \right )\left ( v \right ):=f\left ( v \right )+g\left ( v \right )$ and $\left ( \alpha f \right )\left ( v \right ):=\alpha f\left ( v \right )$

Seemingly, I am unable to progress forward.

Hints are appreciated

Answer 1

When first learning about function structures like your $L(V,W)$, I felt it useful to think of matrices.

Remember all linear transformations $T:V\to W$ where $V$ and $W$ are finite-dimensional have matrix representations. Once you fix bases, you can find a matrix $M$ such that $T(x) = Mx$. If $T$ sends $n-$dimensional vectors to $m-$dimensional vectors then the size of $M$ is size $m\times n$. Specifically, $\dim(L(V,W)) = \dim(M_{m\times n}) = nm$.

So here's what you do. Fix bases $B_V$ and $B_W$ for $V,W$ respectively (you can even use a coordinate transformation to work in $R^k$ if you wish). Then count the number of transformations by counting the number of ways to map $B_V$ to $B_W$. Linear transformations are determined by what they do to a basis.

Answer 2

In case the following words make sense:

The category of finite-dimensional vector spaces over a field $k$ is symmetric monoidal closed and hence enriched over itself.