# Why does this proof that a dual space isomorphic to the original space work?

I am extremely new to the concept of dual space and the notation $$\mathcal{L}(V,W)$$ which denotes the set of all linear transformations from $$V\to W$$. I read that the dual space $$V^*=\mathcal{L}(V,\Bbb{K})$$ is isomorphic to $$V$$, where $$V$$ is some vector space defined over $$\Bbb{K}$$. The “proof” (or perhaps intuition) that was given to me was that $$\dim\mathcal{L}(V,\Bbb{K})=\dim(V)\dim(\Bbb{K})=\dim(V)$$.

Two questions:

1. Rigorously, and I apologise if this is naïve but as I say I have never seen these concepts until approximately five minutes ago, why is $$\dim\mathcal{L}(A,B)=\dim(A)\dim(B)$$?

2. How is it a complete proof to just show that the dimensions of $$V^*$$ and $$V$$ are the same? Surely some more work needs to be done to show isomorphisms? For context, I have never actually seen a proof of isomorphism and do not know what proving an isomorphism in an abstract context like this entails.

It's important that we're talking about finite-dimensional vector spaces here. If $$\dim(A)=n$$ and $$\dim(B)=m$$, then there's a $$1$$-$$1$$ correspondence between linear transformations $$f:A \to B$$ and $$m \times n$$ matrices. That's because any such transformation is determined by its values on a basis for $$A$$, and those values in turn are determined by a column vector with $$m$$ components. The dimension of the space of $$m \times n$$ matrices is $$mn$$.
It's also a theorem that any two finite-dimensional vector spaces over the same field with the same dimension are isomorphic to one another. Choose a basis for each and map element $$k$$ of the basis for $$A$$ to element $$k$$ of the basis for $$B$$. That results in a linear map that is onto a basis of $$B$$, and therefore is onto $$B$$, and that has trivial kernel.