Why $\dim(\hom(V,W))=\dim(V) * \dim(W)$? I have found that the question I want to ask someone had asked, here is the website:
$\hom(V,W)$ is canonic isomorph to $\hom(W^*, V^*)$
Here is my question: Why $\dim(\hom(V,W))=\dim V * \dim W $?

Thanks for explanation~!
 A: Let $V, W$ be finite-dimensional vector spaces over a field $K$ with $\dim V=n$, $\dim W=m$. By picking bases in $V$ and $W$ we obtain $K$ vector space isomorphisms $V\cong K^n$ and $W\cong K^m$. This reduces your question to
$$
\dim(\hom(K^n, K^m)) = n\cdot m.
$$
Now every homomorphism $\varphi:K^n\to K^m$ is uniquely determined by the projections of the images of basis vectors, i.e. $\pi_j(\varphi(e_i))$ where $e_i$ is the $i$-th standard basis vector of $K^n$ and $\pi_j : K^m\to K$ is projection onto the $j$-th coordinate. In fact, if we define the homormorphisms
$$
\delta_{ij} : K^n \to K^m,\ e_k \mapsto
\begin{cases}
0 & k \neq i, \\
e_j & k = i,
\end{cases}
$$
we found a basis of $\hom(K^n, K^m)$, where every $\varphi:K^n \to K^m$ is expressed as
$$
\varphi = \sum_{i=1}^n \sum_{j=1}^m \pi_j(\varphi(e_i))\,\delta_{ij}.
$$
This is just the usual representation of $\varphi$ as an $n\times m$-matrix with entries in $K$, where the entries are the $\pi_j(\varphi(e_i))$.
Thus, we found a basis of $\hom(K^n, K^m)$ with $n\cdot m$ elements, so $\dim(\hom(K^n, K^m)) = n\cdot m$.
