# 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~!

• There is another website also mentioned the truth I want to know:math.stackexchange.com/questions/635039/… – 扁頭科學麵Kevin Lee Apr 22 '14 at 13:26
• Is your question about vector spaces $V, W$ of finite dimension? – Christoph Apr 22 '14 at 13:28
• Do you believe the claim when $V = \mathbb{R}^n$ and $W = \mathbb{R}^m$? In this case, $\mathrm{hom}(V,W)$ is just a space of matrices (and has the dimension that you're looking for). The general (finite-dimensional) case follows by fixing bases. – fuglede Apr 22 '14 at 13:30
• To Christoph: Yes, they are finite dimension. – 扁頭科學麵Kevin Lee Apr 22 '14 at 13:33
• To fuglede: Thanks for your hint, considering R^n and R^m is a way to think. – 扁頭科學麵Kevin Lee Apr 22 '14 at 13:35

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 $\dim(\hom(K^n, K^m)) = n\cdot m$.