Dimension of $\operatorname{Hom}(V, W)$ What is the dimension of $\operatorname{Hom}(V, W)$ if at least one of the two vector spaces $V, W$ is infinite dimensional? In the sense of cardinal numbers. 
Thanks
 A: It depends ;-). If either of $V$ or $W$ is zero-dimensional, we will have $\def\Hom{\mathop{\mathrm{Hom}}}\Hom(V,W) = 0$, regardless of the dimension of the other space. If both of $V$ and $W$ are at least one-dimensional, and one of them is infinite dimensional, $\Hom(V,W)$ is infinite dimensional also:
So suppose $\dim V = \infty$, and $\dim W  \ge 1$ first, let $\{v_n \mid n \in \mathbb N\}$ be a linear independent set in $V$, $w_0 \in W \setminus \{0\}$. Then there are linear maps $f_n \in \Hom(V,W)$ such that 
$$ f_n(v_m) = \delta_{nm}w_0, \quad n,m \in \mathbb N $$
Then $\{f_n \mid n \in \mathbb N\}$ is linear independent, as: Suppose $\sum\alpha_n f_n  = 0$ with only finitely many $ \alpha_n\ne 0$. Applying this to $v_m$ gives 
$$ \alpha_m = \sum \alpha_n f_n(v_m) = 0(v_m) = 0. $$
So $\dim\Hom(V,W) = \infty$.
Suppose now $\dim V \ge 1$, $\dim W = \infty$, let $v_0 \in V \setminus \{0\}$, and $\{w_n \mid n \in \mathbb N\}$ be linear independent in $W$. There are linear maps $g_n \in \Hom(V,W)$ such that 
$$ g_n(v_0) = w_n, \quad n \in \mathbb N $$
Then $\{g_n \mid n \in \mathbb N\}$ is linear independent: Suppose $\sum\alpha_n g_n = 0$. Applying this to $v_0$ gives 
$$ 0 = 0(v_0) = \sum\alpha_n g_n(v_0) = \sum \alpha_n w_n $$
as the $w_n$ are independent, we have $\alpha_n = 0$ for all $n$. So $\dim\Hom(V,W) = \infty$.
A: I have proof for the following in PDF. It's quite lengthy and if I can find a good place to upload it I'll add a link. Here's the answer with an outline.


*

*If $\dim (V) = 0$ or $\dim (W) = 0$ then $\dim (\hom (V, W)) = 0$ This follows as $\hom (V, W)$ only contains the zero transform in either case.

*If $\dim (V)$ is finite then $\dim (\hom (V, W)) = \dim (V).\dim (W)$, whether $\dim (W)$ is finite or infinite. To prove this first establish that $\hom (F, W)$ is isomorphic to $W$, then that $\hom (V, W)$ is isomorphic to $\oplus_{\dim (V)}\hom (F, W)$

*If $\dim (V)$ is infinite and $\dim (W) \ne 0$ then $\dim (\hom (V, W)) = |W|^{\dim (V)}$ (This has a simple corollary that for the dual $V^* = \hom (V, F)$ that $\dim (V^*) = |F|^{dim (V)}$ which confirms for an infinite dimensional space $\dim (V^*) > \dim (V)$). The proof is circuitous. (a) establish for any infinite dimensional space $U$ that $|U| = |F|.\dim (U)$. (b) Establish that $\dim (\hom (V, W))$ is infinite. (c) Establish  for any (non-zero dimensional) $V, W$ that $|\hom (V, W)| = |W|^{\dim (V)}$. (d) Establish that $\dim (\hom (V, W)) \ge |F|$. (e) put all that together so that $|\hom (V, W)| = |W|^{\dim (V)} = |F|.\dim (\hom (V, W)) = \dim (\hom (V, W)$

