Dimension of $\operatorname{Hom}(V, W)$

Question

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

Answer 1

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 $\dim\Hom(V,W) = \infty$.