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

I have a proof for the following in a 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.

1. 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.
2. If $\dim (V)$ is finite then $\dim (\hom (V, W)) = \dim (V)\times\dim (W)$, 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 $\oplus_{\dim (V)}\hom (F, W)$
3. 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 $|\hom (V, W)| = |W|^{\dim (V)} = |F|\times\dim (\hom (V, W)) = \dim (\hom (V, W))$.

Answer 2

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

Answer 3

This solution is meant to fill the gaps in Tom's answer.

Let $F$ be the underlying field, and let $I$, $J$ be bases for $V$, $W$, respectively. We denote a function space by $X \to Y$, and its subspace of finitely supported functions by $X \to_0 Y$. Note that $$\operatorname{Hom}(V, W) \cong I \to (J \to_0 F).$$ We claim that $$\dim(\operatorname{Hom}(V, W)) = \begin{cases} 0, & J = \varnothing, \\ |I| \times |J|, & I \text{ finite}, \\ |\operatorname{Hom}(V, W)|, & \text{otherwise}. \end{cases}$$ The first case is trivial. The second case follows from $$I \to (J \to_0 F) = I \to_0 (J \to_0 F) \cong I \times J \to_0 F,$$ which is just a direct sum of $|I \times J|$ copies of $F$. We focus on the last case.

This proof uses two main lemmas. One of them is the generalized form of Erdős–Kaplansky, stating $\dim(X \to F) = |X \to F|$ for $X$ infinite. The other is the equality $|X \to_0 Y| = \max\{|X|, |Y|\}$ when $X$ is infinite and $|Y| \ge 2$. See this answer for a proof of both.

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If $J \ne \varnothing$ and $I$ is infinite, notice that $$|F^I| = \dim(I \to F) \le \dim(I \to (J \to_0 F)) \le |I \to (J \to_0 F))| = \max\{|J^I|, |F^I|\}.$$ If $|J^I| \le |F^I|$ we're done, so suppose $|J^I| > |F^I|$ and in particular $|I \to (J \to_0 F)| > |F|$. Recall that for a vector space $U$ over $F$, $$|U| = \max\{|F|, \dim(U)\}.$$ This follows from taking a basis $\mathcal B$ and identifying $U \cong \mathcal B \to_0 F$. Substituting $U = I \to (J \to_0 F)$, we find that $|U| = \dim(U)$, as we wanted.