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In Lemma 2 of this answer, it was shown that if $\dim V$ was infinite, then $\dim V^*\ge |F|$. As $V^*=\mathcal{L}(V,F)$, this led me to wonder if the same held for the general space of linear transformations $\mathcal{L}(V,W)$.

More concretely, if either $\dim V$ or $\dim W$ is infinite, does $\dim \mathcal{L}(V,W)\ge |F|$?

As far as I can see, my conjecture holds at least for the case when $\dim V$ is infinite, if we modify the proof for $\dim V^*\ge |F|$ slightly, as follows $-$

Consider an arbitrary non-zero vector $v\in W$. If we re-define $\mathbf{f}_c$ as $\mathbf{f}_c\colon V\to W$, $\mathbf{f}_c(\epsilon_n) = c^n v$ if $n\in\omega$, and $\mathbf{f}_c(\epsilon_i)=0$ if $i\geq\omega$. Then as $\alpha_1\mathbf{f}_{c_1} + \cdots + \alpha_m\mathbf{f}_{c_m} = \mathbf{0}$ implies \begin{align} (\alpha_1 c_1^i + \cdots + \alpha_m c_m^i)v &= 0 \\ \implies \alpha_1 c_1^i + \cdots + \alpha_m c_m^i &= 0, \end{align}

the rest of the proof should proceed just as it did for $\dim V^*\ge |F|$. However, I have not been able to see if my conjecture is valid when $\dim W$ is infinite.

Ishan Deo
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1 Answers1

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If $\dim V$ is infinite and $W$ is not $0$, then by choosing a non-zero vector $x\in W$, you get an inclusion : $$\mathcal L(V,F) \hookrightarrow \mathcal L(V,W)$$

Therefore, $\dim\mathcal L(V,W) \geq |F|$ by the lemma you cite.

If $\dim V= n$ is finite and $\dim W = \kappa$ is infinite, then : $$\mathcal L(V,W) \simeq W^n$$ Therefore $\dim \mathcal L(V,W) = \kappa^n = \kappa$, which can be $<|F|$. (For example if $F = \mathbb R$ and $\kappa = \aleph_0$)

SolubleFish
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