On the grounds that I'm currently teaching a linear algebra class and I enjoy making my students furious, let a **linear algebra** be an algebra $\mathcal{A}$ in the sense of universal algebra equipped with a linear ordering. More interestingly, given a linear algebra $\mathfrak{A}=(\mathcal{A},\le)$, say that a **linear subpower** of $\mathfrak{A}$ is a subalgebra of some power structure $\mathcal{A}^\kappa$ which is linearly ordered by the corresponding product relation $\le^\kappa=\{(a,b): \forall \eta<\kappa, a(\eta)\le b(\eta)\}$ (which is a priori only a partial order).

For example, let ${\bf 2}$ be the two-element Boolean algebra ordered as usual. While (being nontrivial) it has arbitrarily large powers, it has no linear subpowers with more than two elements.

I'm curious how subtle the bounds on sizes of linear subpowers can be:

Suppose $\mathfrak{A}$ is a finite linear algebra which has an infinite linear subpower. Must $\mathfrak{A}$ have arbitrarily large

finitelinear subpowers?

This is related to an old question of mine, which in retrospect seems unlikely to have a good answer.