# Linear algebra with discrete, ordinal, but non-numeric values?

I have a series of statements relating various discrete, ordinal, but non-numeric values. I'd like to reason about those statements intelligently. Looking at the relationships I want to say that it is a linear algebra problem but I'm not sure you can use linear algebra rules with non-numeric coefficients.

vastly oversimplified example case I have the following ordinal relationships

a < b < c
d < e


and the following equations

{a, d} = f
{c, e} = g


in this case, there is no relationship between, say, a and e. It's not that the relationship isn't known, it doesn't exist.

I want a formal way of analyzing those equations to say that f < g given that all the components of f are < all the comparable components of g. Is linear algebra applicable? Is there a better toolset I should be looking at?

I don't see how linear algebra would be relevant here; the "linear" in "linear algebra" doesn't refer to an ordering but rather to linearity in the sense of a function satisfying $$f(ax)=af(x)$$ and $$f(x+y)=f(x)+f(y)$$ for $$x,y$$ vectors and $$a$$ a scalar. Rather, what you're considering here are partial orders (or posets), and the relevant subject is order theory - although partial orders are so ubiquitous in mathematics that you won't need to specifically look at order theory to see them.
Incidentally, your proposed ordering on subsets of a given poset - namely, $$A\trianglelefteq B$$ iff for each $$a\in A$$ and $$b\in B$$, either $$a$$ and $$b$$ are incomparable or $$a - isn't very well-behaved. Specifically:
• It's not antisymmetric: if $$a$$ and $$b$$ are incomparable then $$\{a\}\trianglelefteq \{b\}$$ and $$\{b\}\trianglelefteq \{a\}$$.
• More importantly, it's not transitive: suppose we have a partial order with elements $$a,b,c$$ where $$a and $$c$$ is incomparable with both $$a$$ and $$b$$. Then $$\{b\}\trianglelefteq\{c\}\trianglelefteq\{a\}$$ but $$\{b\}\not\trianglelefteq\{a\}$$.
A better notion is to compare all elements at once: say $$A\preccurlyeq B$$ if for every $$a\in A$$ and every $$b\in B$$ we have $$a. This still isn't entirely great (if $$a then consider $$\{a,c\}$$ versus $$\{a,b,c\}$$) but at least it's transitive.