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?


1 Answer 1


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 $a<b$ - 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 $a<b$ 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 $a<b$. This still isn't entirely great (if $a<b<c$ then consider $\{a,c\}$ versus $\{a,b,c\}$) but at least it's transitive.

  • $\begingroup$ Thank you for the feedback. I wasn't fully outlining the problem so it led to some ambiguity. In my case, a and b and c are things you might compare like say colors (where greater might mean a preference) whereas d and e might be sounds. So the case where c is comparable to both a and d won't occur. If such a comparison made any real sense a, c, and d would all be part of one string of comparatives. $\endgroup$
    – Jason
    Jun 11, 2020 at 23:27
  • $\begingroup$ Thank you. I'll take a look at posets. $\endgroup$
    – Jason
    Jun 11, 2020 at 23:28

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