linearly independent subset

A set of vectors is linearly dependent if one can be written as a linear combination of the others, and linearly independent otherwise. In the latter case, the vectors in the set form a basis of their span.

Let $K$ be a rig, and let $V$ be a (left or right) module over $K$. (Often $K$ is a field so that $V$ is a vector space, but this is unnecessary.) Let $S$ be a subset of the underlying set ${|V|}$ of $V$.

By the adjunction between the underlying-set functor and the free functor, the subset inclusion

$i_S\colon S \to {|V|}$

corresponds to a homomorphism

$\hat{i}_S\colon K[S] \to V .$

Although $i_S$ is (by hypothesis) a monomorphism in $Set$, $\hat{i}_S$ need not be a monomorphism in $K Mod$.

The subset $S$ is **linearly independent** if $\hat{i}_S$ is a monomorphism; otherwise, $S$ is **linearly dependent**.

Conversely, if we start with an (abstract!) set $S$ and a monomorphism from $K[S]$ to $V$, then the corresponding function from $S$ to ${|V|}$ must be a monomorphism (because the underlying-set functor is faithful). Thus, our considering only subsets of ${|V|}$ loses no generality.

Given a linear combination

$\sum_{i=1}^n a_i v_i ,$

this may or may not equal the zero vector? $0_V$. Of course, if every $a_i$ is the zero scalar $0_K$, then the sum must be $0_V$.

The subset $S$ is **linearly independent** if, conversely, for every finite subset $\{v_1, \ldots, v_n\} \subseteq S$, we have $a_i = 0_K$ for all $i$ whenever

$\sum_{i=1}^n a_i v_i = 0_V ;$

otherwise, $S$ is **linearly dependent**.

Observe that the empty set is linearly independent by a vacuous implication.

In constructive mathematics, the definitions above of linear independence are all right (and still equivalent), but the definition of linear dependence as simply the negation of linear independence is unsatisfying. Furthermore, we sometimes want something stronger than mere linear independence.

If $K$ is a Heyting field, then the field structure defines a tight apartness relation $\ne$. Even if $K$ is not a field, we may still suppose that it is equipped with a tight apartness, or at least some inequality relation $\ne$. If we restrict attention to modules with a compatible inequality relation and homomorphisms that preserve this, then we also get an inequality relation $\ne$ on the hom-sets of the category $K Mod$. This allows us to define stronger notions of both linear dependence (which we take to be the default notion) and linear independence (to which we give a new name).

The subset $S$ is **linearly dependent** if $\hat{i}_S$ is non-monic in the strong sense that there exist generalised elements $f, g\colon A \to K[S]$ such that

$A \overset{f}\underset{g}\rightrightarrows K[S] \to V$

are equal but $f \ne g$. Concretely, $S$ is **linearly dependent** if some linear combination

$\sum_{i = 1}^n a_i v_i = 0_V$

but at least one $a_i \ne 0_K$.

The subset $S$ is **linearly free** if $\hat{i}_S$ is a regular monomorphism, or equivalently if it is monic in the strong sense that $f ; \hat{i}_S \ne g ; \hat{i}_S$ whenever $f \ne g$. Concretely, $S$ is **linearly free** if

$\sum_{i = 1}^n a_i v_i \ne 0_V$

whenever at least one $a_i \ne 0_K$.

Then we have the following implications (assuming that $\ne$ is tight, so that $a = b$ holds iff $a \ne b$ fails) but not (in general) their unstated converses:

$LF \Rightarrow LI \Leftrightarrow \neg{LD} ;$

$\neg{LF} \Leftarrow \neg{LI} \Leftarrow LD .$

It may also be instructive to look at the logical structure of each condition: * $LI$: $\forall (a,v),\; \sum a v = 0 \;\Rightarrow\; \forall i,\; a_i = 0$; * $LD$: $\exists (a,v),\; \sum a v = 0 \;\wedge\; \exists i,\; a_i \ne 0$; * $LF$: $\forall (a,v),\; \sum a v \ne 0 \;\Leftarrow\; \exists i,\; a_i \ne 0$.

Revised on October 4, 2015 06:56:29
by Todd Trimble
(67.81.95.215)