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\begin{document}

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\section*{linearly independent subset}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{linear_algebra}{}\paragraph*{{Linear algebra}}\label{linear_algebra}

[[!include homotopy - contents]]

\hypertarget{linearly_independent_subsets}{}\section*{{Linearly independent subsets}}\label{linearly_independent_subsets}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definitions}{Definitions}\dotfill \pageref*{definitions} \linebreak
\noindent\hyperlink{abstractly}{Abstractly}\dotfill \pageref*{abstractly} \linebreak
\noindent\hyperlink{concretely}{Concretely}\dotfill \pageref*{concretely} \linebreak
\noindent\hyperlink{constructively}{Constructively}\dotfill \pageref*{constructively} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A [[concrete set|set]] of [[vector]]s 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 a vector space|basis]] of their [[linear span|span]].

\hypertarget{definitions}{}\subsection*{{Definitions}}\label{definitions}

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 $V$.

\hypertarget{abstractly}{}\subsubsection*{{Abstractly}}\label{abstractly}

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

\begin{displaymath}
i_S\colon S \to {|V|}
\end{displaymath}
corresponds to a [[homomorphism]]

\begin{displaymath}
\hat{i}_S\colon K[S] \to V .
\end{displaymath}
Although $i_S$ is (by hypothesis) a [[monomorphism]] in $Set$, $\hat{i}_S$ need not be a monomorphism in $K Mod$.

\begin{defn}
\label{}\hypertarget{}{}
The subset $S$ is \textbf{linearly independent} if $\hat{i}_S$ is a monomorphism; otherwise, $S$ is \textbf{linearly dependent}.

\end{defn}
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 functor|faithful]]). Thus, our considering only subsets of ${|V|}$ loses no generality.

\hypertarget{concretely}{}\subsubsection*{{Concretely}}\label{concretely}

Given a [[linear combination]]

\begin{displaymath}
\sum_{i=1}^n a_i v_i ,
\end{displaymath}
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 $0_V$.

\begin{defn}
\label{}\hypertarget{}{}
The subset $S$ is \textbf{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

\begin{displaymath}
\sum_{i=1}^n a_i v_i = 0_V ;
\end{displaymath}
otherwise, $S$ is \textbf{linearly dependent}.

\end{defn}
Observe that the empty set is linearly independent by a vacuous implication.

\hypertarget{constructively}{}\subsubsection*{{Constructively}}\label{constructively}

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).

\begin{defn}
\label{}\hypertarget{}{}
The subset $S$ is \textbf{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

\begin{displaymath}
A \overset{f}\underset{g}\rightrightarrows K[S] \to V
\end{displaymath}
are equal but $f \ne g$. Concretely, $S$ is \textbf{linearly dependent} if some linear combination

\begin{displaymath}
\sum_{i = 1}^n a_i v_i = 0_V
\end{displaymath}
but at least one $a_i \ne 0_K$.

\end{defn}
\begin{defn}
\label{}\hypertarget{}{}
The subset $S$ is \textbf{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 \textbf{linearly free} if

\begin{displaymath}
\sum_{i = 1}^n a_i v_i \ne 0_V
\end{displaymath}
whenever at least one $a_i \ne 0_K$.

\end{defn}
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:

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

[[!redirects linearly independent subset]] [[!redirects linearly independent set]] [[!redirects linearly independent]] [[!redirects linear independence]]

[[!redirects linearly dependent subset]] [[!redirects linearly dependent set]] [[!redirects linearly dependent]] [[!redirects linear dependence]]

[[!redirects linearly free subset]] [[!redirects linearly free set]] [[!redirects linearly free]] [[!redirects linear freedom]] [[!redirects linear freeness]]



\end{document}