---- > [!definition] Definition. ([[linearly independent]]) > Let $R$ be an [[integral domain]]. > > Recall that for all sets $I$ there is a canonical [[injection]] $j:I \to F^{R}(I)$, where $F^{R}(I)$ is the [[free module]] on $I$. Moreover, the [[universal property]] of [[free module|free modules]] says any $i:I \to M$ determines a unique $R$-[[module]] [[linear map|homomorphism]] $\varphi:F^{R}(I) \to M$ making the diagram > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRADEA9AJQAoBJAJQgAvqXSZc+QijIBGKrUYs2-UeJAZseAkTnlF9Zq0QgAsqMUwoAc3hFQAMwBOEALZIyIHBCQAmaiMVUwAdEJgADyw4HDgAAgBCOLD6ZzQACyx1J1cPRH1vX0QvIJMQACsQahw6LAY2dIgIAGtskBd3JAKff0DlMqzqBjoAIxgGAAVJHRkQZywbdJxLESA > \begin{tikzcd} > F^R(I) \arrow[r, "\exists ! \varphi"] & M \\ > I \arrow[u, "j", hook] \arrow[ru, "i"'] & > \end{tikzcd} > \end{document} > ``` > commute. > > View $I \xmapsto{i} M$ as an indexed set, assigning an index $\alpha \in I$ to an element $m_{\alpha} \in M$. We call the indexed set $I \xrightarrow{i}{} M$ **linearly independent** if the induced $\varphi$ is an [[injection]]; else it is **linearly dependent**. > > Recalling that $\varphi$ is defined $\varphi\big( (r_{\alpha})_{\alpha \in I} \big):= \sum_{\alpha \in I}r_{\alpha}i(\alpha)= \sum_{\alpha \in I}r_{\alpha}m_{\alpha},$ > where the sums are finite by definition of [[direct sum of modules|direct sum]], this definition is saying an indexed set $S=\{ m_{\alpha} \}_{\alpha \in I}$ of elements in $M$ is [[linearly independent]] [[module homomorphism is injective iff kernel is trivial iff is a monomorphism|if and only if]]$\sum_{\alpha \in I} r_{\alpha}m_{\alpha}=0 \implies r_{\alpha} \equiv 0,$ > or equivalently [^1] if and only if any element $m \in M$ can be written in *at most* one way as a [[linear combination]] $m=\sum_{\alpha \in I} r_{\alpha}m_{\alpha}.$ [^1]: [[free module#^ad5645|Here]] it is discussed that an element of $F^{R}(I)$ can be written uniquely as a [[linear combination]] of elements $j(\alpha)$ for the canonical inclusion $j: I \to F^{R}(I)$. When $I \xrightarrow{i} M$ is [[linearly independent]], $\varphi$ is an [[injection]] and the uniqueness factors through $F^{R}(I)$, thus *if* an element in $M$ can be written as a [[linear combination]] of $m_{\alpha}=i(\alpha)=\varphi \circ j(\alpha)$, that [[linear combination]] must be unique. > [!note] Note. > If $\varphi$ is a [[surjection]], then in light of [[submodule generated by a subset]] we say that $i$ generates $M$. ^note > [!specialization] Specializing to finite dimensional [[vector space|vector spaces]]. > >1. Let $V$ be a [[vector space]]. A set $a_{1}, \dots, a_{n}$ of [[vector]]s in $V$ is said to be [[linearly independent]] if each [[vector]] $x \in V$ can be represented by *at most* a single nontrivial [[linear combination]] of $a_{1}, \dots, a_{n}$. >2. Equivalently, $a_{1}, \dots, a_{n}$ is linearly independent if the only solution to the equation $d_{1}a_{1}+\dots+d_{m}a_{m} = 0$ is $d_{1}=\dots=d_{m}=0$. ^specialization > [!basicexample] > ## 1 > A list $v$ of one [[vector]] $v \in V$ is **linearly independent** if and only if $v \neq 0$. > > ## 2 > A list of two [[vector]]s in $V$ is **linearly independent** if and only if neither vector is a scalar multiple of the other > > ## 3 > $e_1, e_2, e_3, e_4$ are **linearly independent** in $\ff^4$. > > ## 4 > The list $1,z,\dots,z^m$ is **linearly independent** in [[vector space of all polynomials with coefficients in F]] for each nonnegative integer $m$. > > ## 5 > **Digital Signal Processing**. Consider the length-$2$ discrete signals of finite duration (real-valued) $\delta^{(0)}=[1 \ \ 0]^{\top} \and \delta^{(0)}=[0 \ \ 1]^{\top}.$ > To show these are [[linearly independent]] we solve the following linear system: $\beta_{0} \delta^{(0)} + \beta_{1}\delta^{(1)} = [\beta_{0} \ \ 0]^{\top} + [0 \ \ \beta_{1}]^{\top} = [0 \ \ 0]^{\top}.$ > Clearly the unique solution is $\beta_{0}=0=\beta_{1}$. > > [!basicnonexample] > ## 1 > $(2,3,1),(1,-1,2), (7,3,8)$ is **linearly dependent** in $\ff^3$ since $2(2,3,1) + 3(1,-1,2)+(-1)(7,3,8)=(0,0,0).$ > ## 2 > The list $(2,3,1),(1,-1,2),(7,3,c)$ is linearly dependent in $\mathbb{F}^3$ if and only if $c=8$. > > ## 3 > If some [[vector]] in a list of [[vector]]s in $V$ is a [[linear combination]] of the others, then the list is **linearly dependent**. > > ## 4 > As a special case of the above, every list of [[vector]]s in $V$ containing the $0$-[[vector]] is **linearly dependent**. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```