Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i<n} \lambda_i v_i = 0_V$ forces every $\lambda_i = 0_F$, and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent

Let $V$ be a vector space over a field $F$ (Vector space over a field). As in Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$, a finite list of vectors is a function $v : n \to V$ on a von Neumann natural $n = \{0, \dots, n-1\}$ (The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$), written $v_i := v(i)$, and

$\sum_{i<n} \lambda_i v_i$

is the finite sum of The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n = 0$) equal to the identity read additively in the abelian group $(V,+,0_V)$, applied to the list $i \mapsto \lambda_i v_i$. No second notion of finite sum is introduced here.

Independence of a list

A finite list $v : n \to V$ is linearly independent when, for every list of scalars $\lambda : n \to F$,

$\sum_{i<n} \lambda_i v_i = 0_V \quad \Longrightarrow \quad \lambda_i = 0_F \text{ for every } i < n,$

and linearly dependent otherwise, that is, when some $\lambda : n \to F$ has $\sum_{i<n}\lambda_i v_i = 0_V$ while $\lambda_j \ne 0_F$ for at least one $j < n$. Such a $\lambda$ is called a witness to the dependence of $v$.

Independence of a subset

A subset $S \subseteq V$ is linearly independent when every injective finite list $v : n \to S$ (Injection, surjection, bijection) is linearly independent, and linearly dependent otherwise, that is, when some injective finite list into $S$ is linearly dependent.

The injectivity clause is not decoration. A linear combination in Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ is indexed by an arbitrary list $v : n \to S$, which is not required to be injective. If the definition above quantified over all such lists, then for any $w \in S$ the list $v : 2 \to S$ with $v_0 = v_1 = w$ and the scalars $\lambda_0 = 1_F$, $\lambda_1 = -1_F$ would give

$\sum_{i<2}\lambda_i v_i = (0_V + 1_F w) + (-1_F)w = w + (-w) = 0_V$

with $\lambda_0 = 1_F \ne 0_F$ (Field, In any vector space $0_F v = 0_V$, $\lambda 0_V = 0_V$, $(-\lambda)v = -(\lambda v)$, $(-1_F)v = -v$, and $\lambda v = 0_V$ forces $\lambda = 0_F$ or $v = 0_V$), so every nonempty subset of $V$ would be dependent and the notion would be empty. Quantifying over injective lists is what makes the subset notion the intended one. It costs nothing for lists: Finite sums re-indexed along an injection, with a zero term deleted, and concatenated; and the closure properties of linear independence: an independent list is injective and never $0_V$, its sublists are independent, a list is independent exactly when it is injective with linearly independent image, and every subset of a linearly independent set is linearly independent shows that the vanishing condition above already forces a list to be injective, so no injectivity hypothesis has to be carried alongside independence of a list.

The boundary cases are genuine cases

$\mathbb{N}$ contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)), so both of the following are instances of the definitions and neither is a convention.

• The empty list is independent. For $n = 0$ the only list of scalars is the empty one, and the condition "$\lambda_i = 0_F$ for every $i < 0$" holds vacuously.
• $\varnothing \subseteq V$ is independent. The only function $v : n \to \varnothing$ is the empty one, with $n = 0$, and it is independent by the previous point.
• $\{0_V\}$ is dependent. The list $v : 1 \to \{0_V\}$ with $v_0 = 0_V$ is injective, and taking $\lambda_0 = 1_F$ gives $\sum_{i<1}\lambda_i v_i = 0_V + 1_F 0_V = 0_V$ by the recursion of The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n = 0$) equal to the identity and $\lambda 0_V = 0_V$ (In any vector space $0_F v = 0_V$, $\lambda 0_V = 0_V$, $(-\lambda)v = -(\lambda v)$, $(-1_F)v = -v$, and $\lambda v = 0_V$ forces $\lambda = 0_F$ or $v = 0_V$), while $1_F \ne 0_F$ in a field (Field). So $\{0_V\}$, and hence every subset of $V$ containing $0_V$, is linearly dependent.

• Independence is relative to the field, and to the ambient vector space. The scalars range over $F$, so a set of vectors independent over a subfield $K \subseteq F$ may be dependent over $F$; A field is a vector space over itself, and over any subfield $K \subseteq F$ every $F$-vector space is a $K$-vector space by restricting the scalars is what makes both readings available on one set, and the companion page uses the distinction for $\mathbb{R}$ over $\mathbb{Q}$. The ambient space matters only through its addition, its zero and its scalar multiplication, all of which a linear subspace inherits from $V$ (Linear subspace of a vector space); the resulting agreement is recorded in Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis.

• Dependence is a property of a list together with a witness, but of a subset outright. A dependent list carries an explicit vanishing combination with a nonzero coefficient. For a subset, the witness is an injective list drawn from it; which list that is, is not part of the statement that the set is dependent. A subset $S \subseteq V$ is linearly dependent if and only if some $s \in S$ lies in $\operatorname{span}(S \setminus \{s\})$; and $\operatorname{span}(S)$ is already the set of linear combinations of INJECTIVE finite lists into $S$ converts the existential into a statement with no lists in it at all.

• Why the two notions are both kept. Lists carry order, and an ordered list is what a coordinate system is (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i<n} \lambda_i v_i$ for exactly one $\lambda : n \to F$; those scalars are the coordinates of $x$ in that ordered basis); subsets carry no order, and it is subsets that the Zorn argument of Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $\operatorname{span}(S) = V$, there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ runs over. Keeping both, and proving that they agree, is cheaper than translating at every use.