= Linear dependence
{title2=$\sum_i a_i v_i=0,\quad(a_1,\ldots,a_n)\ne0$}
= Linearly dependent
{synonym}
A finite family of <vectors> in a <vector space> over a specified <field> $F$ is <linearly dependent> if it has a nontrivial <linear relation>: $\sum_i a_i v_i=0$ with $a_i\in F$ and at least one <coefficient> nonzero. One can then solve for a <vector> whose <coefficient> is nonzero as a <linear combination> of the others; removing that entry preserves the <linear span>. Repeated <vectors> in a list give <linear dependence>: for example $(v,v)$ has the <linear relation> $1v+(-1)v=0$. A list keeps both entries even though the <set> $\{v,v\}$ has only one element. Strang's notes illustrate redundant columns by expressing one as the sum of two others.https://ocw.mit.edu/courses/18-06sc-linear-algebra-fall-2011/0bbc30e3f1d7933ea07a2d2e9ab050d9_MIT18_06SCF11_Ses1.9sum.pdf[MIT 18.06SC, Strang lecture summary 9, pp. 2–3: dependent columns]{ref}
Keep $v_1=(1,1)$ and $v_2=(1,-1)$, and add $v_3=(2,0)=v_1+v_2$. The three <vectors> still have <linear span> $\mathbb R^2$, but now
$$
v_1+v_2-v_3=0,\qquad (3,1)=2v_1+v_2=v_1+v_3.
$$
The two <coefficient> triples $(2,1,0)$ and $(1,0,1)$ differ by the nontrivial <linear relation> $(-1,-1,1)$. In fact all <linear combination> representations of this target are
$$
(c_1,c_2,c_3)=(2-t,1-t,t),\qquad t\in\mathbb R.
$$
To check both completeness and correctness, use $c_1v_1+c_2v_2+c_3v_3=(c_1+c_3)v_1+(c_2+c_3)v_2$. <Linear independence> of $v_1,v_2$ forces $c_1+c_3=2$ and $c_2+c_3=1$, giving exactly the displayed family. Redundancy therefore preserves attainable targets while allowing their <coefficients> to move. For solving a <linear system>, these alternatives matter; for assigning unique <coordinate vectors>, one instead chooses a <basis>.
\Image[/linear-dependence-redundancy.png]
{title=A redundant third <vector> gives infinitely many representations of (3,1)}
{description=Left: the <linearly independent> pair $v_1,v_2$ gives the unique <coefficients> $(2,1)$. Middle: $v_3=v_1+v_2$ introduces a nontrivial <linear relation>, and $v_1+v_3$ reaches the same target. Right: every $t\in\mathbb R$ gives <coefficients> $(2-t,1-t,t)$ with the same <linear combination>; the dots mark three computed choices.}
The following example shows why <linear independence depends on the scalar field>.
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