= Linear independence
{title2=$\sum_i a_i v_i=0\iff a_1=\cdots=a_n=0$}
{wiki}
= Linearly independent set
{synonym}
= Linearly independent vectors
{synonym}
= Linearly independent
{synonym}
In a <vector space> $V$ over a specified <field> $F$, a finite family of <vectors> $(v_1,\ldots,v_n)$ is <linearly independent> when, for <coefficients> $a_i\in F$,
$$
\sum_{i=1}^n a_i v_i=0\quad\Longleftrightarrow\quad a_1=\cdots=a_n=0.
$$
Here the left-hand zero is the <zero vector>, while the right-hand zeros belong to the <field> $F$. Thus <linear independence> excludes every nontrivial <linear relation>. For an infinite family, require this condition for every finite subfamily.https://ocw.mit.edu/courses/18-06sc-linear-algebra-fall-2011/0bbc30e3f1d7933ea07a2d2e9ab050d9_MIT18_06SCF11_Ses1.9sum.pdf[MIT 18.06SC, Strang lecture summary 9, p. 1: linear independence]{ref}
For $v_1=(1,1)$ and $v_2=(1,-1)$ over the <real numbers>, the <linear relation> $av_1+bv_2=0$ gives $a+b=0$ and $a-b=0$. Adding and subtracting yields $2a=2b=0$, hence $a=b=0$. This verifies <linear independence> directly and explains why the <linear combination> giving $(3,1)$ has exactly the <coefficients> $(2,1)$.
A <linearly independent> family cannot contain the <zero vector>: a <coefficient> of $1$ on that entry and $0$ on every other entry would be a nontrivial <linear relation>. The empty family is <linearly independent>, because its empty sum is the <zero vector> and there are no <coefficients> that could be nonzero. <Linear independence> identifies a collection without redundant <vectors>; combined with a <spanning set>, it gives a <basis>. The <uniqueness of linear combination coefficients> makes its representations unambiguous.
Back to article page
