3.1. LINEAR COMBINATIONS OF VECTORS, INDEPENDENCE 53

Definition 3.0.4 Define Fn ≡{(x1, · · · ,xn) : x j ∈ F for j = 1, · · · ,n

}.

(x1, · · · ,xn) = (y1, · · · ,yn)

if and only if for all j = 1, · · · ,n, x j = y j. When (x1, · · · ,xn)∈Fn, it is conventional to denote(x1, · · · ,xn) by the single bold face letter x. The numbers, x j are called the coordinates.Elements in Fn are called vectors. The set

{(0, · · · ,0, t,0, · · · ,0) : t ∈ R}

for t in the ith slot is called the ith coordinate axis. The point 0 ≡ (0, · · · ,0) is calledthe origin. Note that this can be considered as the set of F valued functions defined on(1,2, · · · ,n) . When the ordered list (x1, · · · ,xn) is considered, it is just a way to say thatf (1) = x1, f (2) = x2 and so forth. Thus it is a case of the typical example of a vectorspace mentioned above.

3.1 Linear Combinations of Vectors, IndependenceThe fundamental idea in linear algebra is the following notion of a linear combination.

Definition 3.1.1 Let x1, · · · ,xn be vectors in a vector space. A finite linear combination ofthese vectors is a vector which is of the form ∑

nj=1 a jx j where the a j are scalars. In short, it

is a sum of scalars times vectors. span(x1, · · · ,xn) denotes the set of all linear combinationsof the vectors x1, · · · ,xn. More generally, if S is any set of vectors, span(S) consists of allfinite linear combinations of vectors from S.

Definition 3.1.2 Let (V,F) be a vector space and its field of scalars. Then S⊆V is said tobe linearly independent if whenever {v1, · · · ,vm} ⊆V with the vi distinct, then there is onlyone way to have a linear combination ∑

ni=1 civi = 0 and this is to have each ci = 0. More

succinctly, if ∑ni=1 civi = 0 then each ci = 0. A set S ⊆ V is linearly dependent if it is not

linearly independent. That is, there is some subset of S {v1, · · · ,vn} and scalars ci not allzero such that ∑

ni=1 civi = 0.

The following is a useful equivalent description of what it means to be independent.

Proposition 3.1.3 A set of vectors S is independent if and only if no vector is a linearcombination of the others.

Proof:⇒ Suppose S is linearly independent. Could you have for some

{u1, · · · ,ur} ⊆ S

ui = ∑ j ̸=i c ju j? No. This is not possible because if the above holds, then you would have0 = (−1)ui +∑ j ̸=i c ju j in contradiction to the assumption that {u1, · · · ,ur} is linearly in-dependent.⇐ Suppose now that no vector in S is a linear combination of the others. Suppose

∑ni=1 ciui = 0 where each ui ∈ S. It is desired to show that whenever this happens, each

ci = 0. Could any of the ci be non zero? No. If ck ̸= 0, then you would have ∑ni=1

cick

ui = 0and so uk = ∑i ̸=k− ci

ckui showing that one can obtain uk as a linear combination of the other

vectors after all. It follows that all ci = 0 and so {u1, · · · ,ur} is linearly independent. ■