8.5. LINEAR INDEPENDENCE AND BASES 161

Corollary 8.5.2 The collection of vectors, {v1, · · · ,vk} is linearly independent if and onlyif none of these vectors is a linear combination of the others.

Proof: If {v1, · · · ,vk} is linearly independent, then every column in(v1 v2 · · · vk

)is a pivot column which requires that the row reduced echelon form is(

e1 e2 · · · ek

).

Now none of the ei vectors is a linear combination of the others. By Lemma 8.2.5 on Page150 none of the vi is a linear combination of the others. Recall this lemma says linearrelationships between the columns are preserved under row operations.

Next suppose none of the vectors {v1, · · · ,vk} is a linear combination of the others.Then none of the columns in (

v1 v2 · · · vk

)is a linear combination of the others. By Lemma 8.2.5 the same is true of the row re-duced echelon form for this matrix. From the description of the row reduced echelon form,it follows that the ith column of the row reduced echelon form must be ei since other-wise, it would be a linear combination of the first i−1 vectors e1,· · · ,ei−1 and by Lemma8.2.5, it follows vi would be the same linear combination of v1, · · · ,vi−1 contrary to theassumption that none of the columns in

(v1 v2 · · · vk

)is a linear combination of

the others. Therefore, each of the k columns in(

v1 v2 · · · vk

)is a pivot column

and so {v1, · · · ,vk} is linearly independent. ■

Corollary 8.5.3 The collection of vectors, {v1, · · · ,vk} is linearly independent if and onlyif whenever

n

∑i=1

civi = 0

it follows each ci = 0.

Proof: Suppose first {v1, · · · ,vk} is linearly independent. Then by Corollary 8.5.2,none of the vectors is a linear combination of the others. Now suppose

n

∑i=1

civi = 0

and not all the ci = 0. Then pick ci which is not zero, divide by it and solve for vi in termsof the other v j, contradicting the fact that none of the vi equals a linear combination of theothers.

Now suppose the condition about the sum holds. If vi is a linear combination of theother vectors in the list, then you could obtain an equation of the form

vi = ∑j ̸=i

c jv j