math111_logo Linear Indpendence

Properties of linear independence

Theorem 1 v1, v2, ..., vk, vk+1 are linearly independent ⇒ v1, v2, ..., vk are linearly independent.

Proof In case k = 2, we assume v1, v2, v3 are linearly independent. Then we have the following implication

c1v1 + c2v2 = 0c1v1 + c2v2 + 0v3 = 0c1 = c2 = 0,

where the second ⇒ follows from the linear independence of v1, v2, v3 (take c3 = 0 in the definition). The implication means exactly that v1, v2 are linearly independent.

The proof for general k is similar.

Theorem 2 v1, v2, ..., vk are linearly independent and vk+1span{v1, v2, ..., vk} ⇒ v1, v2, ..., vk, vk+1 are linearly independent.

Proof In case k = 3, we assume v1, v2, v3 are linearly independent and v4span{v1, v2, v3}.

Assume v1, v2, v3, v4 are linearly dependent, then we have

c1v1 + c2v2 + c3v3 + c4v4 = 0,

with at least one of the coefficients c1, c2, c3, c4 being nonzero. If c4 = 0, then at least one of the coefficients c1, c2, c3 is nonzero, and the equality above becomes c1v1 + c2v2 + c3v3 = 0. This means that v1, v2, v3 are linearly dependent. The contradiction shows that we must have c4 ≠ 0. Thus we may divide c4 and get

v4 = (-c1/c4)v1 + (-c2/c4)v2 + (-c3/c4)v3span{v1, v2, v3},

again contradicting with the assumption.

Therefore we conclude that v1, v2, v3, v4 must be linearly independent. The proof for general k is similar.

Theorem 3 The three operations do not change linear independence.

Proof We only consider special case of the operations on three vectors. The general case is similar.

For the first operation, we will prove

v1, v2, v3 linearly independent ⇔ v1, v2 + rv1, v3 linearly independent.

Assuming left side holds, we have

c1v1 + c2(v2 + rv1) + c3v3 = 0
⇒ (c1 + rc2)v1 + c2v2 + c3v3 = 0
c1 + rc2 = 0, c2 = 0, c3 = 0 (v1, v2, v3 linearly independent)
c1 = 0, c2 = 0, c3 = 0.

The whole implication means exactly the right side. Assume the right side holds. Then

c1v1 + c2v2 + c3v3 = 0
⇒ (c1 - rc2)v1 + c2(v2 + rv1) + c3v3 = 0
c1 - rc2 = 0, c2 = 0, c3 = 0 (v1, v2 + rv1, v3 linearly independent)
c1 = 0, c2 = 0, c3 = 0.

The whole implication means exactly the left side.

For the second operation, we will prove

v1, v2, v3 linearly independent ⇔ v2, v1, v3 linearly independent.

Assume left side holds. Then

c1v2 + c2v1 + c3v3 = 0
c2v1 + c1v2 + c3v3 = 0
c2 = 0, c1 = 0, c3 = 0 (v1, v2, v3 linearly independent)
c1 = 0, c2 = 0, c3 = 0.

The whole implication means exactly the right side. The implication in the other direction can be proved similarly.

For the third operation, we will prove that for d ≠ 0,

v1, v2, v3 linearly independent ⇔ dv1, v2, v3 linearly independent.

Assume left side holds. Then

c1(dv1) + c2v2 + c3v3 = 0
c1d = 0, c2 = 0, c3 = 0 (v1, v2, v3 linearly independent)
c1 = 0, c2 = 0, c3 = 0 (d ≠ 0)

The whole implication means exactly the right side. The implication in the other direction can be proved similarly.