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10.1: Showing Linear Dependence

( \newcommand{\kernel}{\mathrm{null}\,}\)

In the above example we were given the linear combination 3v1+2v2v3+v4 seemingly by magic. The next example shows how to find such a linear combination, if it exists.

Example \(\PageIndex{1}\):

Consider the following vectors in 3:
v1=(001),v2=(121),v3=(123).
Are they linearly independent?

We need to see whether the system
c1v1+c2v2+c3v3=0
has any solutions for c1,c2,c3. We can rewrite this as a homogeneous system by building a matrix whose columns are the vectors v1, v2 and v3:
(v1v2v3)(c1c2c3)=0.
This system has solutions if and only if the matrix M=(v1v2v3) is singular, so we should find the determinant of M:
detM=det(011022113)=det(1122)=0.

Therefore nontrivial solutions exist. At this point we know that the vectors are linearly dependent. If we need to, we can find coefficients that demonstrate linear dependence by solving the system of equations:
(011002201130)(113001100000)(102001100000).
Then c3=c3=:μ, c2=μ, and c1=2μ. Now any choice of μ will produce coefficients c1,c2,c3 that satisfy the linear equation. So we can set μ=1 and obtain:
c1v1+c2v2+c3v3=02v1v2+v3=0.

Theorem (Linear Dependence)

An ordered set of non-zero vectors (v1,,vn) is linearly dependent if and only if one of the vectors vk is expressible as a linear combination of the preceding vectors.

Proof
The theorem is an if and only if statement, so there are two things to show.

(i.) First, we show that if vk=c1v1+ck1vk1 then the set is linearly dependent.

This is easy. We just rewrite the assumption:
c1v1++ck1vk1vk+0vk+1++0vn=0.
This is a vanishing linear combination of the vectors {v1,,vn} with not all coefficients equal to zero, so {v1,,vn} is a linearly dependent set.

(ii.) Now, we show that linear dependence implies that there exists k for which vk is a linear combination of the vectors {v1,,vk1}.

The assumption says that
c1v1+c2v2++cnvn=0.
Take k to be the largest number for which ck is not equal to zero. So:
c1v1+c2v2++ck1vk1+ckvk=0.

(Note that k>1, since otherwise we would have c1v1=0v1=0, contradicting the assumption that none of the vi are the zero vector.)

As such, we can rearrange the equation:
c1v1+c2v2++ck1vk1=ckvk c1ckv1c2ckv2ck1ckvk1=vk.

Therefore we have expressed vk as a linear combination of the previous vectors, and we are done.

Example 10.1.2

Consider the vector space P2(t) of polynomials of degree less than or equal to 2. Set:
v1=1+tv2=1+t2v3=t+t2v4=2+t+t2v5=1+t+t2.
The set {v1,,v5} is linearly dependent, because v4=v1+v2.

Contributor

This page titled 10.1: Showing Linear Dependence is shared under a not declared license and was authored, remixed, and/or curated by David Cherney, Tom Denton, & Andrew Waldron.

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