§ 5. Linear Dependence and Linear Independence

Linear Dependence and Linear Independence

Definition 5.1 : Linearly Dependent

A subset \(S\) of a vector space \(V\) is called linearly dependent if there exist a finite number of distinct vectors \(u_{1}, u_{2}, \ldots, u_{n}\) in \(S\) and scalars \(a_{1}, a_{2}, \ldots, a_{n}\), not all zero, such that

\[ a_{1} u_{1}+a_{2} u_{2}+\cdots+a_{n} u_{n}=0 \]

In this case we also say that the vectors of \(S\) are linearly dependent.

For any vectors \(u_{1}, u_{2}, \ldots, u_{n}\), we have \(a_{1} u_{1}+a_{2} u_{2}+\cdots+a_{n} u_{n}=0\) if \(a_{1}=a_{2}=\cdots=a_{n}=0\). We call this the trivial representation of \(0\) as a linear combination of \(u_{1}, u_{2}, \ldots, u_{n}\). Thus, for a set to be linearly dependent, there must exist a nontrivial representation of \(0\) as a linear combination of vectors in the set.

Definition 5.2 : Linearly Independent

A subset \(S\) of a vector space that is not linearly dependent is called linearly independent. As before, we also say that the vectors of \(S\) are linearly independent.

Concept 5.3 : Basic facts about linear dependence / independence

1. The empty set is linearly independent, for linearly dependent sets must be nonempty.

2. A set consisting of a single nonzero vector is linearly independent. For if \(\{u\}\) is linearly dependent, then \(a u=0\) for some nonzero scalar \(a\). Thus \(u=a^{-1}(a u)=a^{-1} 0=0\).

3. Any subset of a vector space that contains the zero vector is linearly dependent, because \(0=1 \cdot 0\) is a nontrivial representation of \(0\) as a linear combination of vectors in the set.

4. A set is linearly independent if and only if the only representations of \(0\) as linear combinations of its vectors are trivial representations.

The condition in item 4 provides a useful method for determining whether a finite set is linearly independent.

Example 5.4 : Showing that a set is linearly dependent

Consider the set

\[ S=\{(1,3,-4,2),(2,2,-4,0),(1,-3,2,-4),(-1,0,1,0)\} \]

in \(\mathbb{R}^{4}\). We show that \(S\) is linearly dependent and then express one of the vectors in \(S\) as a linear combination of the other vectors in \(S\). To show that \(S\) is linearly dependent, we must find scalars \(a_{1}, a_{2}, a_{3}\), and \(a_{4}\), not all zero, such that

\[ a_{1}(1,3,-4,2)+a_{2}(2,2,-4,0)+a_{3}(1,-3,2,-4)+a_{4}(-1,0,1,0)=0 . \]

Finding such scalars amounts to finding a nonzero solution to the system of linear equations

\[ \begin{alignedat}{5} a_{1} &+2a_{2} &+ a_{3} &- a_{4} &= 0,\\ 3a_{1} &+2a_{2} &-3a_{3} &\phantom{{}-a_{4}} &= 0,\\ -4a_{1}&-4a_{2} &+2a_{3} &+ a_{4} &= 0,\\ 2a_{1} &\phantom{{}+2a_{2}} &-4a_{3} &\phantom{{}-a_{4}} &= 0. \end{alignedat} \]

One such solution is \(a_{1}=4, a_{2}=-3, a_{3}=2\), and \(a_{4}=0\). Thus \(S\) is a linearly dependent subset of \(\mathbb{R}^{4}\), and

\[ 4(1,3,-4,2)-3(2,2,-4,0)+2(1,-3,2,-4)+0(-1,0,1,0)=0 \]

Example 5.5 : Showing that a set is linearly independent

To prove that the set

\[ S=\{(1,0,0,-1),(0,1,0,-1),(0,0,1,-1),(0,0,0,1)\} \]

is linearly independent, we must show that the only linear combination of vectors in \(S\) that equals the zero vector is the one in which all the coefficients are zero. Suppose that \(a_{1}, a_{2}, a_{3}\), and \(a_{4}\) are scalars such that

\[ a_{1}(1,0,0,-1)+a_{2}(0,1,0,-1)+a_{3}(0,0,1,-1)+a_{4}(0,0,0,1)=(0,0,0,0) \]

Equating the corresponding coordinates of the vectors on the left and the right sides of this equation, we obtain the following system of linear equations.

\[ \begin{alignedat}{5} a_{1} &\phantom{{}-a_{2}} &\phantom{{}-a_{3}} &\phantom{{}+a_{4}} &= 0,\\ \phantom{{}-a_{1}} & \phantom{++} a_{2} &\phantom{{}-a_{3}} &\phantom{{}+a_{4}} &= 0,\\ \phantom{{}-a_{1}} &\phantom{{}-a_{2}} & a_{3} &\phantom{{}+a_{4}} &= 0,\\ -a_{1} &-a_{2} &-a_{3} &+a_{4} &= 0. \end{alignedat} \]

Clearly the only solution to this system is \(a_{1}=a_{2}=a_{3}=a_{4}=0\), and so \(S\) is linearly independent.

Minimal Generating Sets and Linear Independence

Suppose that \(V\) is a vector space over an infinite field and that \(W\) is a subspace of \(V\). Unless \(W\) is the zero subspace, \(W\) is an infinite set. It is desirable to find a "small" finite subset \(S\) that generates \(W\) because we can then describe each vector in \(W\) as a linear combination of the finite number of vectors in \(S\).

Concept 5.6 : Minimal generating sets are linearly independent.

Suppose that \(S\) is any linearly dependent set containing two or more vectors. Then some vector \(v \in S\) can be written as a linear combination of the other vectors in \(S\), and the subset obtained by removing \(v\) from \(S\) has the same span as \(S\). It follows that if no proper subset of \(S\) generates the span of \(S\), then \(S\) must be linearly independent.

Another way to view Concept 5.6 is given in Theorem 5.9.

Theorem 5.7 : Superset of a linearly dependent set is linearly dependent.

Let \(V\) be a vector space, and let \(S_{1} \subseteq S_{2} \subseteq V\). If \(S_{1}\) is linearly dependent, then \(S_{2}\) is linearly dependent.

Proof

Since \(S_1\) is linearly dependent, there must exist a nontrivial representation of \(0\) as a linear combination of vectors in \(S_1\). Since \(S_1 \subseteq S_2\), the same set of vectors are a nontrivial representation of \(0\) as a linear combination of vectors in \(S_2\). Therefore, \(S_2\) is linearly dependent.

Corollary 5.8

Let \(V\) be a vector space, and let \(S_{1} \subseteq S_{2} \subseteq V\). If \(S_{2}\) is linearly independent, then \(S_{1}\) is linearly independent.

Proof

If \(S_1\) is linearly dependent, by Theorem 5.7, \(S_2\) is linearly dependent, which immediately leads to a contradiction. Therefore, \(S_1\) is linearly independent.

Theorem 5.9 : Adding a vector creates dependence if and only if it lies in the span.

Let \(S\) be a linearly independent subset of a vector space \(V\), and let \(v\) be a vector in \(V\) that is not in \(S\). Then \(S \cup\{v\}\) is linearly dependent if and only if \(v \in \operatorname{span}(S)\).

Proof

If \(S \cup\{v\}\) is linearly dependent, then there are vectors \(u_{1}, u_{2}, \ldots, u_{n}\) in \(S \cup\{v\}\) such that \(a_{1} u_{1}+a_{2} u_{2}+\cdots+a_{n} u_{n}=0\) for some nonzero scalars \(a_{1}, a_{2}, \ldots, a_{n}\). Because \(S\) is linearly independent, one of the \(u_{i}\) 's, say \(u_{1}\), equals \(v\). Thus \(a_{1} v+a_{2} u_{2}+\cdots+a_{n} u_{n}=0\), and so

\[ v=a_{1}^{-1}\left(-a_{2} u_{2}-\cdots-a_{n} u_{n}\right)=-\left(a_{1}^{-1} a_{2}\right) u_{2}-\cdots-\left(a_{1}^{-1} a_{n}\right) u_{n} \]

Since \(v\) is a linear combination of \(u_{2}, \ldots, u_{n}\), which are in \(S\), we have \(v \in \operatorname{span}(S)\).

Conversely, let \(v \in \operatorname{span}(S)\). Then there exist vectors \(v_{1}, v_{2}, \ldots, v_{m}\) in \(S\) and scalars \(b_{1}, b_{2}, \ldots, b_{m}\) such that \(v=b_{1} v_{1}+b_{2} v_{2}+\cdots+b_{m} v_{m}\). Hence

\[ 0=b_{1} v_{1}+b_{2} v_{2}+\cdots+b_{m} v_{m}+(-1) v \]

Since \(v \neq v_{i}\) for \(i=1,2, \ldots, m\), the coefficient of \(v\) in this linear combination is nonzero, and so the set \(\left\{v_{1}, v_{2}, \ldots, v_{m}, v\right\}\) is linearly dependent. Therefore \(S \cup\{v\}\) is linearly dependent by Theorem 5.7.

Exercise

Exercise 5.9

Let \(u\) and \(v\) be distinct vectors in a vector space \(V\). Show that \(\{u, v\}\) is linearly dependent if and only if \(u\) or \(v\) is a multiple of the other.

Exercise 5.16

Prove that a set \(S\) of vectors is linearly independent if and only if each finite subset of \(S\) is linearly independent.