# 9.3: Linear Independence

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## Outcomes

1. Determine if a set is linearly independent.

In this section, we will again explore concepts introduced earlier in terms of $$\mathbb{R}^n$$ and extend them to apply to abstract vector spaces.

## Definition $$\PageIndex{1}$$: Linear Independence

Let $$V$$ be a vector space. If $$\{\vec{v}_{1},\cdots ,\vec{v}_{n}\} \subseteq V,$$ then it is linearly independent if $\sum_{i=1}^{n}a_{i}\vec{v}_{i}=\vec{0} \;\mbox{implies}\; a_{1}=\cdots =a_{n}=0\nonumber$ where the $$a_i$$ are real numbers.

The set of vectors is called linearly dependent if it is not linearly independent.

Below is a video on linear independent and dependent sets of vectors.

## Example $$\PageIndex{1}$$: Linear Independence

Let $$S \subseteq \mathbb{P}_2$$ be a set of polynomials given by $S = \left\{ x^2 + 2x - 1, 2x^2 - x + 3 \right\}\nonumber$ Determine if $$S$$ is linearly independent.

Solution

To determine if this set $$S$$ is linearly independent, we write $a ( x^2 + 2x -1 ) + b(2x^2 - x + 3) = 0x^2 + 0x + 0\nonumber$ If it is linearly independent, then $$a=b=0$$ will be the only solution. We proceed as follows. \begin{aligned} a ( x^2 + 2x -1 ) + b(2x^2 - x + 3) &= 0x^2 + 0x + 0 \\ ax^2 + 2ax - a + 2bx^2 - bx + 3b &= 0x^2 + 0x + 0 \\ (a+2b)x^2 + (2a -b)x - a + 3b &= 0x^2 + 0x + 0\end{aligned}

It follows that \begin{aligned} a + 2b &= 0 \\ 2a - b &= 0 \\ -a + 3b &= 0\end{aligned}

The augmented matrix and resulting reduced row-echelon form are given by $\left [ \begin{array}{rr|r} 1 & 2 & 0 \\ 2 & -1 & 0 \\ -1 & 3 & 0 \end{array} \right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right ]\nonumber$

Hence the solution is $$a=b=0$$ and the set is linearly independent.

The next example shows us what it means for a set to be dependent.

## Example $$\PageIndex{2}$$: Dependent Set

Determine if the set $$S$$ given below is independent. $S=\left\{ \left [\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right ], \left [\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right ], \left [\begin{array}{c} 1 \\ 3 \\ 5 \end{array}\right ] \right\}\nonumber$

Solution

To determine if $$S$$ is linearly independent, we look for solutions to $a\left [\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right ] +b\left [\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right ] +c\left [\begin{array}{c} 1 \\ 3 \\ 5 \end{array}\right ] =\left [\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right ]\nonumber$ Notice that this equation has nontrivial solutions, for example $$a=2$$, $$b=3$$ and $$c=-1$$. Therefore $$S$$ is dependent.

The following is an important result regarding dependent sets.

## Lemma $$\PageIndex{1}$$: Dependent Sets

Let $$V$$ be a vector space and suppose $$W = \left\{ \vec{v}_1, \vec{v}_2, \cdots, \vec{v}_k \right\}$$ is a subset of $$V$$. Then $$W$$ is dependent if and only if $$\vec{v}_i$$ can be written as a linear combination of $$\left\{ \vec{v}_1, \vec{v}_2, \cdots, \vec{v}_{i-1}, \vec{v}_{i+1}, \cdots, \vec{v}_k \right\}$$ for some $$i \leq k$$.

Revisit Example $$\PageIndex{2}$$ with this in mind. Notice that we can write one of the three vectors as a combination of the others. $\left [\begin{array}{c} 1 \\ 3 \\ 5 \end{array}\right ] = 2\left [\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right ] +3\left [\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right ]\nonumber$

By Lemma $$\PageIndex{1}$$ this set is dependent.

Below is a video on determining if matrix columns are dependent or independent.

If we know that one particular set is linearly independent, we can use this information to determine if a related set is linearly independent. Consider the following example.

Let $$V$$ be a vector space and suppose $$S \subseteq V$$ is a set of linearly independent vectors given by $$S = \left\{ \vec{u}, \vec{v}, \vec{w} \right\}$$. Let $$R \subseteq V$$ be given by $$R = \left\{ 2\vec{u} - \vec{w}, \vec{w} + \vec{v}, 3\vec{v} + \frac{1}{2} \vec{u} \right\}$$. Show that $$R$$ is also linearly independent.

Solution

To determine if $$R$$ is linearly independent, we write $a(2\vec{u} - \vec{w}) + b(\vec{w} + \vec{v}) + c( 3\vec{v} + \frac{1}{2}\vec{u}) = \vec{0}\nonumber$ If the set is linearly independent, the only solution will be $$a=b=c=0$$. We proceed as follows. \begin{aligned} a(2\vec{u} - \vec{w}) + b(\vec{w} + \vec{v}) + c( 3\vec{v} + \frac{1}{2} \vec{u}) &= \vec{0} \\ 2a\vec{u} - a\vec{w} + b\vec{w} + b\vec{v} + 3c\vec{v} + \frac{1}{2}c\vec{u} &= \vec{0}\\ (2a + \frac{1}{2}c) \vec{u} + (b+3c)\vec{v} + (-a + b) \vec{w} &= \vec{0}\end{aligned}

We know that the set $$S = \left\{ \vec{u}, \vec{v}, \vec{w} \right\}$$ is linearly independent, which implies that the coefficients in the last line of this equation must all equal $$0$$. In other words: \begin{aligned} 2a + \frac{1}{2} c &= 0 \\ b + 3c &= 0 \\ -a + b &= 0 \end{aligned}

The augmented matrix and resulting reduced row-echelon form are given by: $\left [ \begin{array}{rrr|r} 2 & 0 & \frac{1}{2} & 0 \\ 0 & 1 & 3 & 0 \\ -1 & 1 & 0 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right ]\nonumber$ Hence the solution is $$a=b=c=0$$ and the set is linearly independent.

The following theorem was discussed in terms in $$\mathbb{R}^n$$. We consider it here in the general case.

## Theorem $$\PageIndex{1}$$: Unique Representation

Let $$V$$ be a vector space and let $$U = \left\{ \vec{v}_1, \cdots, \vec{v}_k \right\} \subseteq V$$ be an independent set. If $$\vec{v} \in \mathrm{span} \;U$$, then $$\vec{v}$$ can be written uniquely as a linear combination of the vectors in $$U$$.

Consider the span of a linearly independent set of vectors. Suppose we take a vector which is not in this span and add it to the set. The following lemma claims that the resulting set is still linearly independent.

## Lemma $$\PageIndex{2}$$: Adding to a Linearly Independent Set

Suppose $$\vec{v}\notin \mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}$$ and $$\left\{ \vec{u}_{1},\cdots , \vec{u}_{k}\right\}$$ is linearly independent. Then the set $\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k},\vec{v} \right\}\nonumber$ is also linearly independent.

Proof

Suppose $$\sum_{i=1}^{k}c_{i}\vec{u}_{i}+d\vec{v}= \vec{0}.$$ It is required to verify that each $$c_{i}=0$$ and that $$d=0.$$ But if $$d\neq 0,$$ then you can solve for $$\vec{v}$$ as a linear combination of the vectors, $$\left\{ \vec{u}_{1},\cdots ,\vec{u} _{k}\right\}$$, $\vec{v}=-\sum_{i=1}^{k}\left( \frac{c_{i}}{d}\right) \vec{u}_{i}\nonumber$ contrary to the assumption that $$\vec{v}$$ is not in the span of the $$\vec{u}_{i}$$. Therefore, $$d=0.$$ But then $$\sum_{i=1}^{k}c_{i} \vec{u}_{i}=\vec{0}$$ and the linear independence of $$\left\{ \vec{u} _{1},\cdots ,\vec{u}_{k}\right\}$$ implies each $$c_{i}=0$$ also.

Below is a video on finding a third vector in $$R^3$$ that makes a set dependent (and then independent).

Consider the following example.

## Example $$\PageIndex{4}$$: Adding to a Linearly Independent Set

Let $$S \subseteq M_{22}$$ be a linearly independent set given by $S = \left\{ \left [ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right ] \right\}\nonumber$ Show that the set $$R \subseteq M_{22}$$ given by $R = \left\{ \left [ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right ] \right\}\nonumber$ is also linearly independent.

Solution

Instead of writing a linear combination of the matrices which equals $$0$$ and showing that the coefficients must equal $$0$$, we can instead use Lemma $$\PageIndex{2}$$.

To do so, we show that $\left [ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right ] \notin \mathrm{span}\left\{ \left [ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right ] \right\}\nonumber$

Write \begin{aligned} \left [ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right ] &= a\left [ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right ] + b\left [ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right ] \\ &= \left [ \begin{array}{rr} a & 0 \\ 0 & 0 \end{array} \right ] + \left [ \begin{array}{rr} 0 & b \\ 0 & 0 \end{array} \right ] \\ &= \left [ \begin{array}{rr} a & b \\ 0 & 0 \end{array} \right ]\end{aligned}

Clearly there are no possible $$a,b$$ to make this equation true. Hence the new matrix does not lie in the span of the matrices in $$S$$. By Lemma $$\PageIndex{2}$$, $$R$$ is also linearly independent.

This page titled 9.3: Linear Independence is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) .