9.3: Linear Independence

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.

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.

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.

Example $\PageIndex{3}$: Related Independent Sets

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.

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.