5.1.3: 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.