Linearly Independent and Dependent Vectors

Understanding whether a set of vectors is linearly independent or dependent is a core concept in linear algebra. It determines if vectors provide unique directions in space or if some are redundant. This page presents the definitions alongside detailed, step-by-step examples.

Table of Contents

Definition of Linearly Independent Vectors

If we can express a vector \( \textbf{u}_1 \) as a linear combination of vectors \( \textbf{u}_2 \) and \( \textbf{u}_3 \), we say that these 3 vectors are linearly dependent. For example, if:

\[ c_1 \textbf{u}_1 + c_2 \textbf{u}_2 + c_3 \textbf{u}_3 = 0 \]

where \( c_1 \ne 0 \), we can rearrange it as:

\[ \textbf{u}_1 = -\frac{c_2}{c_1}\textbf{u}_2 - \frac{c_3}{c_1}\textbf{u}_3 \]

Formal Definition:

Given a set of vectors \( W = \{ \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \} \).

If the vector equation:

\[ r_1 \textbf{u}_1 + r_2 \textbf{u}_2 + \dots + r_n \textbf{u}_n = 0 \]

has only the trivial solution (\( r_1 = r_2 = \dots = r_n = 0 \)), we say that \( W \) is a set of linearly independent vectors.

If the equation has other non-trivial solutions (where at least one \( r_i \ne 0 \)), then \( W \) is a set of linearly dependent vectors.

Tests for Linear Independence

Several methods can be used to determine whether a set of vectors is linearly independent or dependent:

  1. By Inspection: If one vector is a scalar multiple of another, the set is linearly dependent.
  2. Determinant Test: For \( n \) vectors in \( \mathbb{R}^n \), form a square matrix with the vectors as columns. If \( \det(A) \ne 0 \), the vectors are linearly independent. If \( \det(A) = 0 \), they are linearly dependent.
  3. Row Reduction: Form the augmented matrix \( [A | \textbf{0}] \) and row reduce. If there are pivot positions in every column corresponding to a variable, the only solution is trivial (independent). If a column lacks a pivot, there are free variables (dependent).
  4. Rank Test: The vectors are linearly independent if and only if the rank of the matrix equals the number of vectors.

Important Note: Any set containing the zero vector is automatically linearly dependent. Also, in \( \mathbb{R}^n \), any set of more than \( n \) vectors is automatically linearly dependent.

Examples with Solutions

Note: Work through the examples in the order they are presented to fully understand the methods.

Example 1: Identifying Linearly Dependent Vectors in \( \mathbb{R}^2 \)

Show Solution

Show that the vectors:

\[ \textbf{u}_1 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}, \quad \textbf{u}_2 = \begin{bmatrix} -5 \\ -15 \end{bmatrix} \]

are linearly dependent.

Solution

There are two ways to answer this question:

Method 1: By Inspection

There is an obvious scalar relationship between \( \textbf{u}_1 \) and \( \textbf{u}_2 \):

\[ \textbf{u}_2 = -5\textbf{u}_1 \]

Because one vector is a scalar multiple of the other, the two vectors are linearly dependent.

Method 2: Using the Formal Definition

Set up the linearity equation:

\[ r_1 \begin{bmatrix} 1 \\ 3 \end{bmatrix} + r_2 \begin{bmatrix} -5 \\ -15 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

Scalar multiply and add the left side to write a system of equations:

\[ \begin{cases} r_1 - 5r_2 = 0 \\ 3r_1 - 15r_2 = 0 \end{cases} \]

We use the method of elimination. Multiply the top equation by \( -3 \):

\[ \begin{cases} -3r_1 + 15r_2 = 0 \\ 3r_1 - 15r_2 = 0 \end{cases} \]

Add the top and bottom equations together:

\[ 0 = 0 \]

This means the second equation is redundant. Let \( r_2 = t \), where \( t \) is any real number. Substituting this into the first equation yields:

\[ r_1 = 5r_2 = 5t \]

Because there are infinitely many solutions (e.g., \( r_1 = 5, r_2 = 1 \)), the vectors \( \textbf{u}_1 \) and \( \textbf{u}_2 \) are linearly dependent.

Geometric Interpretation: Both vectors lie on the same line through the origin, so they point in the same (or opposite) direction and provide only one independent direction.

Example 2: Testing Linearity using Determinants

Show Solution

Show that the vectors:

\[ \textbf{u}_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \quad \textbf{u}_2 = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, \quad \textbf{u}_3 = \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix} \]

are linearly independent.

Solution

Use the definition to prove the vectors are linearly independent. Set up the equation:

\[ r_1 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + r_2 \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} + r_3 \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

This forms a homogeneous system of linear equations, which can be written in matrix form as:

\[ \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 3 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

Since the coefficient matrix is a square (\( 3 \times 3 \)) matrix, the system has only the trivial solution (\( r_1 = 0, r_2 = 0, r_3 = 0 \)) if and only if the determinant is not equal to zero.

Calculate the determinant using cofactor expansion along the first row:

\[ \text{Det} \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 3 \\ 1 & 2 & 0 \end{bmatrix} = 1 \cdot (1 \cdot 0 - 3 \cdot 2) - 0 \cdot (0 \cdot 0 - 3 \cdot 1) + (-2) \cdot (0 \cdot 2 - 1 \cdot 1) \] \[ \text{Det} = 1(-6) - 0 + (-2)(-1) = -6 + 2 = -4 \]

Since the determinant is \( -4 \ne 0 \), the equation has only the trivial solution. Therefore, the vectors are linearly independent.

Verification by Row Reduction: We can also verify this result by row reducing the matrix to reduced row echelon form (RREF). The RREF is the \( 3 \times 3 \) identity matrix, confirming there are pivot positions in every column and thus only the trivial solution.

Example 3: Finding Unknown Variables for Linearly Dependent Vectors

Show Solution

Find the values of \( m \) for which the vectors:

\[ \textbf{u}_1 = \begin{bmatrix} m \\ 4 \\ 0 \end{bmatrix}, \quad \textbf{u}_2 = \begin{bmatrix} 1 \\ -1 \\ 8 \end{bmatrix}, \quad \textbf{u}_3 = \begin{bmatrix} 0 \\ -1 \\ m \end{bmatrix} \]

are linearly dependent.

Solution

We start with the equation of linearity:

\[ r_1 \begin{bmatrix} m \\ 4 \\ 0 \end{bmatrix} + r_2 \begin{bmatrix} 1 \\ -1 \\ 8 \end{bmatrix} + r_3 \begin{bmatrix} 0 \\ -1 \\ m \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

Express this as a matrix equation:

\[ \begin{bmatrix} m & 1 & 0 \\ 4 & -1 & -1 \\ 0 & 8 & m \end{bmatrix} \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

For the vectors to be linearly dependent, the system must have more than one (non-trivial) solution. This happens when the determinant of the coefficient matrix is exactly \( 0 \).

Calculate the determinant using cofactor expansion along the first row:

\[ \text{Det} \begin{bmatrix} m & 1 & 0 \\ 4 & -1 & -1 \\ 0 & 8 & m \end{bmatrix} = m \cdot ((-1)(m) - (-1)(8)) - 1 \cdot ((4)(m) - (-1)(0)) + 0 \] \[ \text{Det} = m(-m + 8) - 1(4m) = -m^2 + 8m - 4m = -m^2 + 4m \]

Set the determinant to \( 0 \):

\[ -m^2 + 4m = 0 \implies m(-m + 4) = 0 \]

Solving for \( m \):

\[ m = 0 \quad \text{or} \quad m = 4 \]

The vectors are linearly dependent when \( m = 0 \) or \( m = 4 \).

Note: For all other values of \( m \), the determinant is non-zero, and the vectors are linearly independent.

Example 4: Testing Linearity using Row Reduction

Show Solution

Are the vectors:

\[ \textbf{u}_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ -1 \end{bmatrix}, \quad \textbf{u}_2 = \begin{bmatrix} 5 \\ 5 \\ 0 \\ 0 \end{bmatrix}, \quad \textbf{u}_3 = \begin{bmatrix} 0 \\ 0 \\ -1 \\ -1 \end{bmatrix} \]

linearly dependent or independent?

Solution

Write the linearity equation:

\[ r_1 \begin{bmatrix} 1 \\ 1 \\ 0 \\ -1 \end{bmatrix} + r_2 \begin{bmatrix} 5 \\ 5 \\ 0 \\ 0 \end{bmatrix} + r_3 \begin{bmatrix} 0 \\ 0 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \]

Convert this into an augmented matrix for the homogeneous system:

\[ \begin{bmatrix} 1 & 5 & 0 & | & 0 \\ 1 & 5 & 0 & | & 0 \\ 0 & 0 & -1 & | & 0 \\ -1 & 0 & -1 & | & 0 \end{bmatrix} \]

Step 1: Subtract Row 1 from Row 2 to clear the second row:

\[ \begin{bmatrix} 1 & 5 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & -1 & | & 0 \\ -1 & 0 & -1 & | & 0 \end{bmatrix} \]

Step 2: Add Row 1 to Row 4: \((-1+1 = 0, 0+5 = 5, -1+0 = -1)\). Row 4 becomes \([0, 5, -1 | 0]\):

\[ \begin{bmatrix} 1 & 5 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & -1 & | & 0 \\ 0 & 5 & -1 & | & 0 \end{bmatrix} \]

Step 3: Swap Row 2 and Row 4, and multiply Row 2 by \( \frac{1}{5} \):

\[ \begin{bmatrix} 1 & 5 & 0 & | & 0 \\ 0 & 1 & -\frac{1}{5} & | & 0 \\ 0 & 0 & -1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \]

Step 4: Eliminate the remaining entries. Multiply Row 3 by \( -1 \) and use it to eliminate the \( r_3 \) entry in Row 2. Then eliminate the \( r_2 \) entry in Row 1 using Row 2:

\[ \begin{bmatrix} 1 & 0 & 0 & | & 0 \\ 0 & 1 & 0 & | & 0 \\ 0 & 0 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \]

The fully reduced row echelon form (RREF) has pivot positions in the first three columns, corresponding to \( r_1, r_2, r_3 \). This corresponds to the system:

\[ \begin{cases} r_1 = 0 \\ r_2 = 0 \\ r_3 = 0 \end{cases} \]

Because the equation of linearity has only the trivial solution (\( r_1 = 0, r_2 = 0, r_3 = 0 \)), the vectors \( \textbf{u}_1 \), \( \textbf{u}_2 \), and \( \textbf{u}_3 \) are linearly independent.

Observation: Notice that \( \textbf{u}_2 = 5\textbf{u}_1 \) in the first two components, but the third and fourth components break this relationship. If \( \textbf{u}_2 \) were truly \( 5\textbf{u}_1 \), it would need to be \( [5, 5, 0, -5]^T \), but it is \( [5, 5, 0, 0]^T \). This difference is what makes the set linearly independent.

Frequently Asked Questions

What does it mean for vectors to be linearly independent?

A set of vectors is linearly independent if the only solution to the equation \( r_1\textbf{u}_1 + r_2\textbf{u}_2 + \dots + r_n\textbf{u}_n = \textbf{0} \) is the trivial solution where all scalars equal zero (\( r_1 = r_2 = \dots = r_n = 0 \)).

What does it mean for vectors to be linearly dependent?

A set of vectors is linearly dependent if there exists a non-trivial solution to the equation \( r_1\textbf{u}_1 + r_2\textbf{u}_2 + \dots + r_n\textbf{u}_n = \textbf{0} \), meaning at least one scalar is not zero. This means at least one vector can be written as a linear combination of the others.

How do you test if vectors are linearly independent using determinants?

For \( n \) vectors in \( \mathbb{R}^n \), form a square matrix with the vectors as columns. If the determinant is non-zero, the vectors are linearly independent. If the determinant equals zero, the vectors are linearly dependent.

Can a set containing the zero vector be linearly independent?

No. Any set containing the zero vector is always linearly dependent, because we can write \( 1 \cdot \textbf{0} + 0 \cdot \textbf{u}_1 + \dots + 0 \cdot \textbf{u}_n = \textbf{0} \) with a non-zero coefficient on the zero vector.

What is the maximum number of linearly independent vectors in \( \mathbb{R}^n \)?

In \( \mathbb{R}^n \), the maximum number of linearly independent vectors is \( n \). Any set of more than \( n \) vectors in \( \mathbb{R}^n \) is automatically linearly dependent.

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