Linear Combinations of Vectors in Linear Algebra

In linear algebra, combining vectors using scalar multiplication and vector addition produces new vectors. The following guide presents the definition of a linear combination alongside detailed, step-by-step examples.

Table of Contents

Definition of Linear Combinations of Vectors

A vector \( \textbf{v} \) is a linear combination of the vectors \( \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \) if it can be written in the form:

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

where \( r_1, r_2, \dots, r_n \) are scalars (real numbers).

The set of all possible linear combinations of \( \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \) is called the span of these vectors.

Properties of Linear Combinations

If \( \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \) are vectors in \( \mathbb{R}^n \) and \( r_1, r_2, \dots, r_n \) are scalars, then the following properties hold:

  1. Closure under addition: The sum of two linear combinations is also a linear combination.
  2. Closure under scalar multiplication: A scalar multiple of a linear combination is also a linear combination.
  3. Contains the zero vector: Setting all scalars to zero gives the zero vector: \( 0\textbf{u}_1 + 0\textbf{u}_2 + \dots + 0\textbf{u}_n = \textbf{0} \).
  4. Contains each vector: Each \( \textbf{u}_i \) is itself a linear combination (set \( r_i = 1 \) and all other scalars to zero).

Examples with Solutions

Example 1: Expressing a Vector as a Linear Combination

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Express vector \( \textbf{v} = \begin{bmatrix} 1 \\ 7 \end{bmatrix} \) as a linear combination of the vectors \( \textbf{u}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix} -1 \\ 2 \end{bmatrix} \).

Solution

For vector \( \textbf{v} \) to be a linear combination of the vectors \( \textbf{u}_1 \) and \( \textbf{u}_2 \), we need to find scalars \( r_1 \) and \( r_2 \) such that:

\[ \textbf{v} = r_1\textbf{u}_1 + r_2\textbf{u}_2 \]

Substitute \( \textbf{v}, \textbf{u}_1, \) and \( \textbf{u}_2 \) with their components and perform scalar multiplication to obtain the equation:

\[ \begin{bmatrix} 1 \\ 7 \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + r_2 \begin{bmatrix} -1 \\ 2 \end{bmatrix} \] \[ \begin{bmatrix} 1 \\ 7 \end{bmatrix} = \begin{bmatrix} r_1 - r_2 \\ r_1 + 2r_2 \end{bmatrix} \]

Equality of the components yields a system of linear equations:

\[ \begin{cases} r_1 - r_2 = 1 \\ r_1 + 2r_2 = 7 \end{cases} \]

From the first equation, we get \( r_1 = r_2 + 1 \). Substituting this into the second equation:

\[ (r_2 + 1) + 2r_2 = 7 \implies 3r_2 + 1 = 7 \implies 3r_2 = 6 \implies r_2 = 2 \]

Now substitute \( r_2 = 2 \) back into the first equation:

\[ r_1 - 2 = 1 \implies r_1 = 3 \]

Hence, the linear combination is:

\[ \textbf{v} = 3\textbf{u}_1 + 2\textbf{u}_2 \]

Verification: \( 3\begin{bmatrix} 1 \\ 1 \end{bmatrix} + 2\begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} + \begin{bmatrix} -2 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 7 \end{bmatrix} \) ✓

Example 2: When a Linear Combination Cannot Be Formed

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Show that the vector \( \textbf{v} = \begin{bmatrix} 1 \\ 3 \end{bmatrix} \) cannot be expressed as a linear combination of the vectors \( \textbf{u}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix} 2 \\ 4 \end{bmatrix} \).

Solution

We need to show that there are no scalars \( r_1 \) and \( r_2 \) that satisfy the equation:

\[ \begin{bmatrix} 1 \\ 3 \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + r_2 \begin{bmatrix} 2 \\ 4 \end{bmatrix} \]

Scalar multiply and add the vectors on the right side of the equation:

\[ \begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} r_1 + 2r_2 \\ 2r_1 + 4r_2 \end{bmatrix} \]

This provides the following system of linear equations:

\[ \begin{cases} r_1 + 2r_2 = 1 \\ 2r_1 + 4r_2 = 3 \end{cases} \]

We can attempt to solve this system using the method of elimination. Multiply the first equation by \( -2 \):

\[ \begin{cases} -2r_1 - 4r_2 = -2 \\ 2r_1 + 4r_2 = 3 \end{cases} \]

Add the two equations together:

\[ 0 = 1 \]

This is a contradiction. The system has no solution; therefore, vector \( \textbf{v} \) cannot be expressed as a linear combination of the given vectors \( \textbf{u}_1 \) and \( \textbf{u}_2 \).

Geometric Interpretation: Since \( \textbf{u}_2 = 2\textbf{u}_1 \), both vectors lie on the same line through the origin. Their span is only that line, and \( \textbf{v} \) does not lie on it.

Example 3: Finding Unknown Variables for Valid Linear Combinations

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For what values of \( k \) can the vector \( \textbf{v} = \begin{bmatrix} -1 \\ 5 \end{bmatrix} \) be expressed as a linear combination of the vectors \( \textbf{u}_1 = \begin{bmatrix} -3 \\ 2 \end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix} k \\ 8 \end{bmatrix} \)?

Solution

We need to find values for \( r_1 \) and \( r_2 \) such that:

\[ \begin{bmatrix} -1 \\ 5 \end{bmatrix} = r_1 \begin{bmatrix} -3 \\ 2 \end{bmatrix} + r_2 \begin{bmatrix} k \\ 8 \end{bmatrix} \]

Scalar multiply and add the vectors on the right side:

\[ \begin{bmatrix} -1 \\ 5 \end{bmatrix} = \begin{bmatrix} -3r_1 + kr_2 \\ 2r_1 + 8r_2 \end{bmatrix} \]

Rewrite this as a system of equations:

\[ \begin{cases} -3r_1 + kr_2 = -1 \\ 2r_1 + 8r_2 = 5 \end{cases} \]

For this system to have a unique solution for \( r_1 \) and \( r_2 \), the determinant of the coefficient matrix must be non-zero:

\[ \text{Det} \begin{bmatrix} -3 & k \\ 2 & 8 \end{bmatrix} \ne 0 \]

Calculate the determinant:

\[ (-3)(8) - (k)(2) = -24 - 2k \]

Set the determinant to not equal zero and solve for \( k \):

\[ -24 - 2k \ne 0 \implies 2k \ne -24 \implies k \ne -12 \]

Conclusion: Vector \( \textbf{v} \) can be expressed as a linear combination of the given vectors \( \textbf{u}_1 \) and \( \textbf{u}_2 \) for any value of \( k \ne -12 \).

What happens when \( k = -12 \)? In that case, \( \textbf{u}_2 = \begin{bmatrix} -12 \\ 8 \end{bmatrix} = 4\begin{bmatrix} -3 \\ 2 \end{bmatrix} = 4\textbf{u}_1 \), so the two vectors are linearly dependent and span only a line. Whether \( \textbf{v} \) lies on that line determines if a solution exists.

Frequently Asked Questions

What is a linear combination of vectors?

A vector \( \textbf{v} \) is a linear combination of vectors \( \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \) if it can be written as \( \textbf{v} = r_1\textbf{u}_1 + r_2\textbf{u}_2 + \dots + r_n\textbf{u}_n \), where \( r_1, r_2, \dots, r_n \) are scalars.

When can a vector NOT be expressed as a linear combination?

A vector cannot be expressed as a linear combination of given vectors when the resulting system of linear equations is inconsistent (has no solution), such as when elimination leads to a contradiction like \( 0 = 1 \).

How do you determine if a vector is a linear combination of other vectors?

Set up the vector equation \( \textbf{v} = r_1\textbf{u}_1 + r_2\textbf{u}_2 + \dots + r_n\textbf{u}_n \), convert it to a system of linear equations by equating components, and solve for the scalars. If a solution exists, \( \textbf{v} \) is a linear combination.

What is the span of a set of vectors?

The span of vectors \( \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \) is the set of all possible linear combinations of these vectors. It forms a subspace of \( \mathbb{R}^n \).

What does it mean if two vectors are linearly dependent?

Two vectors are linearly dependent if one is a scalar multiple of the other. In that case, they span only a line, and their linear combinations cannot reach every vector in the plane.

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