Question

Determine whether the given set of vectors is linearly independent for \(-\infty

Short answer

Question: Determine if the given set of vectors is linearly independent or dependent, and find the linear relation among them, if any: Vectors: $$ \mathbf{x}^{(1)}(t) = (e^{-t}, 2 e^{-t}) $$ $$ \mathbf{x}^{(2)}(t) = (e^{-t}, e^{-t}) $$ $$ \mathbf{x}^{(3)}(t) = (3 e^{-t}, 0) $$ Answer: The given set of vectors is linearly dependent for \(-\infty<t<\infty\). The linear relation among them is: $$ k\left(\frac{1}{6}\mathbf{x}^{(1)}(t) - \frac{1}{3} \mathbf{x}^{(2)}(t) + \frac{1}{6} \mathbf{x}^{(3)}(t)\right) = (0, 0) $$

Step-by-step solution

Write the linear combination

Consider the linear combination of the vectors: $$ c_1 \mathbf{x}^{(1)}(t) + c_2 \mathbf{x}^{(2)}(t) + c_3 \mathbf{x}^{(3)}(t) $$ where \(c_1, c_2, c_3\) are scalar constants.

Substitute given vectors and set the linear combination to the zero vector

Write the linear combination using the given vectors and set it equal to the zero vector: $$ c_1 (e^{-t}, 2 e^{-t}) + c_2 (e^{-t}, e^{-t}) + c_3 (3 e^{-t}, 0) = (0, 0) $$

Distribute the coefficients and find the componentwise equation

Distribute the coefficients and write the equation in component form: $$ \begin{cases} c_1 e^{-t} + c_2 e^{-t} + 3 c_3 e^{-t} = 0 \\ 2 c_1 e^{-t} + c_2 e^{-t} = 0 \end{cases} $$ To make it homogeneous, divide components by \(e^{-t}\): $$ \begin{cases} c_1 + c_2 + 3 c_3 = 0 \\ 2 c_1 + c_2 = 0 \end{cases} $$

Solve the system of equations and determine linear independence or dependence

We have a system of 2 equations with 3 unknowns. Solve the system to find the relations between \(c_1\), \(c_2\), and \(c_3\): $$ \begin{cases} c_1 + c_2 + 3 c_3 = 0 \\ 2 c_1 + c_2 = 0 \end{cases} $$ From the second equation, we get \(c_1 = -\frac{1}{2} c_2\). Substitute this into the first equation: $$ -\frac{1}{2}c_2 + c_2 + 3c_3 = 0 \implies \frac{1}{2}c_2 + 3c_3 = 0 $$ Now, let's introduce a new constant \(k = -2c_2\). With this redefinition, the equation becomes: $$ k + 6c_3 = 0 $$ So, the only solution for \(c_2\) and \(c_3\) is when \(c_2 = -\frac{1}{3}k\) and \(c_3 = \frac{1}{6}k\), where \(k\) can be any constant value. Since there is a non-trivial solution, the given vectors are linearly dependent.

Determine the linear relation

We have found the relation between \(c_2\) and \(c_3\). Substitute back the value of \(c_1\) in terms of \(c_2\): $$ c_1 = -\frac{1}{2}(-\frac{1}{3}k) = \frac{1}{6}k $$ The linear relation among the vectors is given by: $$ \frac{1}{6}k \mathbf{x}^{(1)}(t) - \frac{1}{3}k \mathbf{x}^{(2)}(t) + \frac{1}{6}k \mathbf{x}^{(3)}(t) = (0, 0) $$ This can also be written as: $$ k\left(\frac{1}{6}\mathbf{x}^{(1)}(t) - \frac{1}{3} \mathbf{x}^{(2)}(t) + \frac{1}{6} \mathbf{x}^{(3)}(t)\right) = (0, 0) $$ So, the given set of vectors is linearly dependent for \(-\infty<t<\infty\) with the linear relation provided above.

Key concepts

Linear Combination

A linear combination involves combining vectors using scalar coefficients. For a set of vectors like \( \mathbf{x}^{(1)}(t) \), \( \mathbf{x}^{(2)}(t) \), and \( \mathbf{x}^{(3)}(t) \), a linear combination would look like \( c_1 \mathbf{x}^{(1)}(t) + c_2 \mathbf{x}^{(2)}(t) + c_3 \mathbf{x}^{(3)}(t) \), where \( c_1, c_2, \) and \( c_3 \) are scalars. When we talk about linear combinations, we are asking whether these scalars can be set to make the combination equal to another vector, such as the zero vector.
This idea is pivotal in determining if sets of vectors are linearly dependent or independent. If only the trivial solution (where all coefficients are zero) exists, the vectors are independent. If there are other non-zero solutions, they are dependent.

System of Equations

A system of equations is a collection of equations that we solve together. In our example, after setting up the linear combination to equal the zero vector, we derive two equations:

• \( c_1 + c_2 + 3c_3 = 0 \)
• \( 2c_1 + c_2 = 0 \)

These equations form a system because they both need to be satisfied at the same time. Solving this system involves finding values for \( c_1, c_2, \) and \( c_3 \) that make both equations true simultaneously.
This problem-solving technique is essential in linear algebra, as it allows us to explore the relationships between different vectors, potentially revealing whether they are linearly dependent or independent.

Linear Relation

A linear relation describes a dependency between vectors expressed through a linear combination that equals zero. When the vectors are linearly dependent, as in our example, there exists a non-trivial solution that involves at least one of the scalar coefficients being non-zero. In this case, we found that:

• \( c_1 = \frac{1}{6}k \)
• \( c_2 = -\frac{1}{3}k \)
• \( c_3 = \frac{1}{6}k \)

where \( k \) is any constant. Such a linear relation proves that the initial vectors aren't linearly independent. This concept is significant because finding a linear relation helps identify redundancy among the vectors, which has practical implications in many fields, like physics or computer science. Linear relations help simplify complex systems by reducing unnecessary dimensions in vector spaces.

Differential Equations

In this exercise, although differential equations aren't directly tackled, they are conceptually linked due to the involvement of exponentials and time parameter \( t \). Differential equations often deal with functions and their derivatives, and when vectors depend on \( t \), as with \( e^{-t} \) in our example, they are sometimes part of solutions in such equations.
Understanding the relationship between differential equations and vectors is crucial in fields such as engineering and physics. Systems of differential equations often need to be solved using linear algebra techniques, particularly when determining the system's behavior over time. Linear dependence among vectors can reflect dependencies in solutions to differential equations, affecting stability and long-term predictions in dynamic systems.