Problem 3 - Differential Equations and Linear Algebra

Question

Show that the given vector functions are linearly independent on \((-\infty, \infty)\). $$\begin{array}{l} \mathbf{x}_{1}(t)=\left[\begin{array}{c} t+1 \\ t-1 \\ 2 t \end{array}\right], \quad \mathbf{x}_{2}(t)=\left[\begin{array}{c} e^{t} \\ e^{2 t} \\ e^{3 t} \end{array}\right] \\ \mathbf{x}_{3}(t)=\left[\begin{array}{c} 1 \\ \sin t \\ \cos t \end{array}\right] \end{array}$$

Step-by-Step Solution

Answer

The given vector functions \(\mathbf{x}_1(t)\), \(\mathbf{x}_2(t)\), and \(\mathbf{x}_3(t)\) are linearly independent on \((-\infty, \infty)\) because the only solution to the equation \(c_1\mathbf{x}_1(t) + c_2\mathbf{x}_2(t) + c_3\mathbf{x}_3(t) = \mathbf{0}\) for all \(t\) is the trivial solution where \(c_1 = c_2 = c_3 = 0\).

Step 1: Set up the linear combination equation

Create the linear combination of the given vector functions using constants \(c_1\), \(c_2\), and \(c_3\). $$c_1\left[\begin{array}{c} t+1\\ t-1\\ 2t \end{array}\right] + c_2\left[\begin{array}{c} e^t\\ e^{2t}\\ e^{3t} \end{array}\right] + c_3\left[\begin{array}{c} 1\\ \sin t\\ \cos t \end{array}\right] = \left[\begin{array}{c} 0\\ 0\\ 0 \end{array}\right]$$

Step 2: Write the linear combination as a system of equations

The linear combination equation can be represented as a system of three scalar equations. $$\begin{cases} c_1(t+1) + c_2e^t + c_3 = 0 \\ c_1(t-1) + c_2e^{2t} + c_3\sin t = 0 \\ c_1(2t) + c_2e^{3t} + c_3\cos t = 0 \end{cases}$$

Step 3: Check if the trivial solution is the only solution

Observe the system of equations, we see that there is no clear method for solving for the constants. Instead, let's try to determine if the trivial solution is the only solution. For \(t=0\), the system of equations becomes: $$\begin{cases} c_1 + c_2 + c_3 = 0 \\ -c_1 + c_2 + c_3 = 0 \\ c_3 = 0 \end{cases}$$ From the third equation, we can conclude that \(c_3\) must be 0. Substituting this value into the first and second equations, we get: $$\begin{cases} c_1 + c_2 = 0 \\ -c_1 + c_2 = 0 \end{cases}$$ Subtracting the second equation from the first, we find that \(2c_1 = 0\), thus \(c_1 = 0\). Substituting this value back into the first equation, we get \(c_2 = 0\). Thus, the only solution is the trivial solution where \(c_1 = c_2 = c_3 = 0\).

Step 4: Conclusion

Since the trivial solution is the only solution for the system of equations, we can conclude that the given vector functions \(\mathbf{x}_1(t)\), \(\mathbf{x}_2(t)\), and \(\mathbf{x}_3(t)\) are linearly independent on \((-\infty, \infty)\).

Key Concepts

Vector Functions

Vector functions are an essential part of understanding vector calculus and differential equations. A vector function is a function that takes one or more variables and returns a vector as the output. In our case, the variable is time, denoted as \( t \), and the vector functions \( \mathbf{x}_1(t) \), \( \mathbf{x}_2(t) \), and \( \mathbf{x}_3(t) \) produce three-dimensional vectors based on varying values of \( t \).

Each of these functions describes a different kind of motion or transformation in three-dimensional space. For instance:

• \( \mathbf{x}_1(t) \) is a linear combination of \( t \) that results in straight lines being traced out in space as \( t \) changes.
• \( \mathbf{x}_2(t) \) involves exponential functions, which can grow or decay at different rates for each component of the vector, leading to curves that grow exponentially.
• \( \mathbf{x}_3(t) \) involves trigonometric functions like sine and cosine, which usually describe oscillatory or circular motion.

Recognizing how each vector function behaves is crucial for exploring concepts like linear independence.

Trivial Solution

The concept of a trivial solution is central to understanding linear independence. A trivial solution refers to a solution where every variable or constant in an equation equals zero. In the context of our exercise, we have the equation formed by a linear combination of vector functions:\[c_1 \mathbf{x}_1(t) + c_2 \mathbf{x}_2(t) + c_3 \mathbf{x}_3(t) = \mathbf{0}\] This equation asks whether the only possible way to form a zero vector \( \mathbf{0} \) using constants \( c_1 \), \( c_2 \), and \( c_3 \) is by setting each constant to zero, which is the trivial solution.

It is critical to verify that no other combination of \( c_1 \), \( c_2 \), and \( c_3 \) can lead to zero. By showing that only when \( c_1 = c_2 = c_3 = 0 \) do the equations hold true, we demonstrate linear independence of the vector functions.

System of Equations

The setup or transformation of a linear combination into a system of equations is a classic method to test for linear independence among vector functions. In this instance, our vector equation becomes:\[\begin{cases} c_1(t+1) + c_2e^t + c_3 = 0 \ c_1(t-1) + c_2e^{2t} + c_3\sin t = 0 \ c_1(2t) + c_2e^{3t} + c_3\cos t = 0 \end{cases}\]This system represents three scalar equations derived from the components of the vector functions. Solving the system involves finding values for \( c_1 \), \( c_2 \), and \( c_3 \) that satisfy all equations.

By focusing on specific values of \( t \), such as \( t=0 \), the system simplifies, allowing us to solve more easily. We then check if solutions beyond the trivial \( c_1 = c_2 = c_3 = 0 \) could possibly exist. If not, as shown in our example, the vector functions are indeed linearly independent.