Problem 2 - Differential Equations and Linear Algebra

Question

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

Step-by-Step Solution

Answer

The given vector functions \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\) are linearly independent on the interval \((-\infty, \infty)\) because we can only find the trivial solution \(C_1 = 0\) and \(C_2 = 0\) when trying to satisfy the equation \(C_1 \mathbf{x}_{1}(t) + C_2 \mathbf{x}_{2}(t) = \mathbf{0}\).

Step 1: Write down the equation we want to investigate

We first write down the equation that we wish to investigate: $$C_1 \mathbf{x}_{1}(t) + C_2 \mathbf{x}_{2}(t) = \mathbf{0}$$ Substitute the given vector functions into the equation: $$C_1 \left[\begin{array}{l} t \\\ t \end{array}\right] + C_2 \left[\begin{array}{l} t \\\ t^{2} \end{array}\right] = \left[\begin{array}{l} 0 \\\ 0 \end{array}\right]$$

Step 2: Equate the components

Now we have to equate the components on both sides of the equation. This will give us two separate scalar equations: Equation 1 (first component): $$C_1 t + C_2 t = 0$$ Equation 2 (second component): $$C_1 t + C_2 t^2 = 0$$

Step 3: Investigate solutions for constants

We have two scalar equations to analyze possible solutions for \(C_1\) and \(C_2\): Equation 1: \(C_1 t + C_2 t = 0\) Factor out \(t\): $$t(C_1 + C_2) = 0$$ Here, either \(t = 0\) or \(C_1 + C_2 = 0\). Since \(t\) ranges from \((-\infty, \infty)\), we cannot conclude that \(t = 0\). So, we are left with: $$C_1 + C_2 = 0 \Rightarrow C_2 = -C_1$$ Equation 2: \(C_1 t + C_2 t^2 = 0\) Substitute \(C_2 = -C_1\): $$C_1 t - C_1 t^2 = 0$$ Factor out \(C_1\): $$C_1(t - t^2) = 0$$ There are two potential cases for this equation: Either \(C_1 = 0\) or \(t = t^2\).

Step 4: Determine whether the vector functions are linearly independent or dependent

Recall that if there are non-trivial solutions for \(C_1\) and \(C_2\), the vector functions are linearly dependent. In other words, if we can only find the trivial solution, i.e., \(C_1 = 0\) and \(C_2 = 0\), the vector functions are linearly independent. Case 1: \(C_1 = 0\) If \(C_1 = 0\), then from the relation found previously, \(C_2 = -C_1\), we have: $$C_2 = -0 = 0$$ So, we have found the solution \(C_1 = 0\) and \(C_2 = 0\). Case 2: \(t = t^2\) This condition holds for \(t = 0, 1\). However, we cannot conclude that \(t = 0, 1\) for the entire range of \((-\infty, \infty)\). Since we only found trivial solutions for the coefficients \(C_1\) and \(C_2\) (i.e., \(C_1 = 0\) and \(C_2 = 0\)), we can conclude that the given vector functions are linearly independent on the interval \((-\infty, \infty)\).

Key Concepts

Vector Functions

Vector functions are crucial in understanding the dynamics of vectors that change over time or across dimensions. In this particular problem, we have two vector functions:

• \(\mathbf{x}_{1}(t) = \begin{bmatrix} t \ t \end{bmatrix}\)
• \(\mathbf{x}_{2}(t) = \begin{bmatrix} t \ t^{2} \end{bmatrix}\)

These functions map each real number \(t\) in the domain \((-\infty, \infty)\) to a two-dimensional vector. Each component of the vectors is expressed as a function of \(t\). For example, in the vector function \(\mathbf{x}_{2}(t)\), the first component increases linearly with \(t\), while the second component follows a quadratic relationship. This contrast forms a basis for investigating linear independence, as changes in scalar multiples of these vectors will behave distinctly over their domain.
Vector functions often represent path or transformations in various scientific and engineering fields. They help in visualizing the path through the space the vector covers over time.

Scalar Equations

Scalar equations play a fundamental role in determining whether vector functions are linearly independent or dependent. In the given exercise, after substituting the vector functions into the equation:

\[C_1 \mathbf{x}_{1}(t) + C_2 \mathbf{x}_{2}(t) = \mathbf{0}\],

we equate the components of the vectors to form scalar equations:

• First component: \(C_1 t + C_2 t = 0\)
• Second component: \(C_1 t + C_2 t^2 = 0\)

Both these equations help us break down complex vector operations into simpler, more manageable steps. By analyzing these scalar equations separately, we can explore solutions for \(C_1\) and \(C_2\).
Such equations allow us to isolate specific terms and examine conditions under which the overall expression is equal to zero. Solving these scalar equations provides insights into the relationship between the vectors, essential for understanding their independence or dependence.

Trivial Solutions

Finding trivial solutions is crucial when analyzing linear independence. Trivial solutions imply that the only solution to the vector equation is when all scalar coefficients are zero; in this problem, \(C_1 = 0\) and \(C_2 = 0\). Here, solving the scalar equations yielded such a trivial solution:

• From \(C_1 t + C_2 t = 0\), we deduced that \(C_1 + C_2 = 0\), leading to \(C_2 = -C_1\).
• Substituting \(C_2 = -C_1\) into the second equation provides \(C_1(t - t^2) = 0\).
• The potential solutions are \(C_1 = 0\) or \(t = t^2\), where \(t = t^2\) only holds for \(t = 0, 1\).

Since the only viable solution over the entire interval \((-\infty, \infty)\) is the trivial one, the vector functions are linearly independent. Recognizing trivial solutions helps establish the uniqueness of vector relations, ensuring meaningful interpretations in application scenarios.