Question

In this problem we indicate how to show that \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\), as given by Eqs. (9), are linearly independent. Let \(r_{1}=\lambda+i \mu\) and \(\bar{r}_{1}=\lambda-i \mu\) be a pair of conjugate eigenvalues of the coefficient matrix \(\mathbf{A}\) of \(\mathrm{Fq}(1)\); let \(\xi^{(1)}=\mathbf{a}+i \mathbf{b}\) and \(\bar{\xi}^{(1)}=\mathbf{a}-i \mathbf{b}\) be the corresponding eigenvectors. Recall that it was stated in Section 7.3 that if \(r_{1} \neq \bar{r}_{1},\) then \(\boldsymbol{\xi}^{(1)}\) and \(\bar{\xi}^{(1)}\) are linearly independent. (a) First we show that a and b are linearly independent. Consider the equation \(c_{1} \mathrm{a}+\) \(c_{2} \mathrm{b}=0 .\) Express a and \(\mathrm{b}\) in terms of \(\xi^{(1)}\) and \(\bar{\xi}^{(1)},\) and then show that \(\left(c_{1}-i c_{2}\right) \xi^{(1)}+\) \(\left(c_{1}+i c_{2}\right) \bar{\xi}^{(1)}=0\) (b) Show that \(c_{1}-i c_{2}=0\) and \(c_{1}+i c_{2}=0\) and then that \(c_{1}=0\) and \(c_{2}=0 .\) Consequently, a and b are linearly independent. (c) To show that \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent consider the equation \(c_{1} \mathbf{u}\left(t_{0}\right)+\) \(c_{2} \mathbf{v}\left(t_{0}\right)=\mathbf{0}\), where \(t_{0}\) is an arbitrary point. Rewrite this equation in terms of a and \(\mathbf{b}\), and then proceed as in part (b) to show that \(c_{1}=0\) and \(c_{2}=0 .\) Hence \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent at the arbitrary point \(t_{0}\). Therefore they are linearly independent at every point and on every interval.

Short answer

Explain your answer. Answer: Yes, \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent at every point and on every interval. This is because we showed that their coefficients \(c_{1}\) and \(c_{2}\) are zero for an arbitrary point \(t_{0}\), and as a result, their linear combination is equal to the zero vector. Since this holds for any point, \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent on every interval.

Step-by-step solution

Linear Independence of a and b

To show that a and b are linearly independent, consider the equation: $$c_{1} \mathrm{a} + c_{2} \mathrm{b} = 0$$ We know that: $$\mathrm{a} = \frac{\xi^{(1)} + \bar{\xi}^{(1)}}{2}$$ $$\mathrm{b} = \frac{\xi^{(1)} - \bar{\xi}^{(1)}}{2i}$$ Substitute the expressions of a and b into the equation: $$c_{1}\frac{\xi^{(1)} + \bar{\xi}^{(1)}}{2} + c_{2}\frac{\xi^{(1)} - \bar{\xi}^{(1)}}{2i} =0$$ Multiplying by 2 and pulling the constants \(c_{1}\) and \(c_{2}\) inside: $$(c_{1}-i c_{2})\xi^{(1)} + (c_{1}+i c_{2})\bar{\xi}^{(1)}=0$$

Determine the Preconditions for Linear Independence of \(u(t)\) and \(v(t)\)

First, show that \(c_{1}-i c_{2}=0\) and \(c_{1}+i c_{2}=0\). This can be implied for linearly independent eigenvectors \(\xi^{(1)}\) and \(\bar{\xi}^{(1)}\). Since their linear combination is zero, we must have: $$c_{1}-i c_{2}=0$$ $$c_{1}+i c_{2}=0$$ This means that both \(c_{1}=0\) and \(c_{2}=0\). Therefore, a and b are linearly independent.

Demonstrate the Linear Independence of \(u(t)\) and \(v(t)\)

To show that \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent, consider the equation: $$c_{1} \mathbf{u}\left(t_{0}\right) + c_{2} \mathbf{v}\left(t_{0}\right)=\mathbf{0}$$ where \(t_{0}\) is an arbitrary point. Recall that: $$\mathbf{u}(t) = \mathrm{a} \cos{\mu t} - \mathrm{b} \sin{\mu t}$$ $$\mathbf{v}(t) = \mathrm{a} \sin{\mu t} + \mathrm{b} \cos{\mu t}$$ Substitute the expressions for \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) into the equation: $$c_{1}(\mathrm{a} \cos{\mu t_{0}} - \mathrm{b} \sin{\mu t_{0}}) + c_{2} (\mathrm{a} \sin{\mu t_{0}} + \mathrm{b} \cos{\mu t_{0}}) =0$$ Rewrite this equation in terms of a and b: $$[(c_{1}\cos{\mu t_{0}}\: +\: c_{2}\sin{\mu t_{0}})]\mathrm{a}+ [(c_{2}\cos{\mu t_{0}}\: -\: c_{1}\sin{\mu t_{0}})]\mathrm{b} =0$$ We already know a and b are linearly independent, which means the coefficients of a and b must be zero: $$c_{1}\cos{\mu t_{0}}\: +\: c_{2}\sin{\mu t_{0}}=0$$ $$c_{2}\cos{\mu t_{0}}\: -\: c_{1}\sin{\mu t_{0}}=0$$ Since \(\cos{\mu t_{0}}\) and \(\sin{\mu t_{0}}\) are not equal to zero, these equations imply that \(c_{1} = 0\) and \(c_{2} = 0\). This shows that \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent at an arbitrary point \(t_{0}\). As a result, \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent at every point and on every interval.

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial for analyzing linear transformations, especially in the context of matrix theory. Given a square matrix \( \mathbf{A} \), an eigenvalue \( \lambda \) is a scalar such that \( \mathbf{A}\mathbf{v} = \lambda\mathbf{v} \), where \( \mathbf{v} \) is a non-zero vector referred to as an eigenvector. The essence of this relationship is that the action of \( \mathbf{A} \) merely scales the eigenvector \( \mathbf{v} \) without changing its direction.

In the exercise, eigenvalues come in complex conjugate pairs, indicated as \( r_1 = \lambda + i \mu \) and \( \bar{r}_1 = \lambda - i \mu \). Their corresponding eigenvectors \( \xi^{(1)} \) and \( \bar{\xi}^{(1)} \) reflect a similar property, and they are the key players in determining the linear independence necessary for various proofs in linear algebra. When eigenvalues are distinct, their associated eigenvectors are naturally linearly independent.

• This concept helps in solving differential equations where the coefficient matrix \( \mathbf{A} \) is involved in generating eigenvector solutions.
• Key applications of eigenvalues and eigenvectors extend to stability analysis in systems, quantum mechanics, and the diagonalization of matrices.

Complex Conjugate

In mathematics, a complex conjugate is formed by changing the sign of the imaginary part of a complex number. For any complex number \( z = a + bi \), its complex conjugate is \( \bar{z} = a - bi \). Complex conjugates are imperative in several areas of mathematics, particularly in relation to eigenvalues and eigenvectors.

For instance, when dealing with eigenvalues that emerge as complex numbers, their conjugates often appear as eigenvalues of real matrices. These conjugate pairs play a role in ensuring certain symmetries and behaviors, particularly in fields like differential equations and signal processing. The exercise highlights the eigenvalues \( r_1 = \lambda + i \mu \) and \( \bar{r}_1 = \lambda - i \mu \). These are complex conjugates, and their linear independence is deduced from the distinct nature of these values.

• Complex conjugates simplify calculations, such as finding the magnitude of a complex number.
• They are used to simplify the roots of polynomials and in other algebraic manipulations.

Differential Equations

Differential equations involve equations containing derivatives of one or more functions. They are pivotal in mathematical modeling, describing how a quantity changes in relation to another. In analyzing linear systems, particularly involving matrices, eigenvalues and eigenvectors facilitate the solutions to differential equations.

The exercise shows solutions in terms of eigenvalues being used to break down complex systems into simpler parts, aiding in predicting system behavior over time. Solutions \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \), represented using sine and cosine functions, showcase solving linear differential equations using a method involving eigenvalues. These are particularly useful in systems with oscillatory behavior, governed by complex eigenvalues.

• Understanding differential equations is fundamental in physics, engineering, and economics where they describe change and flow relationships.
• They are classified mainly into ordinary (ODEs) and partial differential equations (PDEs).

Vector Spaces

Vector spaces are a foundational concept in linear algebra and involve a collection of vectors that can be scaled and added together while still remaining within the space. Key properties include vector addition and scalar multiplication. In the context of the exercise, vector spaces allow us to discuss linear independence, a core concept when proving the uniqueness of solutions to equations involving vectors, such as eigenvectors.

Linear independence implies that no vector in the set is a linear combination of others. It's crucial for vector spaces, ensuring no redundancy among vectors. The exercise showcases this by establishing the linear independence of the solutions to a differential equation, affirming the distinctiveness and necessity of each vector or solution.

• Vector spaces support operations, solutions, and transformations fundamental to fields such as computer graphics, data analysis, and quantum computing.
• Dimension, bases, span, and subspaces are key aspects in understanding the structure and behavior of vector spaces.