Question

Under the assumptions of Theorem 10.5.2, let $$ \begin{array}{l} \mathbf{y}_{1}=\mathbf{x} e^{\lambda_{1} t} \\ \mathbf{y}_{2}=\mathbf{u} e^{\lambda_{1} t}+\mathbf{x} t e^{\lambda_{1} t}, \text { and } \\ \mathbf{y}_{3}=\mathbf{v} e^{\lambda_{1} t}+\mathbf{u} t e^{\lambda_{1} t}+\mathbf{x} \frac{t^{2} e^{\lambda_{1} t}}{2} \end{array} $$ Complete the proof of Theorem 10.5 .2 by showing that \(\mathbf{y}_{3}\) is a solution of \(\mathbf{y}^{\prime}=A \mathbf{y}\) and that \(\left\\{\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\\}\) is linearly independent.

Short answer

In conclusion, we were able to verify that each of the given vector functions \(\mathbf{y}_1\), \(\mathbf{y}_2\), and \(\mathbf{y}_3\) is a solution to the system of equations represented by \(\mathbf{y'} = A\mathbf{y}\). Furthermore, we also proved that the set \(\left\{\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\}\) is linearly independent. As a result, we successfully proved Theorem 10.5.2.

Step-by-step solution

Verify that \(\mathbf{y}_3\) is a solution of the given equation

To verify that \(\mathbf{y}_3\) is a solution to the given system, we will first find the derivative \(\mathbf{y}_3'\) and then confirm that it matches the equation \(\mathbf{y}' = A\mathbf{y}\). Differentiate \(\mathbf{y}_3\) with respect to \(t\), using the product and chain rules: $$ \begin{aligned} \mathbf{y}_3' &= \mathbf{v}e^{\lambda_1t}\lambda_1 + \mathbf{u}e^{\lambda_1t}\lambda_1 + \mathbf{u}te^{\lambda_1t}\lambda_1 + \mathbf{x}\frac{t^2e^{\lambda_1t}\lambda_1}{2} + \mathbf{x}te^{\lambda_1t} \\ &= \lambda_1(\mathbf{v} + \mathbf{u}t) e^{\lambda_1t} + (2\mathbf{x}t + \mathbf{u}) e^{\lambda_1t}\lambda_1 \end{aligned} $$ Now, compute \(A\mathbf{y}_3\): $$ A\mathbf{y}_3 = A(\mathbf{v}e^{\lambda_1t}+\mathbf{u}te^{\lambda_1t}+\mathbf{x}\frac{t^{2}e^{\lambda_1 t}}{2}) = A\mathbf{v}e^{\lambda_1t} + A\mathbf{u}te^{\lambda_1t} + A\mathbf{x}\frac{t^{2}e^{\lambda_1 t}}{2} $$ Since \(\mathbf{x}\), \(\mathbf{u}\), and \(\mathbf{v}\) are eigenvectors with eigenvalue \(\lambda_1\): $$ A\mathbf{x} = \lambda_1\mathbf{x} \\ A\mathbf{u} = \lambda_1\mathbf{u} \\ A\mathbf{v} = \lambda_1\mathbf{v} $$ Thus, $$ A\mathbf{y}_3 = (\lambda_1\mathbf{v})e^{\lambda_1t}+(\lambda_1\mathbf{u})te^{\lambda_1t}+\frac{t^{2}(\lambda_1\mathbf{x})e^{\lambda_1 t}}{2} = \lambda_1(\mathbf{v}e^{\lambda_1t}+\mathbf{u}te^{\lambda_1t}+\mathbf{x}\frac{t^{2}e^{\lambda_1 t}}{2}) = \mathbf{y}_3' $$ Since \(\mathbf{y}_3' = A\mathbf{y}_3\), \(\mathbf{y}_3\) is a solution to the given system.

Prove that the set \(\left\{\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\}\) is linearly independent

To show that the set is linearly independent, assume that there exists coefficients \(c_1\), \(c_2\), and \(c_3\) such that: $$ c_1\mathbf{y}_1 + c_2\mathbf{y}_2 + c_3\mathbf{y}_3 = \mathbf{0} $$ We will show that \(c_1 = c_2 = c_3 = 0\). Substitute the given expressions for \(\mathbf{y}_1\), \(\mathbf{y}_2\), and \(\mathbf{y}_3\): $$ c_1\mathbf{x}e^{\lambda_1t} + c_2(\mathbf{u}e^{\lambda_1t}+\mathbf{x}te^{\lambda_1t}) + c_3(\mathbf{v}e^{\lambda_1t}+\mathbf{u}te^{\lambda_1t}+\mathbf{x}\frac{t^{2}e^{\lambda_1 t}}{2}) = \mathbf{0} $$ Combine terms and factor out \(e^{\lambda_1t}\): $$ e^{\lambda_1t}((c_1\mathbf{x} + c_2\mathbf{u} + c_3\mathbf{v}) + (c_2\mathbf{x}t+c_3\mathbf{u}t^2) + (c_3\mathbf{x}\frac{t^2}{2})) = \mathbf{0} $$ Since the vector can only equal zero if each component is equal to zero, we have: 1. \(c_1\mathbf{x} + c_2\mathbf{u} + c_3\mathbf{v} = \mathbf{0}\) 2. \(c_2\mathbf{x} + c_3\mathbf{u} = \mathbf{0}\) 3. \(c_3\mathbf{x} = \mathbf{0}\) From equation (3), it follows that \(c_3 = 0\). Then, substituting into equation (2), we find that \(c_2 = 0\). Finally, substituting both values into equation (1), we get \(c_1 = 0\). Since \(c_1 = c_2 = c_3 = 0\), the set \(\left\{\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\}\) is linearly independent. By showing that each vector is a solution to the given system and proving that the set is linearly independent, we have completed the proof of Theorem 10.5.2.

Key concepts

Eigenvectors

An eigenvector is a special vector associated with a linear transformation represented by a matrix. When the matrix transforms the eigenvector, the direction of the vector does not change, although it might be stretched or shrunk by a certain factor. This factor is known as the eigenvalue.

For a matrix \(A\), if there exists a non-zero vector \(\mathbf{x}\) such that \(A\mathbf{x} = \lambda \mathbf{x}\), then \(\mathbf{x}\) is an eigenvector and \(\lambda\) is its corresponding eigenvalue.

Eigenvectors are fundamental in various mathematical concepts, especially in solving systems of linear equations. They simplify the matrices and help in understanding significant characteristics of linear transformations.

Eigenvalues

Eigenvalues are the factors by which an eigenvector is stretched or shrunk during a transformation by a matrix. Essentially, they are the scalars \(\lambda\) that satisfy the equation \(A\mathbf{x} = \lambda \mathbf{x}\).

To find eigenvalues, one generally solves the characteristic equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix of the same size as \(A\). Determining the eigenvalues helps in understanding various properties of the matrix, such as its stability and behavior in dynamic systems.

For example, when analyzing differential equations, eigenvalues can provide insights into system stability. Each eigenvalue corresponds to an eigenvector, and together, they offer a comprehensive picture of what impact a matrix will have on vectors within its space.

Linear Independence

Vectors are said to be linearly independent if none of the vectors in the set can be written as a linear combination of the others.

In simpler terms, if you can't find coefficients \(c_1, c_2, ..., c_n\) (not all zero) such that \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + ... + c_n\mathbf{v}_n = \mathbf{0}\), then the vectors are linearly independent. This concept is crucial in understanding how vectors span a vector space.

Linear independence ensures that each vector contributes something unique to the space. For instance, in the given exercise, proving that \(\{\mathbf{y}_1, \mathbf{y}_2, \mathbf{y}_3\} \) is linearly independent confirms that these vectors span a dimensionally optimal subset of the vector space influenced by the matrix \(A\).

Theorem Proof

Proving a theorem, like in the exercise, involves verifying all conditions and showing that each statement logically follows from the assumptions. The process typically involves these steps:

• Identifying the given information, assumptions, and what needs to be proved.
• Using known formulas, identities, and theorems to build logical steps towards the proof.
• Ensuring every step follows logically from the previous, maintaining rigor in assumptions and results.

The exercise highlights this through verifying the equation \(\mathbf{y}' = A\mathbf{y}\). It carefully checks whether \(\mathbf{y}_3\) meets this requirement by differentiation and substitution, confirming it independently solves the equation.

Similarly, linear independence is established by assuming a linear combination of the vectors equals zero, then proving the only solution is the trivial one. This methodical approach ensures all steps are logically sound, guaranteeing the theorem holds true under given conditions.