Question

In each exercise, an eigenvalue \(\lambda\) is given for the matrix \(A\). (a) Find an eigenvector corresponding to the given eigenvalue \(\lambda\). (b) Find the other two eigenvalues of the matrix \(A\). (c) Find eigenvectors corresponding to the eigenvalues found in part (b). (d) Do the three solutions of \(\mathbf{y}^{\prime}=A \mathbf{y}\) formed from the eigenpairs make up a fundamental set of solutions? $$ A=\left[\begin{array}{rrr} 1 & 2 & 0 \\ -4 & 7 & 0 \\ 0 & 0 & 1 \end{array}\right], \quad \lambda=1 $$

Short answer

Answer: No, the three solutions cannot make up a fundamental set of solutions for the given matrix \(A\), as we were unable to find a valid eigenvector for the eigenvalue \(\lambda_2 = 7\).

Step-by-step solution

Find an eigenvector corresponding to the given eigenvalue \(\lambda = 1\)

First, we need to find \(\mathbf{x}\) such that \(A\mathbf{x} = \lambda\mathbf{x}\), which can be written as \((A - \lambda I)\mathbf{x} = \mathbf{0}\) where \(I\) is the identity matrix. Since the \(\lambda\) is already given, we will apply it directly: $$ (A-\lambda I)\mathbf{x} = \begin{bmatrix} 1-1 & 2 & 0 \\ -4 & 7-1 & 0 \\ 0 & 0 & 1-1 \end{bmatrix}\mathbf{x} = \begin{bmatrix} 0 & 2 & 0 \\ -4 & 6 & 0 \\ 0 & 0 & 0 \end{bmatrix}\mathbf{x} = \mathbf{0} $$ To solve the system of linear equations, we can use row reduction: $$ \begin{bmatrix} 0 & 2 & 0 \\ -4 & 6 & 0 \\ 0 & 0 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} -4 & 6 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} -4 & 6 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -\dfrac{3}{2} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ We get the system of equations: $$ \begin{cases} x_1 - \dfrac{3}{2} x_2 = 0 \\ 2x_2 = 0 \end{cases} $$ From the second equation, we have \(x_2 = 0\). Substituting it into the first equation, we get \(x_1 = 0\). Therefore, the eigenvector \(\mathbf{x}\) corresponding to \(\lambda = 1\) is: $$ \mathbf{x} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$

Find the other two eigenvalues of the matrix \(A\)

Using the characteristic equation, \(\det(A - \lambda I) = 0\), to find the eigenvalues of \(A\): $$ \det \begin{bmatrix} 1-\lambda & 2 & 0 \\ -4 & 7-\lambda & 0 \\ 0 & 0 & 1-\lambda \end{bmatrix} = (1-\lambda)((7-\lambda)(1-\lambda)) = 0 $$ Since we already know that \(\lambda = 1\) is an eigenvalue, the other two eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 7\).

Find eigenvectors corresponding to the other eigenvalues

Next, we need to find eigenvectors corresponding to the eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = 7\). However, we already found the eigenvector for \(\lambda_1 = 1\) in Step 1, so we only need to find the eigenvector for \(\lambda_2 = 7\). For \(\lambda_2 = 7\): $$ (A - 7I)\mathbf{x} = \begin{bmatrix} -6 & 2 & 0 \\ -4 & 0 & 0 \\ 0 & 0 & -6 \end{bmatrix}\mathbf{x} = \mathbf{0} $$ Using row reduction again: $$ \begin{bmatrix} -6 & 2 & 0 \\ -4 & 0 & 0 \\ 0 & 0 &-6 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -\dfrac{1}{3} & 0 \\ -4 & 0 & 0 \\ 0 & 0 & -6 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -\dfrac{1}{3}& 0 \\ 0 & \dfrac{4}{3}& 0 \\ 0 & 0 & -6 \end{bmatrix} $$ We get the system of equations: $$ \begin{cases} x_1 - \dfrac{1}{3} x_2 = 0 \\ \dfrac{4}{3}x_2 = 0 \\ -6x_3 = 0 \end{cases} $$ From the second equation, we get \(x_2 = 0\). Substituting it into the first equation, we get \(x_1 = 0\). From the third equation, we can determine \(x_3 = 0\). Therefore, the eigenvector \(\mathbf{x}\) corresponding to \(\lambda_2 = 7\) is: $$ \mathbf{x} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

Check whether the three solutions formed from the eigenpairs make up a fundamental set of solutions

It is important to note that the eigenvector we found for \(\lambda_2 = 7\) is the zero vector, which is not a valid eigenvector. Since we cannot find a valid eigenvector for \(\lambda_2 = 7\), we cannot find a fundamental set of solutions using the eigenpairs for this matrix \(A\).

Key concepts

Differential Equations

Differential equations are mathematical equations that involve unknown functions and their derivatives. In many applications, these equations are used to describe the behavior of dynamical systems. A typical example of a differential equation in a real-world context could be describing the growth of a population over time, or the cooling process of an object.

When we encounter a system of linear differential equations, \(\mathbf{y}'=A\mathbf{y}\), where \(A\) is a matrix and \(\mathbf{y}\) is a vector of functions, we're looking into a situation where each function's derivative in the vector \(\mathbf{y}\) is a linear combination of all the functions in \(\mathbf{y}\). Solving such systems often requires us to find eigenvectors and eigenvalues, as they provide a straightforward way to solve for 'modes' of the system that can describe the entire solution space.

Matrix Algebra

Matrix algebra involves operations with matrices, which are rectangular arrays of numbers, expressions, or symbols arranged in rows and columns. The significance of matrices lies in their powerful utility across various mathematical disciplines, including linear algebra, engineering, physics, and computer science.

In our context of solving eigenvalue problems, matrix manipulation techniques like row reductions become critical in determining eigenvectors. These matrix operations are part of matrix algebra and are essential tools for simplifying matrices to their simplest forms to solve linear equations arising from eigenvalue problems.

Characteristic Equation

The characteristic equation of a matrix is a polynomial equation derived from the determinant of the matrix after subtracting a scalar \(\lambda\) times the identity matrix, \(\det(A - \lambda I) = 0\). This equation is fundamental in finding eigenvalues, which are scalar values that, when substituted back into the matrix, produce a nontrivial solution for eigenvectors.

For instance, the characteristic equation helps us find the specific values that \(\lambda\) must take so that the matrix \(A - \lambda I\) becomes non-invertible, or when its determinant is zero. These specific values, \(\lambda\)'s, are direct indicators of inherent properties of the matrix \(A\) and give rise to the eigenvectors associated with them.

Fundamental Set of Solutions

A fundamental set of solutions refers to a collection of solutions to a system of differential equations that are linearly independent and span the solution space. In the context of systems defined by \(\mathbf{y}'=A\mathbf{y}\), each solution can often be constructed from an eigenpair (eigenvalue and eigenvector) of the matrix \(A\). To form a fundamental set, we need as many linearly independent solutions as the dimension of the vector \(\mathbf{y}\).

If the matrix \(A\) has enough distinct eigenvalues, with corresponding linearly independent eigenvectors, these can be used to construct the fundamental set of solutions. However, if there are not enough valid eigenpairs, as can happen when an eigenvalue has higher multiplicity and not enough associated independent eigenvectors, we must delve into more advanced methods such as the generalized eigenvector approach, to build a complete solution set for the differential equations.