Problem 25 - Differential Equations and Linear Algebra

Question

Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{rrr}0 & 1 & -1 \\\0 & 2 & 0 \\\2 & -1 & 3\end{array}\right]$$.

Step-by-Step Solution

Answer

The eigenvalues of the matrix are \(0, 1, 5\). The corresponding eigenvectors are \(\mathbf{v}=\begin{bmatrix}1\\0\\2\end{bmatrix}\) for \(\lambda = 0\), \(\mathbf{v}=\begin{bmatrix}1\\1\\1\end{bmatrix}\) for \(\lambda = 1\), and \(\mathbf{v}=\begin{bmatrix}1\\0\\-1\end{bmatrix}\) for \(\lambda = 5\).

Step 1: Find the Characteristic Equation

Subtract the eigenvalue multiplied by the identity matrix from the given matrix and compute the determinant of that matrix. The matrix then becomes: \[ \left[\begin{array}{rrr} 0-\lambda &1 & -1\\ 0 & 2-\lambda & 0\\ 2 & -1 & 3-\lambda \end{array}\right] \] Now, compute the determinant of the matrix: \[ \begin{vmatrix} 0-\lambda & 1 & -1\\ 0 & 2-\lambda & 0 \\ 2 & -1 & 3-\lambda \end{vmatrix} \]

Step 2: Solve the Characteristic Equation

Compute the determinant: \(D(\lambda) = (0-\lambda)\left((2-\lambda)(3-\lambda) - (0)\right) - 2 \((0)(3-\lambda) - (1)(-1)\) \) \(D(\lambda) = -\lambda((2-\lambda)(3-\lambda) - 1) \) Now, solve the equation D(λ) = 0 to find eigenvalues: \(-\lambda((2-\lambda)(3-\lambda) - 1) = 0\) \(\lambda = 0\) or \(((2-\lambda)(3-\lambda) - 1) = 0\) Upon solving the second equation, we find the eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = 5\). Thus, the eigenvalues of the matrix are \(0, 1, 5\).

Step 3: Find the Eigenvectors for Each Eigenvalue

Next, we will find the eigenvectors corresponding to each eigenvalue by solving the equation \((A-\lambda I) \mathbf{v} = 0\). 1. Eigenvector for \(\lambda = 0\): \[ \left(\begin{array}{rrr} 0 & 1 & -1\\ 0 & 2 & 0\\ 2 & -1 & 3 \end{array}\right) \mathbf{v} = \mathbf{0} \] By solving this system of linear equations, we get the eigenvector as \(\mathbf{v}=\begin{bmatrix} 1\\0\\ 2\end{bmatrix}.\) 2. Eigenvector for \(\lambda = 1\): \[ \left(\begin{array}{rrr} -1 & 1 & -1\\ 0 & 1 & 0\\ 2 & -1 & 2 \end{array}\right) \mathbf{v} = \mathbf{0} \] By solving this system of linear equations, we get the eigenvector as \(\mathbf{v}=\begin{bmatrix} 1\\1\\ 1\end{bmatrix}.\) 3. Eigenvector for \(\lambda = 5\): \[ \left(\begin{array}{rrr} -5 & 1 & -1\\ 0 & -3 & 0\\ 2 & -1 & -2 \end{array}\right) \mathbf{v} = \mathbf{0} \] By solving this system of linear equations, we get the eigenvector as \(\mathbf{v}=\begin{bmatrix} 1\\0\\ -1\end{bmatrix}.\) The eigenvalue \(\lambda = 0\) has eigenvector \(\mathbf{v}=\begin{bmatrix}1\\0\\2\end{bmatrix}\), the eigenvalue \(\lambda = 1\) has eigenvector \(\mathbf{v}=\begin{bmatrix}1\\1\\1\end{bmatrix}\), and the eigenvalue \(\lambda = 5\) has eigenvector \(\mathbf{v}=\begin{bmatrix}1\\0\\-1\end{bmatrix}\).

Key Concepts

Characteristic Equation

The characteristic equation is a fundamental concept in linear algebra used to find the eigenvalues of a matrix. To derive the characteristic equation, you begin by taking the matrix you are working with, let's call it matrix \( A \). You subtract \( \lambda \) times the identity matrix \( I \) from the original matrix \( A \). This operation results in a new matrix \( A - \lambda I \). The goal is to find the values of \( \lambda \) that make the determinant of this new matrix equal to zero. These values of \( \lambda \) are the eigenvalues of the matrix.

For example, given a matrix \( A \):

• Subtract \( \lambda I \) from \( A \): \[ \begin{bmatrix} 0-\lambda & 1 & -1 \ 0 & 2-\lambda & 0 \ 2 & -1 & 3-\lambda \end{bmatrix} \]
• Next, compute the determinant of this matrix, which leads to the characteristic equation.

In our exercise, solving the determinant yields the characteristic equation which helps us find the eigenvalues \( \lambda = 0, 1, 5 \). This process is critical as it provides the necessary \( \lambda \) values that help determine eigenvectors.

Determinant Computation

The computation of a determinant is crucial when dealing with matrices and characteristic equations. The determinant of a matrix is essentially a special number that can be calculated from its elements and is used to solve systems of linear equations and analyze linear transformations. Given a square matrix, there are different methods to compute its determinant depending on its size.

For a given 3x3 matrix:

• The determinant \( det \) can be calculated by choosing a row or column and applying the rule of Sarrus or cofactor expansion for larger matrices.
• In our example, the matrix was: \[\begin{vmatrix}0-\lambda & 1 & -1\0 & 2-\lambda & 0 \2 & -1 & 3-\lambda\end{vmatrix}\]and involved quite a bit of computation using cofactor expansion to find the polynomial in \( \lambda \).
• The key is finding the value(s) of \( \lambda \) for which this resultant polynomial equals zero.

Understanding and computing the determinant accurately is crucial, as it directly leads us to the eigenvalues of the matrix.

Linear Algebra Concepts

Linear algebra is an expansive field that includes several key concepts relevant to our task of finding eigenvalues and eigenvectors. At its core, linear algebra studies vectors, matrices, and linear transformations, which are fundamental in various applications across science and engineering.

Some critical concepts involved in our exercise include:

• Matrix: An arrangement of numbers into rows and columns capable of representing linear transformations.
• Eigenvalues and Eigenvectors: Eigenvalues \( \lambda \) are scalars that transform an eigenvector only by a factor of itself when a matrix is applied. The equation \( A\mathbf{v} = \lambda \mathbf{v} \) defines this relationship, where \( A \) is the matrix, and \( \mathbf{v} \) is the eigenvector.
• Characteristic Equation: Introduced to find \( \lambda \) values (i.e., eigenvalues) by setting the determinant of \( A - \lambda I \) to zero.
• Determinant: A value calculated from a matrix that provides insights into the matrix properties, such as invertibility, and facilitates eigenvalue computation.

By mastering these fundamental linear algebra concepts, students can solve a wide array of problems in mathematics and related fields.