Problem 23 - Differential Equations and Linear Algebra

Question

Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{rrr}6 & 3 & -4 \\\\-5 & -2 & 2 \\\0 & 0 & -1\end{array}\right]$$.

Step-by-Step Solution

Answer

The eigenvalues of the given matrix are \(\lambda_1 = 6, \lambda_2 = -2, \lambda_3 = -1\) and the corresponding eigenvectors are \(v_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}, v_3 = \begin{pmatrix} -2 \\ -5 \\ 7 \end{pmatrix}\).

Step 1: Find the determinant of the matrix subtracted by lambda times the identity matrix

First, we construct the matrix A - lambda * I: $$ \left[\begin{array}{ccc} 6-\lambda & 3 & -4 \\ -5 & -2-\lambda & 2 \\ 0 & 0 & -1-\lambda \end{array}\right] $$ Now, we find the determinant of the matrix: $$ \text{det}(A - \lambda I) = (6-\lambda) \left(((-2-\lambda)(-1-\lambda)\right) - 3(0) - 4(0)$$

Step 2: Solve the characteristic equation for lambda (eigenvalues)

Simplifying the determinant expression, we get the following characteristic equation: $$(6-\lambda)((-2-\lambda)(-1-\lambda))=0$$ By solving, we get three eigenvalues: \(\lambda_1 = 6, \lambda_2 = -2, \lambda_3 = -1\).

Step 3: For each eigenvalue, find the eigenvectors by solving the following system of equations: (A - lambda * I) * v = 0

We will find the eigenvectors for each eigenvalue separately: For \(\lambda_1 = 6\), we have the matrix: $$ \left[\begin{array}{ccc} 0 & 3 & -4 \\ -5 & -8 & 2 \\ 0 & 0 & -7 \end{array}\right] $$ The system of equations is: $$ \begin{cases} 0v_{1} + 3v_{2} - 4v_{3} = 0 \\ -5v_{1} - 8v_{2} + 2v_{3} = 0 \\ -7v_{3} = 0 \end{cases} $$ Simplifying and solving, we get the eigenvector for \(\lambda_1\): \(v_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\). For \(\lambda_2 = -2\), we have the matrix: $$ \left[\begin{array}{ccc} 8 & 3 & -4 \\ -5 & 0 & 2 \\ 0 & 0 & 1 \end{array}\right] $$ The system of equations is: $$ \begin{cases} 8v_{1} + 3v_{2} - 4v_{3} = 0 \\ -5v_{1} + 2v_{3} = 0 \\ v_{3} = 0 \end{cases} $$ Simplifying and solving, we get the eigenvector for \(\lambda_2\): \(v_2 = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}\). For \(\lambda_3 = -1\), we have the matrix: $$ \left[\begin{array}{ccc} 7 & 3 & -4 \\ -5 & -1 & 2 \\ 0 & 0 & 0 \end{array}\right] $$ The system of equations is: $$ \begin{cases} 7v_{1} + 3v_{2} - 4v_{3} = 0 \\ -5v_{1} - v_{2} + 2v_{3} = 0 \\ 0 = 0 \end{cases} $$ Simplifying and solving, we get the eigenvector for \(\lambda_3\): \(v_3 = \begin{pmatrix} -2 \\ -5 \\ 7 \end{pmatrix}\). The eigenvalues are \(\lambda_1 = 6, \lambda_2 = -2, \lambda_3 = -1\) and the corresponding eigenvectors are \(v_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}, v_3 = \begin{pmatrix} -2 \\ -5 \\ 7 \end{pmatrix}\).

Key Concepts

Characteristic Equation

In linear algebra, when we talk about finding the eigenvalues of a square matrix, we are often referring to solving the characteristic equation. The characteristic equation is derived from the determinant of the matrix subtracted by \( \lambda \) times the identity matrix, which is symbolically represented as \( \text{det}(A - \lambda I) \). Solving this equation helps us find the eigenvalues \( \lambda \) of the matrix. In simple terms, these eigenvalues are special numbers that give us insight into the transformation properties of the matrix.

• To formulate the characteristic equation, you create a new matrix by subtracting \( \lambda \) times the identity matrix from the given matrix.
• Next, calculate the determinant of this newly formed matrix.
• Set the determinant expression equal to zero, which gives you a polynomial equation — this is your characteristic equation.
• By solving this polynomial for \( \lambda \), you determine the eigenvalues of the matrix.

The roots of the characteristic polynomial (solutions for \( \lambda \)) are the eigenvalues. In our original problem, solving the characteristic equation resulted in three specific eigenvalues \( \lambda_1 = 6 \), \( \lambda_2 = -2 \), and \( \lambda_3 = -1 \). These eigenvalues are crucial as they form the basis for finding corresponding eigenvectors and give insights into the nature of linear transformations described by the matrix.

Determinant of a Matrix

The determinant is a special number that can be calculated from a square matrix. In the context of eigenvalues and eigenvectors, the determinant is an essential tool used to solve the characteristic equation. To find the determinant, we use a set computation based on the size of the matrix. For a 3x3 matrix like in our example, it involves cross multiplication and summation of the products of the matrix's diagonal elements, taking into account the signs.

• The formula to compute the determinant of a 3x3 matrix \( \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \end{pmatrix} \) is:
• \( \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \)
• This results in a scalar value that is crucial for determining whether the matrix has unique solutions or is invertible.

When applying the above concepts to the characteristic matrix \( A - \lambda I \), calculating the determinant involves expressing it as a polynomial involving \( \lambda \). Solving \( \text{det}(A - \lambda I) = 0 \) provides the characteristic polynomial equation which, in turn, helps us identify the eigenvalues.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, matrices, and systems of linear equations. It provides the foundational framework for dealing with many mathematical problems in multiple dimensions. Concepts like eigenvalues and eigenvectors are central to linear algebra as they connect to many applications in physics, computer science, and more.

• Linear algebra studies linear equations and matrices, which can be seen as tools for transforming vectors in a space.
• Eigenvalues and eigenvectors play a significant role because they indicate how these transformations scale and rotate within this space.
• Eigenvalues reveal how much, and eigenvectors show in which direction, a vector is stretched or compressed under a linear transformation.

Linear algebra's techniques are widely used for problem solving in various scientific fields, including engineering, statistics, and data science. This makes understanding components like eigenvalues and eigenvectors essential. By solving the characteristic equation and calculating eigenvectors, linear algebra provides a rich language for describing the geometric interpretation of matrix transformations.