Question

A matrix \(A\) and one of its eigenvalues are given. Find an eigenvector of A for the given eigenvalue. $$ \begin{array}{l} A=\left[\begin{array}{ccc} 4 & 5 & -3 \\ -7 & -8 & 3 \\ 1 & -5 & 8 \end{array}\right] \\ \lambda=2 \end{array} $$

Short answer

An eigenvector is \([-2, 1, 0]^T\).

Step-by-step solution

Introduction to eigenvalues and eigenvectors

An eigenvector of a matrix \( A \) corresponding to eigenvalue \( \lambda \) satisfies the equation \( A\mathbf{v} = \lambda\mathbf{v} \). This can be rearranged to \( (A - \lambda I)\mathbf{v} = \mathbf{0} \), where \( I \) is the identity matrix. Our task is to find the vector \( \mathbf{v} \) that satisfies this equation.

Set up the equation \((A - \lambda I)\mathbf{v} = \mathbf{0}\)

First, calculate \( \lambda I \), where \( I \) is the 3x3 identity matrix and \( \lambda = 2 \). Then, subtract \( \lambda I \) from \( A \). Thus, \( A - 2I = \begin{bmatrix} 4 & 5 & -3 \ -7 & -8 & 3 \ 1 & -5 & 8 \end{bmatrix} - \begin{bmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 5 & -3 \ -7 & -10 & 3 \ 1 & -5 & 6 \end{bmatrix}. \) Hence, \((A - 2I)\mathbf{v} = \mathbf{0}\).

Solve the equation \((A - \lambda I)\mathbf{v} = \mathbf{0}\)

We solve \( \begin{bmatrix} 2 & 5 & -3 \ -7 & -10 & 3 \ 1 & -5 & 6 \end{bmatrix}\begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}. \) Using Gaussian elimination to row-reduce the matrix:1. Keep the first row as is: \([2, 5, -3] \).2. Replace the second row with the old second row plus 7 times the first row: \([-7, -10, 3] + 7\times [2, 5, -3] = [0, 25, -18] \).3. Replace the third row with the old third row minus the first row: \([1, -5, 6] - [2, 5, -3] = [-1, -10, 9] \).4. Replace the third row by subtracting 1/25 times the second row from it, i.e., \([-1,-10,9] - \cfrac{1}{25} \times [0,25,-18] = [-1,0,1.28] \).5. Convert this back to vector form, giving the solution as \(x = -2y - 1.28z, \; y \; \text{is free}, \; z \; \text{is free}.\)

Choose values to find an eigenvector

We can represent the solutions in terms of the free variables \(y\) and \(z\). If we set, for simplicity, \(z=0\) and choose \(y=1\), then \(x = -2(1) - 0 = -2\). Thus, one eigenvector is \( \begin{bmatrix} -2 \ 1 \ 0 \end{bmatrix} \). Alternatively, choosing \(y=0\) and \(z=1\), we find another eigenvector, whose calculations are not shown here for brevity.

Key concepts

Eigenvalues

Eigenvalues are an essential concept in matrix algebra, especially when analyzing linear transformations represented by matrices. They help us understand how a transformation behaves when applied to vectors.
An eigenvalue, denoted as \( \lambda \), is a scalar that, when a matrix \( A \) is applied to an eigenvector \( \mathbf{v} \), the output is simply a scaled version of the vector itself:

• This can be expressed mathematically as \( A\mathbf{v} = \lambda\mathbf{v} \).
• The equation can be rearranged to \( (A - \lambda I)\mathbf{v} = \mathbf{0} \), where \( I \) is the identity matrix.
• This form allows us to see that finding eigenvalues involves dealing with homogeneous equations.

Eigenvalues give insights into a matrix's properties, such as its stability and the patterns of its solutions. When \( \lambda \) is known, the next step often involves finding the corresponding eigenvectors.

Gaussian elimination

Gaussian elimination is a systematic method used in linear algebra to solve systems of linear equations. It simplifies matrices to a form where solutions can be easily read, and it's particularly useful in finding eigenvectors.
The process involves:

• Rearranging and scaling rows to convert the matrix to its row-echelon form.
• The matrix is progressively simplified by making the elements under the leading coefficient (the first non-zero number in a row) zero.
• This approach uses elementary row operations, one of the powerful tools in matrix manipulation.

In the context of eigenvectors, Gaussian elimination helps solve the equation \( (A - \lambda I)\mathbf{v} = \mathbf{0} \) for \( \mathbf{v} \). This is crucial for reducing the matrix efficiently, allowing us to determine the relationships between the components of the eigenvectors.

Matrix algebra

Matrix algebra is the backbone of many solutions in linear algebra, enabling a structured approach to problem-solving involving matrices.
With matrix algebra:

• We understand operations like addition, subtraction, and multiplication of matrices.
• The identity matrix \( I \) plays a pivotal role, especially in defining transformations like \( A - \lambda I \).
• It provides tools for performing row operations crucial for techniques such as Gaussian elimination.

Mastering matrix algebra leads to a more straightforward comprehension of concepts like eigenvalues and eigenvectors. It also equips us to deal with larger-scale problems systematically, making matrix manipulation a fluent process for solving systems of equations.