Question

A matrix \(A\) and one of its eigenvalues are given. Find an eigenvector of A for the given eigenvalue. $$ \begin{array}{l} A=\left[\begin{array}{cc} -2 & 6 \\ -9 & 13 \end{array}\right] \\ \lambda=7 \end{array} $$

Short answer

The eigenvector corresponding to the eigenvalue 7 is \(\begin{bmatrix} 2 \\ 3 \end{bmatrix}\).

Step-by-step solution

Define Eigenvalue Equation

To find an eigenvector for the matrix \(A\) corresponding to the eigenvalue \(\lambda=7\), we need to solve 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.

Set Up the Equation

Subtract \(\lambda \times I = 7 \times I\) from \(A\):\[A - \lambda I = \begin{bmatrix} -2 & 6 \ -9 & 13 \end{bmatrix} - \begin{bmatrix} 7 & 0 \ 0 & 7 \end{bmatrix} = \begin{bmatrix} -2-7 & 6 \ -9 & 13-7 \end{bmatrix} = \begin{bmatrix} -9 & 6 \ -9 & 6 \end{bmatrix}\]

Solve the Homogeneous System

We need to find a non-zero vector \(\mathbf{v} = \begin{bmatrix} x \ y \end{bmatrix}\) such that \((A - \lambda I)\mathbf{v} = \mathbf{0}\):\[\begin{bmatrix} -9 & 6 \ -9 & 6 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\]Which results in the system:\[\begin{align*}-9x + 6y &= 0 \-9x + 6y &= 0\end{align*}\]Since these are the same equation, we only need one.

Solve for Eigenvector Components

The equation \(-9x + 6y = 0\) implies:\[6y = 9x \implies y = \frac{9}{6}x = \frac{3}{2}x\]Choose an arbitrary value for \(x\), say \(x = 2\), then \(y = 3\). So an eigenvector is \(\mathbf{v} = \begin{bmatrix} 2 \ 3 \end{bmatrix}\).

Verify the Eigenvector

Check that \(A\mathbf{v} = 7\mathbf{v}\):\[\begin{bmatrix} -2 & 6 \ -9 & 13 \end{bmatrix} \begin{bmatrix} 2 \ 3 \end{bmatrix} = \begin{bmatrix} -2(2) + 6(3) \ -9(2) + 13(3) \end{bmatrix} = \begin{bmatrix} 14 \ 21 \end{bmatrix} = 7 \begin{bmatrix} 2 \ 3 \end{bmatrix}\]This confirms \(\mathbf{v} = \begin{bmatrix} 2 \ 3 \end{bmatrix}\) is an eigenvector for the eigenvalue \(\lambda = 7\).

Key concepts

Eigenvalues

Eigenvalues are a fundamental concept in matrix algebra often appearing in linear transformations. An eigenvalue, denoted as \(\lambda\), signifies by how much the transformation scales the original vector.

Every square matrix has eigenvalues, which can be real or complex numbers. The process involves solving the characteristic polynomial, obtained from the equation \(\det(A - \lambda I) = 0\), which helps identify eigenvalues as the roots. In our case, the matrix \(A\) provided has \(\lambda = 7\) as one of its eigenvalues.

Identifying these values is crucial in various applications like stability analysis, quantum mechanics, and finding modes in vibration problems. Eigenvalues help determine whether a system expands, compresses, or rotates in space.

Matrix Algebra

Matrix algebra is the language of linear equations and transformations, providing a structured way to organize and manipulate data.

Matrices are arrays of numbers that can represent systems of linear equations. Operations such as addition, subtraction, and multiplication allow for efficient calculations. One of the critical aspects of matrix algebra is understanding whether matrices can transform vectors in particular ways, such as using eigenvalues.

In our exercise, matrix subtraction enabled us to form the matrix \(A - \lambda I\), simplifying to \(\begin{bmatrix} -9 & 6 \ -9 & 6 \end{bmatrix}\). Simplifying matrices like this is vital as it transforms complex linear transformations into a system that can easily be manipulated or solved.

Linear Equations

Linear equations form the core of many scientific and engineering computations. A linear equation defines a straight-line relationship between variables and can be represented in vector form.

In the eigenvector context, finding such an equation means discovering a vector that satisfies the transformation: \((A - \lambda I)\mathbf{v} = \mathbf{0}\). It reduces a two-dimensional problem into a form where both rows of the matrix, \(\begin{bmatrix} -9 & 6 \ -9 & 6 \end{bmatrix}\), reduce to a single equation.

The heart of solving these equations is to find the relationships between variables. Here, we found that \(y = \frac{3}{2}x\), leading to a dependency that helps in calculating the eigenvector when driven by constants.

Homogeneous Systems

Homogeneous systems are special linear systems where all equations sum to zero, often forming the basis for eigenvector calculations.

These systems only have the trivial solution (zero vector) or multiple solutions defined by spaces formed by linear combinations of vectors. Homogeneous equations are represented by \((A - \lambda I)\mathbf{v} = \mathbf{0}\).

In our example, we aimed to solve \(-9x + 6y = 0\). The simplicity of such systems is that they can be handled by reducing variables into expressions of one another, as shown with \(y = \frac{3}{2}x\). Homogeneous systems dictate that the non-trivial solutions like \(\mathbf{v} = \begin{bmatrix} 2 \ 3 \end{bmatrix}\) help represent the real-world phenomenon they model.