Question

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{rr}{5} & {-1} \\ {3} & {1}\end{array}\right) $$

Short answer

Based on the given solution, the matrix has two eigenvalues, λ = 2 and λ = 4. The eigenvector corresponding to eigenvalue λ = 2 is v = (1, 3), and the eigenvector corresponding to eigenvalue λ = 4 is v = (1, 1).

Step-by-step solution

Determine the characteristic equation

To find the characteristic equation, we need to compute the determinant of the matrix (A - λI): $$ \begin{vmatrix} 5- \lambda & -1 \\ 3 & 1 - \lambda \end{vmatrix} $$ Now calculate the determinant: $$ (5 - \lambda)(1 - \lambda) - (-1)(3) = \lambda^2 - 6\lambda + 8 $$

Find the eigenvalues

Solve the quadratic equation we found in Step 1 to determine the eigenvalues: $$ \lambda^2 - 6\lambda + 8 = 0 $$ Factoring the quadratic, we get: $$ (\lambda - 2)(\lambda - 4) = 0 $$ So, the eigenvalues are λ = 2 and λ = 4.

Find the eigenvectors

Let's find the eigenvectors corresponding to each eigenvalue. First, for λ = 2, plug this value into the equation (A - λI)v = 0: $$ \begin{pmatrix} 5 - 2 & -1 \\ 3 & 1 - 2 \end{pmatrix} v = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ Which is equivalent to the following system of linear equations: $$ \begin{cases} 3v_1 - v_2 = 0\\ 3v_1 - v_2 = 0 \end{cases} $$ This system has infinitely many solutions, but we look for a basic eigenvector. So we can set v₁ = 1 and obtain v₂ = 3. The eigenvector for λ = 2 is v = (1, 3). Now, for λ = 4, plug this value into the equation (A - λI)v = 0: $$ \begin{pmatrix} 5 - 4 & -1 \\ 3 & 1 - 4 \end{pmatrix} v = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $$ Which is equivalent to the following system of linear equations: $$ \begin{cases} v_1 - v_2 = 0\\ 3v_1 + 3v_2 = 0 \end{cases} $$ This system has infinitely many solutions, but we look for a basic eigenvector. So we can set v₁ = 1 and obtain v₂ = 1. The eigenvector for λ = 4 is v = (1, 1).

State the eigenvalues and eigenvectors

In conclusion, the eigenvalues and their corresponding eigenvectors of the given matrix are: Eigenvalue λ = 2, Eigenvector v = (1, 3) Eigenvalue λ = 4, Eigenvector v = (1, 1)

Key concepts

Characteristic Equation

The characteristic equation is essential for finding eigenvalues of a matrix. This equation is derived from the matrix equation \[ \text{det}(A - \lambda I) = 0 \] where \(A\) is the matrix in question, \(\lambda\) represents a scalar (the eigenvalue), and \(I\) is the identity matrix of the same dimension as \(A\).

• The determinant of \(A - \lambda I\) leads to a polynomial equation.
• The roots of this polynomial are the eigenvalues of the matrix.

By determining these values, we can dig deeper into the behavior of the matrix, as eigenvalues reveal important properties like stability and oscillatory behavior. In our worked example, the characteristic equation was obtained as \[ \lambda^2 - 6\lambda + 8 = 0. \] This tells us we need to find the roots of this equation to identify the eigenvalues.

Quadratic Equation

Quadratic equations are involved in the determination of eigenvalues when the characteristic equation results in a second-degree polynomial. The general form of a quadratic equation is \[ ax^2 + bx + c = 0. \]To find the roots of this equation, we have several methods, such as:

• Factoring, if possible.
• Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
• Completing the square.

In our example, the characteristic equation \[ \lambda^2 - 6\lambda + 8 = 0 \] was factored into \[ (\lambda - 2)(\lambda - 4) = 0, \] yielding the eigenvalues \(\lambda = 2\) and \(\lambda = 4.\)

System of Linear Equations

A system of linear equations emerges when we seek the eigenvectors associated with each eigenvalue of a matrix. This involves solving the equation \[ (A - \lambda I)v = 0 \] for the vector \(v\).

• For each eigenvalue, substitute \(\lambda\) into the matrix equation.
• Convert the resulting matrix into a system of linear equations.
• Solve this system to find the eigenvectors, often by choosing a parameter for free variables.

In the provided exercise, when solving for eigenvectors:- For \(\lambda = 2\), the equations simplify to \(3v_1 - v_2 = 0\).- For \(\lambda = 4\), to \(v_1 - v_2 = 0\).These simplifications demonstrate how such systems can produce multiple solutions, hinting at the direction of the vector space spanned by eigenvectors.

Matrix Algebra

Matrix algebra is a crucial tool in linear algebra. It allows us to perform various operations such as addition, subtraction, multiplication, and determination of determinants. In the context of eigenvalues and eigenvectors:

• The operation \(A - \lambda I\) is fundamental, as it is used to find eigenvalues and eigenvectors.
• Understanding matrix operations helps in manipulating and simplifying complex systems of equations.
• Matrix algebra also provides insights into the ways matrices can transform vector spaces.

In this exercise, matrix subtraction (forming \(A - \lambda I\)), followed by computation of the determinant and solving for eigenvectors, demonstrated the powerful application of matrix algebra in solving real-world mathematical problems. It gives insights not only into theoretical applications but also practical computations, showing the structure and properties of matrices.