Find The Characteristic Polynomial And The Eigenvalues Of The Matrix

Find The Characteristic Polynomial And The Eigenvalues Of The Matrix

In linear algebra, matrices play a crucial role in representing linear transformations and solving systems of linear equations. One of the key concepts associated with matrices is the characteristic polynomial and its relationship to the eigenvalues of the matrix. In this article, we will explore what the characteristic polynomial is, how to find it, and how it helps us determine the eigenvalues of a matrix.

What is the Characteristic Polynomial?

The characteristic polynomial of a square matrix A, denoted as p(λ), is a polynomial in the variable λ (lambda) defined by the equation:

\[ p(\lambda) = \text{det}(A – \lambda I), \]

where det() denotes the determinant of a matrix, A is the given square matrix, λ is the scalar variable, and I is the identity matrix of the same size as A. The roots of the characteristic polynomial are the eigenvalues of the matrix A.

How to Find the Characteristic Polynomial and Eigenvalues:

To find the characteristic polynomial and eigenvalues of a matrix, follow these steps:

1. Subtract λI from A: Start by subtracting λ times the identity matrix I from the matrix A. The result is a new matrix, let’s call it B, defined as:

\[ B = A – \lambda I. \]

2. Calculate the Determinant: Calculate the determinant of the matrix B. This determinant will be a polynomial in λ, which is the characteristic polynomial p(λ).

3. Find the Eigenvalues: The roots of the characteristic polynomial p(λ) are the eigenvalues of the matrix A. These eigenvalues represent the scalar values λ for which the matrix A – λI is singular, meaning that it does not have an inverse.

Example:

Let’s consider a 2×2 matrix A:

\[ A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \]

To find the characteristic polynomial and eigenvalues of A:

1. Subtract λI from A:

\[ B = A – \lambda I = \begin{bmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{bmatrix}. \]

2. Calculate the determinant of B:

\[ \text{det}(B) = (2-\lambda)(2-\lambda) – 1 \cdot 1 = (2-\lambda)^2 – 1 = 4 – 4\lambda + \lambda^2 – 1 = \lambda^2 – 4\lambda + 3. \]

So, the characteristic polynomial is p(λ) = λ² – 4λ + 3.

3. Find the eigenvalues:

To find the roots of the characteristic polynomial, we set p(λ) = 0:

\[ λ^2 – 4λ + 3 = 0. \]

Factoring the polynomial, we get:

\[ (λ – 3)(λ – 1) = 0. \]

Therefore, the eigenvalues of matrix A are λ₁ = 3 and λ₂ = 1.

The characteristic polynomial and eigenvalues of a matrix are fundamental concepts in linear algebra. By finding the characteristic polynomial and its roots, we can determine the eigenvalues of a matrix, which provide important information about the behavior of linear transformations associated with the matrix. Understanding these concepts is essential for various applications in mathematics, physics, engineering, and computer science.