Finding Characteristic Polynomials: An Example
Let us consider the following square matrix:
A = [ 2 1 ]
[ 3 4 ]
To find the characteristic polynomial, we need to calculate the determinant of the matrix A consisting of the identity matrix I multiplied by the scalar lambda (λ).
Step 1: Calculate A – λI
A – λI =
[ 2-λ 1 ]
[ 3 4-λ ]
Step 2: Calculate the determinant of A – λI
det(A - λI) = (2-λ)(4-λ) - 1 * 3
= λ^2 - 6λ + 5
Therefore, the characteristic polynomial of the matrix A is:
f(λ) = λ^2 - 6λ + 5
Analysis of characteristic polynomial:
The characteristic polynomial represents the equation whose roots are the eigenvalues of the matrix A.
In this case, the characteristic polynomial has two roots (eigenvalues) that can be found by solving the equation:
λ^2 – 6λ + 5 = 0
Using the quadratic formula, we find the eigenvalues:
λ1 = (6 + √(36 - 4*5))/2 = 5
λ2 = (6 - √(36 - 4*5))/2 = 1
The characteristic polynomial of matrix A is f(λ) = λ^2 - 6λ + 5.
The eigenvalues of matrix A are 5 and 1.
Comments
Post a Comment