Math Problem Statement

Eigenvalue finding steps

Solution

Finding eigenvalues of a matrix involves determining the values of λ\lambda that satisfy the characteristic equation of the matrix. Here are the detailed steps to find eigenvalues:

Step 1: Define the matrix

Let AA be an n×nn \times n matrix. The goal is to find eigenvalues λ\lambda such that there exists a non-zero vector vv (eigenvector) that satisfies the equation:

Av=λvA v = \lambda v

Step 2: Set up the characteristic equation

Rearranging the equation Av=λvA v = \lambda v, we get:

AvλIv=0A v - \lambda I v = 0

where II is the identity matrix of the same dimension as AA. This can be rewritten as:

(AλI)v=0(A - \lambda I) v = 0

For non-zero eigenvectors vv, this equation only has non-trivial solutions if the matrix (AλI)(A - \lambda I) is singular. The matrix is singular if and only if its determinant is zero, so we get the characteristic equation:

det(AλI)=0\det(A - \lambda I) = 0

Step 3: Compute the determinant

Calculate the determinant of the matrix AλIA - \lambda I. This will result in a polynomial in terms of λ\lambda, known as the characteristic polynomial.

For example, if AA is a 2×22 \times 2 matrix:

A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

then the characteristic equation is:

det(aλbcdλ)=(aλ)(dλ)bc=0\det \begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix} = (a - \lambda)(d - \lambda) - bc = 0

This simplifies to a quadratic equation in λ\lambda:

λ2(a+d)λ+(adbc)=0\lambda^2 - (a+d)\lambda + (ad - bc) = 0

Step 4: Solve the characteristic polynomial

Solve the characteristic polynomial for λ\lambda. The roots of this polynomial are the eigenvalues of the matrix AA.

For the 2×22 \times 2 example above, solve the quadratic equation:

λ2(a+d)λ+(adbc)=0\lambda^2 - (a+d)\lambda + (ad - bc) = 0

using the quadratic formula:

λ=(a+d)±(a+d)24(adbc)2\lambda = \frac{-(a+d) \pm \sqrt{(a+d)^2 - 4(ad - bc)}}{2}

Step 5: Verify the results

After solving for the eigenvalues, verify by plugging them back into the matrix equation Av=λvA v = \lambda v to find the corresponding eigenvectors, or use numerical methods for large matrices.


Would you like a specific example to clarify this, or do you have any questions?

Relative Questions:

  1. How do you find the eigenvectors after determining the eigenvalues?
  2. What is the significance of eigenvalues in real-world applications?
  3. Can there be complex eigenvalues, and how are they interpreted?
  4. How does the process change for larger matrices (e.g., 3×33 \times 3)?
  5. What are the practical computational methods for finding eigenvalues in large matrices?

Tip:

Eigenvalues help to understand matrix behavior in linear transformations, such as scaling, rotating, and shearing of vectors.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Eigenvalues
Eigenvectors
Determinants
Characteristic Equation

Formulas

A v = λ v
(A - λ I) v = 0
det(A - λ I) = 0
Quadratic formula: λ = [-(a+d) ± √((a+d)^2 - 4(ad - bc))] / 2

Theorems

Eigenvalue Theorem
Characteristic Polynomial

Suitable Grade Level

Undergraduate