Eigenvalues and eigenvectors are fundamental concepts in linear algebra, particularly useful in the context of 3D mathematics and computer graphics. They provide insights into the properties of linear transformations and are widely used in various applications, including 3D graphics, stability analysis, and more.

Key Concepts

  1. Eigenvalues and Eigenvectors

  • Eigenvalue (λ): A scalar that indicates how much the eigenvector is stretched or compressed during the linear transformation.
  • Eigenvector (v): A non-zero vector that only changes by a scalar factor when a linear transformation is applied.

  1. Eigenvalue Equation

The relationship between a matrix \(A\), an eigenvector \(v\), and its corresponding eigenvalue \(λ\) is given by: \[ A \mathbf{v} = \lambda \mathbf{v} \]

  1. Characteristic Equation

To find the eigenvalues of a matrix \(A\), we solve the characteristic equation: \[ \det(A - \lambda I) = 0 \] where \(I\) is the identity matrix of the same dimension as \(A\).

  1. Properties of Eigenvalues and Eigenvectors

  • Eigenvectors corresponding to distinct eigenvalues are linearly independent.
  • The sum of the eigenvalues of a matrix equals the trace of the matrix.
  • The product of the eigenvalues of a matrix equals the determinant of the matrix.

Practical Example

Let's consider a 2x2 matrix \(A\) and find its eigenvalues and eigenvectors.

Example Matrix

\[ A = \begin{pmatrix} 4 & 1 \ 2 & 3 \end{pmatrix} \]

Step-by-Step Solution

Step 1: Find the Characteristic Equation

\[ \det(A - \lambda I) = 0 \] \[ \det \begin{pmatrix} 4 - \lambda & 1 \ 2 & 3 - \lambda \end{pmatrix} = 0 \] \[ (4 - \lambda)(3 - \lambda) - (2 \cdot 1) = 0 \] \[ \lambda^2 - 7\lambda + 10 = 0 \]

Step 2: Solve for Eigenvalues

\[ \lambda^2 - 7\lambda + 10 = 0 \] \[ (\lambda - 5)(\lambda - 2) = 0 \] So, the eigenvalues are: \[ \lambda_1 = 5 \] \[ \lambda_2 = 2 \]

Step 3: Find Eigenvectors

For each eigenvalue, solve \( (A - \lambda I) \mathbf{v} = 0 \).

Eigenvector for \( \lambda_1 = 5 \)

\[ (A - 5I) \mathbf{v} = 0 \] \[ \begin{pmatrix} 4 - 5 & 1 \ 2 & 3 - 5 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \] \[ \begin{pmatrix} -1 & 1 \ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \] \[ -v_1 + v_2 = 0 \] \[ v_2 = v_1 \] So, an eigenvector corresponding to \( \lambda_1 = 5 \) is: \[ \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \]

Eigenvector for \( \lambda_2 = 2 \)

\[ (A - 2I) \mathbf{v} = 0 \] \[ \begin{pmatrix} 4 - 2 & 1 \ 2 & 3 - 2 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \] \[ \begin{pmatrix} 2 & 1 \ 2 & 1 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \] \[ 2v_1 + v_2 = 0 \] \[ v_2 = -2v_1 \] So, an eigenvector corresponding to \( \lambda_2 = 2 \) is: \[ \mathbf{v}_2 = \begin{pmatrix} 1 \ -2 \end{pmatrix} \]

Practical Exercises

Exercise 1

Find the eigenvalues and eigenvectors of the following matrix: \[ B = \begin{pmatrix} 3 & 2 \ 2 & 3 \end{pmatrix} \]

Solution 1

Step 1: Characteristic Equation

\[ \det(B - \lambda I) = 0 \] \[ \det \begin{pmatrix} 3 - \lambda & 2 \ 2 & 3 - \lambda \end{pmatrix} = 0 \] \[ (3 - \lambda)^2 - 4 = 0 \] \[ \lambda^2 - 6\lambda + 5 = 0 \] \[ (\lambda - 5)(\lambda - 1) = 0 \] Eigenvalues: \[ \lambda_1 = 5 \] \[ \lambda_2 = 1 \]

Step 2: Eigenvectors

For \( \lambda_1 = 5 \)

\[ (B - 5I) \mathbf{v} = 0 \] \[ \begin{pmatrix} 3 - 5 & 2 \ 2 & 3 - 5 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \] \[ \begin{pmatrix} -2 & 2 \ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \] \[ -2v_1 + 2v_2 = 0 \] \[ v_2 = v_1 \] Eigenvector: \[ \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \]

For \( \lambda_2 = 1 \)

\[ (B - I) \mathbf{v} = 0 \] \[ \begin{pmatrix} 3 - 1 & 2 \ 2 & 3 - 1 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \] \[ \begin{pmatrix} 2 & 2 \ 2 & 2 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \] \[ 2v_1 + 2v_2 = 0 \] \[ v_2 = -v_1 \] Eigenvector: \[ \mathbf{v}_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \]

Summary

In this section, we explored the concepts of eigenvalues and eigenvectors, learned how to derive the characteristic equation, and solved for eigenvalues and eigenvectors with practical examples. Understanding these concepts is crucial for manipulating and interpreting linear transformations in 3D space, which is foundational for advanced topics in 3D graphics and simulations.

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