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Eigenvalues and Eigenvectors

  • Understand the definition of eigenvalues and eigenvectors.
  • Learn how to compute eigenvalues and eigenvectors.
  • Explore the characteristic equation.
  • Interpret eigenvalues and eigenvectors in transformations.
  • Understand diagonalization and its applications.

Definition of Eigenvalues and Eigenvectors

In a linear transformation, an eigenvector is a nonzero vector that only changes by a scalar factor when the transformation is applied.

For a square matrix \( A \), an eigenvector \( \mathbf{v} \) and its corresponding eigenvalue \( \lambda \) satisfy:

\[ A\mathbf{v} = \lambda \mathbf{v} \]
  • \( A \) is an \( n \times n \) matrix.
  • \( \mathbf{v} \neq 0 \) is an eigenvector.
  • \( \lambda \) is a scalar eigenvalue.

Characteristic Equation

To find eigenvalues, solve the **characteristic equation**:

\[ \det(A - \lambda I) = 0 \]

where \( I \) is the identity matrix.

Example: Finding Eigenvalues

For \( A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \), solve:

\[ \det \left( \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = 0 \] \[ \det \begin{bmatrix} 4-\lambda & 2 \\ 1 & 3-\lambda \end{bmatrix} = 0 \] \[ (4-\lambda)(3-\lambda) - (2)(1) = 0 \] \[ \lambda^2 - 7\lambda + 10 = 0 \] \[ (\lambda - 5)(\lambda - 2) = 0 \] \[ \lambda = 5, 2 \]

Proof

From \( A\mathbf{v} = \lambda \mathbf{v} \), rewrite as:

\[ (A - \lambda I) \mathbf{v} = 0 \]

For a nontrivial solution, the determinant must be zero:

\[ \det(A - \lambda I) = 0 \]

This gives the characteristic equation.

Diagonalization

A matrix \( A \) is **diagonalizable** if it can be written as:

\[ A = PDP^{-1} \]

where:

  • \( P \) is a matrix of eigenvectors.
  • \( D \) is a diagonal matrix of eigenvalues.

Examples

Example 1: Compute eigenvalues and eigenvectors for:

\[ A = \begin{bmatrix} 6 & -2 \\ 2 & 3 \end{bmatrix} \]

Step 1: Find eigenvalues by solving \( \det(A - \lambda I) = 0 \).

\[ \det \begin{bmatrix} 6-\lambda & -2 \\ 2 & 3-\lambda \end{bmatrix} = 0 \] \[ (6-\lambda)(3-\lambda) - (-2)(2) = 0 \] \[ \lambda^2 - 9\lambda + 16 = 0 \] \[ (\lambda - 4)(\lambda - 5) = 0 \] \[ \lambda = 4, 5 \]

Exercises

  • Question 1: Find the eigenvalues of \( A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \).
  • Question 2: Compute the eigenvectors for \( A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} \).
  • Question 3: Determine if \( A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \) is diagonalizable.

  • Answer 1: \( \lambda = 3, 1 \).
  • Answer 2: Eigenvectors: \( \mathbf{v}_1 = (1,0) \), \( \mathbf{v}_2 = (0,1) \).
  • Answer 3: Yes, because it has two distinct eigenvalues.

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