Eigenvalue Decomposition

Eigenvalue decomposition is a fundamental concept in linear algebra that expresses a square matrix as a product of its eigenvectors and a diagonal matrix of its eigenvalues. This decomposition allows a matrix to be diagonalized and simplifies computations involving matrix powers, linear transformations, and analysis of system stability.

Definitions

Eigenvalue: For a square matrix \( A \), a scalar \( \lambda \) is called an eigenvalue if there exists a nonzero vector \( \vec{v} \) such that:

\[ A \vec{v} = \lambda \vec{v} \]

Intuitively, this means that multiplying \( A \) by \( \vec{v} \) only scales the vector by \(\lambda\) without changing its direction. Eigenvalues provide important information about the matrix, such as scaling factors in transformations, stability in dynamical systems, and the modes of linear transformations.

Eigenvector: An eigenvector corresponding to the eigenvalue \( \lambda \) is a nonzero vector \( \vec{v} \) that satisfies the above equation. Eigenvectors indicate the directions along which the linear transformation represented by \( A \) acts purely by scaling, without rotation. Eigenvectors are always nonzero because the zero vector has no direction and trivially satisfies the equation for any scalar.

Eigenvalues and eigenvectors can also be determined by solving the characteristic equation:

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

Here, \( I \) is the identity matrix of the same size as \( A \). Each solution \( \lambda \) of this equation is an eigenvalue of \( A \), and the corresponding eigenvectors are the nonzero solutions \( \vec{v} \) of \((A - \lambda I)\vec{v} = 0\).

Example

Consider the matrix:

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

Step 1: Compute Eigenvalues

Solve the characteristic equation:

\[ \det(A - \lambda I) = \begin{vmatrix} 1 - \lambda & 2 \\ 3 & 4 - \lambda \end{vmatrix} = (1 - \lambda)(4 - \lambda) - 6 = \lambda^2 - 5\lambda - 2 \]

The eigenvalues are:

\[ \lambda_1 \approx 5.372, \quad \lambda_2 \approx -0.372 \]

Step 2: Compute Eigenvectors

For each eigenvalue \( \lambda_i \), solve \((A - \lambda_i I)\vec{v}_i = 0\) to find the corresponding eigenvector (up to a nonzero scalar multiple).

Eigenvectors for this matrix are approximately:

\[ \vec{v}_1 \approx \begin{bmatrix} -0.416 \\ -0.909 \end{bmatrix}, \quad \vec{v}_2 \approx \begin{bmatrix} -0.825 \\ 0.566 \end{bmatrix} \]

Step 3: Construct Eigenvalue Decomposition

Form the matrix of eigenvectors \( P \) and the diagonal matrix of eigenvalues \( \Lambda \):

\[ P = [\vec{v}_1 \ \vec{v}_2] = \begin{bmatrix} -0.416 & -0.825 \\ -0.909 & 0.566 \end{bmatrix}, \quad \Lambda = \begin{bmatrix} 5.372 & 0 \\ 0 & -0.372 \end{bmatrix} \]

The eigenvalue decomposition of \( A \) is then:

\[ A = P \Lambda P^{-1} \]

This decomposition is valid because \( A \) has two linearly independent eigenvectors, making it diagonalizable.