Eigenvalue Decomposition

Eigenvalue decomposition expresses a matrix as a product of its eigenvectors and eigenvalues, enabling diagonalization and simplifying the analysis of matrix powers and system stability.

Let’s assume a square matrix \( A \).

The characteristic equation is:

where \( I \) is the identity matrix and \( \lambda \) are eigenvalues.

For eigenvalue \( \lambda = \lambda_1 \), the corresponding eigenvector \( x \) satisfies:

Let \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \). The eigenvalue decomposition of a square matrix \( A \) is given by:

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

Where:

• \( \Lambda \) is a diagonal matrix of eigenvalues
• \( P \) is a matrix whose columns are the eigenvectors corresponding to the eigenvalues

Step 1: Find the eigenvalues

We solve the characteristic equation \( \det(A - \lambda I) = 0 \).

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

Solving: \( \lambda = \frac{5 \pm \sqrt{33}}{2} \approx -0.372, \ 5.372 \)

Step 2: Find eigenvectors

For each eigenvalue \( \lambda \), solve \( (A - \lambda I)\vec{v} = 0 \) to get the corresponding eigenvector \( \vec{v} \).

For example, for \( \lambda_1 = 5.372 \), the corresponding eigenvector is approximately:

\[ \vec{v}_1 = \begin{bmatrix} -0.416 \\ -0.909 \end{bmatrix} \]

And for \( \lambda_2 = -0.372 \):

\[ \vec{v}_2 = \begin{bmatrix} -0.825 \\ 0.566 \end{bmatrix} \]

Step 3: Construct the decomposition

Using:

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

You can verify that \( A = P \Lambda P^{-1} \) using matrix multiplication.