Electrical engineers frequently encounter systems represented by matrices, especially when analyzing circuits, networks, and control systems. Understanding eigenvalues and eigenvectors is crucial for solving problems involving system stability, vibration modes, and signal processing. This chapter extends matrix concepts to eigenvalues and eigenvectors, enabling engineers to analyze linear transformations and system behaviours effectively.

5.1 Definition of Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra applied widely in electrical engineering, such as in analyzing system stability in power grids or modal analysis of circuits. They describe special scalars and vectors associated with a square matrix that reveal intrinsic properties of linear transformations.

5.1.1 Eigenvalues: Concept and Calculation

Eigenvalues are scalars \(\lambda\) such that for a given square matrix \(A\), there exists a non-zero vector \(x\) satisfying the equation

$$ A x = \lambda x $$

The eigenvalue \(\lambda\) represents a factor by which the eigenvector \(x\) is scaled during the linear transformation represented by \(A\). To find eigenvalues, solve the characteristic polynomial equation

$$ \det(A - \lambda I) = 0 $$

where \(I\) is the identity matrix of the same order as \(A\).

Worked Examples

Example 1: Find the eigenvalues of the matrix$$ A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} $$

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

Characteristic polynomial:$$ \det(A - \lambda I) = 0 $$$$ \det \begin{bmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{bmatrix} = 0 $$

Calculate determinant:$$ (4 - \lambda)(3 - \lambda) - 2 \times 1 = 0 $$$$ (4 - \lambda)(3 - \lambda) - 2 = 0 $$$$ (12 - 4\lambda - 3\lambda + \lambda^2) - 2 = 0 $$$$ \lambda^2 - 7\lambda + 10 = 0 $$

Solve quadratic:$$ \lambda^2 - 7\lambda + 10 = 0 $$$$ (\lambda - 5)(\lambda - 2) = 0 $$

Answer: \(\lambda_1 = 5, \lambda_2 = 2\)

Example 2: Find eigenvalues of$$ B = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & 4 \\ 0 & 0 & 1 \end{bmatrix} $$

Given:
\(B = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 3 & 4 \\ 0 & 0 & 1 \end{bmatrix}\)

Calculate \(\det(B - \lambda I) = 0\):$$ \det \begin{bmatrix} 2 - \lambda & 0 & 0 \\ 1 & 3 - \lambda & 4 \\ 0 & 0 & 1 - \lambda \end{bmatrix} = 0 $$

Since the matrix is block triangular, determinant is product of diagonal elements:$$ (2 - \lambda) \times \det \begin{bmatrix} 3 - \lambda & 4 \\ 0 & 1 - \lambda \end{bmatrix} = 0 $$$$ (2 - \lambda) \times (3 - \lambda)(1 - \lambda) = 0 $$

Eigenvalues satisfy:$$ (2 - \lambda)(3 - \lambda)(1 - \lambda) = 0 $$

Answer: \(\lambda_1 = 2, \lambda_2 = 3, \lambda_3 = 1\)

Example 3: Find eigenvalues of$$ C = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $$

Given:
\(C = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\)

Characteristic equation:$$ \det(C - \lambda I) = \det \begin{bmatrix} -\lambda & -1 \\ 1 & -\lambda \end{bmatrix} = 0 $$$$ (-\lambda)(-\lambda) - (-1)(1) = \lambda^2 + 1 = 0 $$

Solve:$$ \lambda^2 = -1 $$$$ \lambda = \pm i $$ (imaginary eigenvalues)

Answer: \(\lambda_1 = i, \lambda_2 = -i\)

5.1.2 Eigenvectors: Concept and Determination

An eigenvector \(x\) corresponding to an eigenvalue \(\lambda\) of matrix \(A\) satisfies

$$ (A - \lambda I) x = 0 $$

The eigenvector \(x\) is a non-zero vector that, when transformed by \(A\), results in a scalar multiple \(\lambda x\). Finding eigenvectors involves solving the homogeneous system above.

Worked Examples

Example 1: Find eigenvectors of$$ A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} $$ corresponding to \(\lambda = 5\).

Given:
\(A\), \(\lambda = 5\)

Calculate:$$ (A - 5I) x = 0 $$$$ \begin{bmatrix} 4 - 5 & 1 \\ 2 & 3 - 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

Write system:$$ -x_1 + x_2 = 0 $$$$ 2x_1 - 2x_2 = 0 $$

From first equation:$$ x_2 = x_1 $$

Choose \(x_1 = 1\), then$$ x = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Answer: Eigenvector corresponding to \(\lambda=5\) is any scalar multiple of \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\)

Example 2: Find eigenvectors of$$ A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} $$ corresponding to \(\lambda = 2\).

Calculate:$$ (A - 2I) x = 0 $$$$ \begin{bmatrix} 4 - 2 & 1 \\ 2 & 3 - 2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} $$

System:$$ 2x_1 + x_2 = 0 $$$$ 2x_1 + x_2 = 0 $$ (same equation twice)

From first:$$ x_2 = -2x_1 $$

Choose \(x_1 = 1\), then$$ x = \begin{bmatrix} 1 \\ -2 \end{bmatrix} $$

Answer: Eigenvector corresponding to \(\lambda=2\) is any scalar multiple of \(\begin{bmatrix} 1 \\ -2 \end{bmatrix}\)

Example 3: Find eigenvectors of$$ D = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{bmatrix} $$ corresponding to \(\lambda = 4\).

Calculate:$$ (D - 4I) x = 0 $$$$ \begin{bmatrix} 3 - 4 & 0 & 0 \\ 0 & 4 - 4 & 1 \\ 0 & 0 & 4 - 4 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} $$

System:$$ -x_1 = 0 $$$$ 0 \cdot x_2 + 1 \cdot x_3 = 0 $$$$ 0 = 0 $$

From first:$$ x_1 = 0 $$

From second:$$ x_3 = 0 $$

\(x_2\) is free. Choose \(x_2 = 1\), then$$ x = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $$

Answer: Eigenvector corresponding to \(\lambda=4\) is any scalar multiple of \(\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\)

5.1.3 Characteristic Polynomial: Formation and Use

The characteristic polynomial of a square matrix \(A\) is a polynomial in \(\lambda\) defined by

$$ p(\lambda) = \det(A - \lambda I) $$

It is used to determine the eigenvalues of \(A\) by solving the equation \(p(\lambda) = 0\). This polynomial is of degree \(n\) for an \(n \times n\) matrix.

Worked Examples

Example 1: Form the characteristic polynomial of$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $$

Calculate:$$ p(\lambda) = \det \begin{bmatrix} 1 - \lambda & 2 \\ 3 & 4 - \lambda \end{bmatrix} $$$$ = (1 - \lambda)(4 - \lambda) - 3 \times 2 $$$$ = (1 - \lambda)(4 - \lambda) - 6 $$$$ = 4 - \lambda - 4\lambda + \lambda^2 - 6 $$$$ = \lambda^2 - 5\lambda - 2 $$

Answer: Characteristic polynomial is \(p(\lambda) = \lambda^2 - 5\lambda - 2\)

Example 2: Find characteristic polynomial of$$ B = \begin{bmatrix} 2 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & 2 \end{bmatrix} $$

Calculate:$$ p(\lambda) = \det(B - \lambda I) = \det \begin{bmatrix} 2 - \lambda & 0 & 1 \\ 0 & 3 - \lambda & 0 \\ 1 & 0 & 2 - \lambda \end{bmatrix} $$

Using cofactor expansion along second row:$$ (3 - \lambda) \times \det \begin{bmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{bmatrix} $$

Calculate minor determinant:$$ (2 - \lambda)(2 - \lambda) - 1 \times 1 = (2 - \lambda)^2 - 1 $$

Hence,$$ p(\lambda) = (3 - \lambda) \times ((2 - \lambda)^2 - 1) $$

Answer: \(p(\lambda) = (3 - \lambda)((2 - \lambda)^2 - 1)\)

Example 3: Find characteristic polynomial of$$ C = \begin{bmatrix} 0 & 1 \\ -2 & 3 \end{bmatrix} $$

Calculate:$$ p(\lambda) = \det \begin{bmatrix} -\lambda & 1 \\ -2 & 3 - \lambda \end{bmatrix} $$$$ = (-\lambda)(3 - \lambda) - (-2)(1) $$$$ = -3\lambda + \lambda^2 + 2 $$$$ = \lambda^2 - 3\lambda + 2 $$

Answer: Characteristic polynomial is \(p(\lambda) = \lambda^2 - 3\lambda + 2\)

5.1.4 Distinct Eigenvalues and Normalized Eigenvectors

Distinct eigenvalues are eigenvalues with different numerical values. For each distinct eigenvalue, there is at least one eigenvector. Normalizing an eigenvector means scaling it to have unit length, which is often required in electrical engineering for standardization, such as in modal analysis.

The normalized eigenvector \(u\) of vector \(x\) is

$$ u = \frac{x}{\|x\|} $$

where \(\|x\|\) is the Euclidean norm (length) of \(x\).

Worked Examples

Example 1: Normalize eigenvector$$ x = \begin{bmatrix} 3 \\ 4 \end{bmatrix} $$

Calculate norm:$$ \|x\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$

Normalize:$$ u = \frac{1}{5} \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} \frac{3}{5} \\ \frac{4}{5} \end{bmatrix} $$

Answer: Normalized eigenvector is \(\begin{bmatrix} 0.6 \\ 0.8 \end{bmatrix}\)

Example 2: Given eigenvalues \(\lambda_1 = 2\), \(\lambda_2 = 3\) of$$ A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} $$ find normalized eigenvectors.

For \(\lambda_1 = 2\):

$$ (A - 2I) x = 0 $$$$ \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0 $$

From first row:$$ 0 \times x_1 + 1 \times x_2 = 0 \Rightarrow x_2 = 0 $$

Choose \(x_1 = 1\), vector:$$ x = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$

Normalize:$$ \|x\| = \sqrt{1^2 + 0^2} = 1 $$ Normalized vector:$$ u = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$

For \(\lambda_2 = 3\):

$$ (A - 3I) x = 0 $$$$ \begin{bmatrix} -1 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0 $$

From first row:$$ -x_1 + x_2 = 0 \Rightarrow x_2 = x_1 $$

Choose \(x_1 = 1\), vector:$$ x = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Normalize:$$ \|x\| = \sqrt{1^2 + 1^2} = \sqrt{2} $$$$ u = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix} $$

Answer:
Normalized eigenvectors:
For \(\lambda_1 = 2\), \(u_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\)
For \(\lambda_2 = 3\), \(u_2 = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix}\)

Example 3: Normalize eigenvector$$ x = \begin{bmatrix} -2 \\ 2 \\ 1 \end{bmatrix} $$

Calculate norm:$$ \|x\| = \sqrt{(-2)^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 $$

Normalize:$$ u = \frac{1}{3} \begin{bmatrix} -2 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{bmatrix} $$

Answer: Normalized eigenvector is \(\begin{bmatrix} -0.6667 \\ 0.6667 \\ 0.3333 \end{bmatrix}\)

Chapter Summary

This chapter introduced the concepts of eigenvalues and eigenvectors, explaining how they are derived from characteristic polynomials and emphasizing the importance of distinct eigenvalues and normalized eigenvectors. It detailed the step-by-step calculation methods for finding eigenvalues and their corresponding eigenvectors. The chapter then explored matrix functions, including similarity transformations and their properties, as well as the exponential and logarithmic functions of matrices and their interpretations. The definition and significance of the Jordan form of a matrix were presented, highlighting its role in simplifying matrix structures. Finally, the chapter covered the state transition matrix, focusing on its properties in continuous-time linear time-varying systems, which is essential for analyzing system dynamics. Overall, the chapter provided a comprehensive understanding of advanced matrix concepts and their applications in engineering mathematics.