Eigenvalues and eigenvectors are fundamental concepts in electronics engineering, particularly in analyzing systems such as signal processing, control systems, and circuit stability. Understanding these concepts allows engineers to simplify complex linear transformations and solve differential equations that model electronic circuits. In Kenya’s growing electronics industry, mastery of eigenvalues and eigenvectors supports advanced tasks like filter design, system stability analysis, and modal analysis of electrical networks.

7.1 Definition of Eigen Values and Eigen Vectors

7.1.1 Concept of Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors describe special vectors that remain in the same direction after a linear transformation, only scaled by a certain factor called the eigenvalue. For a square matrix \(A\), an eigenvector \(\mathbf{v}\) and its corresponding eigenvalue \(\lambda\) satisfy the equation:

$$ A \mathbf{v} = \lambda \mathbf{v} $$

where \(\mathbf{v} \neq \mathbf{0}\). This means applying the transformation \(A\) to \(\mathbf{v}\) stretches or compresses it by \(\lambda\) but does not change its direction. Eigenvalues can be real or complex numbers, and eigenvectors are non-zero vectors associated with these values.

Worked Examples

Example 1: Given matrix \(A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}\), find eigenvalues and eigenvectors.

Given:
\(A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}\)

Eigenvalues satisfy \(\det(A - \lambda I) = 0\):

$$ \det \begin{bmatrix} 2 - \lambda & 0 \\ 0 & 3 - \lambda \end{bmatrix} = (2 - \lambda)(3 - \lambda) = 0 $$

$$ \lambda_1 = 2, \quad \lambda_2 = 3 $$

For \(\lambda_1 = 2\):

$$ (A - 2I)\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

The first row gives no restriction, the second row gives \(v_2 = 0\). So eigenvector is:

$$ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$

For \(\lambda_2 = 3\):

$$ (A - 3I)\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

Second row no restriction, first row gives \(-v_1 = 0\), so \(v_1 = 0\). Eigenvector:

$$ \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$

Answer: Eigenvalues are 2 and 3 with eigenvectors \(\begin{bmatrix}1 \\ 0 \end{bmatrix}\) and \(\begin{bmatrix}0 \\ 1 \end{bmatrix}\).

Example 2: Find eigenvalues of \(A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\).

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

Calculate characteristic polynomial:

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

$$ = (4 - \lambda)(3 - \lambda) - 2 = (12 - 4\lambda - 3\lambda + \lambda^2) - 2 = \lambda^2 - 7\lambda + 10 = 0 $$

Solve quadratic:

$$ \lambda = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm 3}{2} $$

$$ \lambda_1 = \frac{7 + 3}{2} = 5, \quad \lambda_2 = \frac{7 - 3}{2} = 2 $$

Answer: Eigenvalues are 5 and 2.

Example 3: Find eigenvectors for eigenvalue \(\lambda = 5\) of matrix \(A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\).

Given:
\(A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\), \(\lambda = 5\)

Solve:

$$ (A - 5I)\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

From first row:

$$ -1 \times v_1 + 1 \times v_2 = 0 \implies v_2 = v_1 $$

Second row is a multiple of the first, so no new info.

Eigenvector:

$$ \mathbf{v} = k \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad k \neq 0 $$

Answer: Eigenvector is any scalar multiple of \(\begin{bmatrix}1 \\ 1\end{bmatrix}\).

Example 4: Find eigenvalues of \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\).

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

Characteristic polynomial:

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

Solve:

$$ \lambda^2 = -1 \implies \lambda = \pm i $$

Answer: Eigenvalues are \(i\) and \(-i\) (complex conjugates).

Example 5: Find eigenvector corresponding to \(\lambda = i\) for matrix \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\).

Given:
\(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\), \(\lambda = i\)

Solve:

$$ (A - iI)\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} -i & -1 \\ 1 & -i \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

From first row:

$$ -i v_1 - v_2 = 0 \implies v_2 = -i v_1 $$

Second row:

$$ v_1 - i v_2 = 0 $$

Substitute \(v_2\):

$$ v_1 - i(-i v_1) = v_1 - i^2 v_1 = v_1 - (-1) v_1 = v_1 + v_1 = 2 v_1 = 0 $$

Implies \(v_1 = 0\), but zero vector not allowed. Instead, first equation suffices.

Eigenvector:

$$ \mathbf{v} = k \begin{bmatrix} 1 \\ -i \end{bmatrix}, \quad k \neq 0 $$

Answer: Eigenvector is any scalar multiple of \(\begin{bmatrix}1 \\ -i\end{bmatrix}\).

7.1.2 Characteristic Equation and Its Role in Finding Eigenvalues

The characteristic equation is the polynomial equation derived from the determinant of \((A - \lambda I)\) set equal to zero. It provides the eigenvalues of a matrix \(A\). Formally:

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

This is a polynomial equation in \(\lambda\) of degree \(n\) for an \(n \times n\) matrix. Solving this equation yields all eigenvalues, which may be real or complex. The characteristic equation is crucial in electronics engineering for analyzing system behavior such as resonance frequencies and stability.

Worked Examples

Example 1: Find the characteristic equation of matrix \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\).

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

Calculate:

$$ \det(A - \lambda I) = \det \begin{bmatrix} 1 - \lambda & 2 \\ 3 & 4 - \lambda \end{bmatrix} = (1 - \lambda)(4 - \lambda) - 6 $$

$$ = (1 \times 4) - (1 \times \lambda) - (4 \times \lambda) + \lambda^2 - 6 = 4 - 5\lambda + \lambda^2 - 6 = \lambda^2 - 5\lambda - 2 $$

Set equal to zero:

$$ \lambda^2 - 5\lambda - 2 = 0 $$

Answer: Characteristic equation is \(\lambda^2 - 5\lambda - 2 = 0\).

Example 2: Find eigenvalues from the characteristic equation \(\lambda^2 - 4\lambda + 3 = 0\).

Given: \(\lambda^2 - 4\lambda + 3 = 0\)

Solve quadratic:

$$ \lambda = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2} $$

$$ \lambda_1 = \frac{4 + 2}{2} = 3, \quad \lambda_2 = \frac{4 - 2}{2} = 1 $$

Answer: Eigenvalues are 3 and 1.

Example 3: Write the characteristic equation for matrix \(A = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{bmatrix}\).

Given:
\(A = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{bmatrix}\)

Calculate:

$$ \det(A - \lambda I) = \det \begin{bmatrix} 5 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 2 \\ 0 & 2 & 1 - \lambda \end{bmatrix} $$

Since first row and column are block diagonal:

$$ = (5 - \lambda) \times \det \begin{bmatrix} 1 - \lambda & 2 \\ 2 & 1 - \lambda \end{bmatrix} $$

Calculate the 2x2 determinant:

$$ (1 - \lambda)(1 - \lambda) - 4 = (1 - \lambda)^2 - 4 $$

Hence:

$$ \det(A - \lambda I) = (5 - \lambda) [(1 - \lambda)^2 - 4 ] = 0 $$

Expand \((1 - \lambda)^2 - 4\):

$$ (1 - 2\lambda + \lambda^2) - 4 = \lambda^2 - 2\lambda - 3 $$

Final characteristic equation:

$$ (5 - \lambda)(\lambda^2 - 2\lambda - 3) = 0 $$

Answer: Characteristic equation is \((5 - \lambda)(\lambda^2 - 2\lambda - 3) = 0\).

Example 4: Find eigenvalues from \((5 - \lambda)(\lambda^2 - 2\lambda - 3) = 0\).

Given:

$$ (5 - \lambda)(\lambda^2 - 2\lambda - 3) = 0 $$

Solve each factor:

$$ 5 - \lambda = 0 \implies \lambda = 5 $$

Solve quadratic:

$$ \lambda^2 - 2\lambda - 3 = 0 $$

$$ \lambda = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm 4}{2} $$

$$ \lambda_1 = \frac{2 + 4}{2} = 3, \quad \lambda_2 = \frac{2 - 4}{2} = -1 $$

Answer: Eigenvalues are 5, 3, and -1.

Example 5: Find the characteristic equation of \(A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -11 & -6 \end{bmatrix}\).

Given:

\(A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -11 & -6 \end{bmatrix}\)

Calculate:

$$ \det(A - \lambda I) = \det \begin{bmatrix} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ -6 & -11 & -6 - \lambda \end{bmatrix} $$

Expand determinant by first row:

$$ (-\lambda) \times \det \begin{bmatrix} -\lambda & 1 \\ -11 & -6 - \lambda \end{bmatrix} - 1 \times \det \begin{bmatrix} 0 & 1 \\ -6 & -6 - \lambda \end{bmatrix} + 0 $$

Calculate first minor:

$$ (-\lambda) [(-\lambda)(-6 - \lambda) - (-11)(1) ] = (-\lambda) [\lambda (6 + \lambda) + 11 ] $$

$$ = (-\lambda)(\lambda^2 + 6\lambda + 11) = -\lambda^3 - 6 \lambda^2 - 11 \lambda $$

Calculate second minor:

$$ -1 \times (0 \times (-6 - \lambda) - (-6) \times 1) = -1 \times (0 + 6) = -6 $$

Sum:

$$ -\lambda^3 - 6 \lambda^2 - 11 \lambda - 6 = 0 $$

Rearranged:

$$ \lambda^3 + 6 \lambda^2 + 11 \lambda + 6 = 0 $$

Answer: Characteristic equation is \(\lambda^3 + 6 \lambda^2 + 11 \lambda + 6 = 0\).

7.1.3 Procedure to Find Eigenvalues and Eigenvectors of a Matrix

Finding eigenvalues and eigenvectors is a stepwise process essential for analyzing electronic systems such as filter designs or state-space models of circuits. The procedure involves calculating the characteristic equation, solving it for eigenvalues, and then determining eigenvectors for each eigenvalue.

$$ A \mathbf{v} = \lambda \mathbf{v} $$

Worked Examples

Example 1: Find eigenvalues and eigenvectors of \(A = \begin{bmatrix} 3 & 1 \\ 0 & 2 \end{bmatrix}\).

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

Step 1: Calculate characteristic equation:

$$ \det(A - \lambda I) = (3 - \lambda)(2 - \lambda) - 0 = 0 $$

$$ (3 - \lambda)(2 - \lambda) = 0 $$

Step 2: Solve for \(\lambda\):

$$ \lambda_1 = 3, \quad \lambda_2 = 2 $$

Step 3: Find eigenvector for \(\lambda_1 = 3\):

$$ (A - 3I)\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} 0 & 1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

From first row:

$$ 1 \times v_2 = 0 \implies v_2 = 0 $$

From second row:

$$ -1 \times v_2 = 0 \implies v_2 = 0 $$

No restriction on \(v_1\), so eigenvector:

$$ \mathbf{v}_1 = k \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$

Step 4: Find eigenvector for \(\lambda_2 = 2\):

$$ (A - 2I)\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

From first row:

$$ v_1 + v_2 = 0 \implies v_2 = -v_1 $$

Second row no restriction.

Eigenvector:

$$ \mathbf{v}_2 = k \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$

Answer: Eigenvalues are 3 and 2 with eigenvectors \(\begin{bmatrix}1 \\ 0\end{bmatrix}\) and \(\begin{bmatrix}1 \\ -1\end{bmatrix}\).

Example 2: Find eigenvalues and eigenvectors of \(A = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}\).

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

Step 1: Characteristic equation:

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

$$ \lambda(3 + \lambda) + 2 = \lambda^2 + 3\lambda + 2 = 0 $$

Step 2: Solve quadratic:

$$ \lambda = \frac{-3 \pm \sqrt{9 - 8}}{2} = \frac{-3 \pm 1}{2} $$

$$ \lambda_1 = -1, \quad \lambda_2 = -2 $$

Step 3: Find eigenvector for \(\lambda_1 = -1\):

$$ (A + I) \mathbf{v} = \mathbf{0} \implies \begin{bmatrix} 1 & 1 \\ -2 & -2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

From first row:

$$ v_1 + v_2 = 0 \implies v_2 = -v_1 $$

Second row is a multiple of first.

Eigenvector:

$$ \mathbf{v}_1 = k \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$

Step 4: Find eigenvector for \(\lambda_2 = -2\):

$$ (A + 2I)\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} 2 & 1 \\ -2 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

From first row:

$$ 2 v_1 + v_2 = 0 \implies v_2 = -2 v_1 $$

Second row is multiple of first.

Eigenvector:

$$ \mathbf{v}_2 = k \begin{bmatrix} 1 \\ -2 \end{bmatrix} $$

Answer: Eigenvalues \(-1\), \(-2\) with eigenvectors \(\begin{bmatrix}1 \\ -1\end{bmatrix}\) and \(\begin{bmatrix}1 \\ -2\end{bmatrix}\).

Example 3: Find eigenvalues and eigenvectors of \(A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}\).

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

Step 1: Find characteristic equation:

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

Since matrix is upper triangular:

$$ = (2 - \lambda)(2 - \lambda)(3 - \lambda) = (2 - \lambda)^2 (3 - \lambda) $$

Set to zero:

$$ (2 - \lambda)^2 (3 - \lambda) = 0 $$

Eigenvalues:

$$ \lambda = 2 \text{ (multiplicity 2)}, \quad \lambda = 3 $$

Step 2: Find eigenvectors for \(\lambda = 2\):

$$ (A - 2I) \mathbf{v} = \mathbf{0} \implies \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \mathbf{0} $$

From first row:

$$ 1 \times v_2 = 0 \implies v_2 = 0 $$

From third row:

$$ 1 \times v_3 = 0 \implies v_3 = 0 $$

No restriction on \(v_1\):

$$ \mathbf{v} = k \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$

Step 3: Find eigenvector for \(\lambda = 3\):

$$ (A - 3I) \mathbf{v} = \mathbf{0} \implies \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \mathbf{0} $$

Second row:

$$ -1 \times v_2 = 0 \implies v_2 = 0 $$

First row:

$$ -1 \times v_1 + 1 \times v_2 = 0 \implies -v_1 = 0 \implies v_1 = 0 $$

No restriction on \(v_3\):

$$ \mathbf{v} = k \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$

Answer: Eigenvalues 2 (with eigenvector \(\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\)) and 3 (with eigenvector \(\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\)).

Example 4: Find eigenvalues and eigenvectors of \(A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\).

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

Step 1: Characteristic equation:

$$ \det(A - \lambda I) = (1 - \lambda)^2 - 4 = 0 $$

$$ (1 - \lambda)^2 = 4 $$

$$ 1 - \lambda = \pm 2 $$

Eigenvalues:

$$ \lambda_1 = 1 - 2 = -1, \quad \lambda_2 = 1 + 2 = 3 $$

Step 2: Eigenvector for \(\lambda = -1\):

$$ (A + I) \mathbf{v} = \mathbf{0} \implies \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

First row:

$$ 2 v_1 + 2 v_2 = 0 \implies v_1 = -v_2 $$

Eigenvector:

$$ \mathbf{v} = k \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$

Step 3: Eigenvector for \(\lambda = 3\):

$$ (A - 3I) \mathbf{v} = \mathbf{0} \implies \begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

First row:

$$ -2 v_1 + 2 v_2 = 0 \implies v_2 = v_1 $$

Eigenvector:

$$ \mathbf{v} = k \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Answer: Eigenvalues \(-1\) and \(3\) with eigenvectors \(\begin{bmatrix}1 \\ -1\end{bmatrix}\) and \(\begin{bmatrix}1 \\ 1\end{bmatrix}\).

Example 5: Find eigenvalues and eigenvectors of \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\).

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

Step 1: Characteristic equation:

$$ \det(A - \lambda I) = \lambda^2 + 1 = 0 $$

Eigenvalues:

$$ \lambda = \pm i $$

Step 2: Eigenvector for \(\lambda = i\):

$$ (A - iI)\mathbf{v} = \mathbf{0} \implies \begin{bmatrix} -i & -1 \\ 1 & -i \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \mathbf{0} $$

From first row:

$$ - i v_1 - v_2 = 0 \implies v_2 = -i v_1 $$

Eigenvector:

$$ \mathbf{v} = k \begin{bmatrix} 1 \\ -i \end{bmatrix} $$

Answer: Eigenvalues are \(i\), \(-i\) with eigenvectors \(\begin{bmatrix}1 \\ -i\end{bmatrix}\) and \(\begin{bmatrix}1 \\ i\end{bmatrix}\).

Chapter Summary

This chapter introduced the concept of eigenvalues and eigenvectors, fundamental in understanding matrix transformations. It detailed the step-by-step process for calculating eigenvalues by solving the characteristic equation and then finding the corresponding eigenvectors. The chapter also explained the function of matrices, showing how matrix functions extend scalar functions to matrix arguments, which is essential in solving differential equations and system analyses. The Jordan form of a matrix was defined as a special canonical form that simplifies matrices into blocks, facilitating easier computations especially when matrices are not diagonalizable. Finally, the chapter covered the state transition matrix, a key tool in systems engineering that describes the evolution of state variables over time in linear dynamic systems. Together, these topics build a strong foundation for advanced matrix applications in engineering problems.