Eigenvalues & Eigenvectors

Short Notes

For a square matrix A, a scalar λ is called an eigenvalue if there exists a non-zero vector v such that Av = λv. The vector v is called the corresponding eigenvector.

Characteristic Equation: \(\det(A - \lambda I) = 0\)
Eigenspace: The set of all eigenvectors corresponding to an eigenvalue λ, along with the zero vector, forms a subspace.
Algebraic Multiplicity: The number of times λ appears as a root of the characteristic equation.
Geometric Multiplicity: The dimension of the eigenspace corresponding to λ.
Diagonalizability: A matrix is diagonalizable if and only if algebraic multiplicity = geometric multiplicity for all eigenvalues.

Key Theories & Formulas

1. Properties of Eigenvalues

Sum of eigenvalues = Trace(A) = \(\sum_{i} a_{ii}\)
Product of eigenvalues = det(A)
Eigenvalues of \(A^T\) are same as eigenvalues of \(A\)
Eigenvalues of \(A^{-1}\) are \(\frac{1}{\lambda}\) (if A is invertible)
Eigenvalues of \(A^k\) are \(\lambda^k\)
Eigenvalues of \(A + kI\) are \(\lambda + k\)
Eigenvalues of \(kA\) are \(k\lambda\)

2. Special Matrices

Matrix Type	Eigenvalue Properties
Symmetric (\(A = A^T\))	All eigenvalues are real
Skew-Symmetric (\(A = -A^T\))	All eigenvalues are purely imaginary or zero
Orthogonal (\(A^T A = I\))	All eigenvalues have magnitude 1
Idempotent (\(A^2 = A\))	Eigenvalues are 0 or 1
Nilpotent (\(A^k = 0\))	All eigenvalues are 0
Involutory (\(A^2 = I\))	Eigenvalues are 1 or -1

3. Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic equation.

If characteristic equation is \(\lambda^2 - 5\lambda + 6 = 0\), then \(A^2 - 5A + 6I = 0\)
Useful for computing \(A^{-1}\) and higher powers of \(A\)

4. Diagonalization

If \(A\) is diagonalizable: \(A = PDP^{-1}\)

\(P\) = matrix of eigenvectors (column-wise)
\(D\) = diagonal matrix of eigenvalues
\(A^n = PD^nP^{-1}\) (efficient for computing powers)

Example Problems

Problem 1: Find eigenvalues of \(A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\)

Characteristic equation: \(\det(A - \lambda I) = 0\)
\((4-\lambda)(3-\lambda) - 2 = 0\)
\(\lambda^2 - 7\lambda + 10 = 0\)
\((\lambda - 5)(\lambda - 2) = 0\)
Eigenvalues: λ = 5, 2

Verification:

Trace = 4 + 3 = 7 = 5 + 2 ✓
Det = 12 - 2 = 10 = 5 × 2 ✓

Problem 2: Find \(A^{100}\) if eigenvalues of \(A\) are 1 and -1.

Using diagonalization: Eigenvalues of \(A^{100}\) are \(1^{100} = 1\) and \((-1)^{100} = 1\).

Hardest GATE Questions

Topic: Cayley-Hamilton & Matrix Powers

Question (GATE 2017 Style): Let \(A\) be a 3×3 matrix with eigenvalues 1, 2, 3. Find the value of \(\det(A^2 - 4A + 3I)\).

Solution:

The eigenvalues of \(A^2 - 4A + 3I\) are obtained by substituting eigenvalues of \(A\) into \(f(\lambda) = \lambda^2 - 4\lambda + 3\)
For \(\lambda_1 = 1\): \(f(1) = 1 - 4 + 3 = 0\)
For \(\lambda_2 = 2\): \(f(2) = 4 - 8 + 3 = -1\)
For \(\lambda_3 = 3\): \(f(3) = 9 - 12 + 3 = 0\)
Eigenvalues of \((A^2 - 4A + 3I)\) are: 0, -1, 0
\(\det(A^2 - 4A + 3I) = 0 \times (-1) \times 0 = \mathbf{0}\)

Why it's hard: Requires understanding that polynomial functions of matrices preserve eigenvalue relationships.

Topic: Geometric vs Algebraic Multiplicity

Question (GATE 2015 Variant): For which value of \(k\) is the matrix \(A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & k \\ 0 & 0 & 2 \end{pmatrix}\) diagonalizable?