Eigenvalues, Eigenvectors & Cayley-Hamilton Theorem
“Characteristic polynomials, eigenspaces, algebraic vs geometric multiplicity, and matrix powers.”
1.Eigenspaces, Characteristic Polynomials, and Invariants
§ 1.1.For an n x n real or complex matrix A:
Definition: A non-zero vector x in C^n is an eigenvector of A with associated eigenvalue lambda in C if:
A x = lambda x <=> (A - lambda I) x = 0
Since x != 0, the matrix (A - lambda * I) must be singular, meaning:
p(lambda) = Det(A - lambda * I) = 0 (Characteristic Equation)
Cayley-Hamilton Theorem: Every square matrix satisfies its own characteristic equation. That is, if p(lambda) = lambda^n + c_{n-1}lambda^{n-1} + ... + c_1lambda + c_0 = 0, then:
p(A) = A^n + c_{n-1}A^{n-1} + ... + c_1A + c_0*I = 0
Applications of Cayley-Hamilton:
1. Expressing high powers of A (e.g., A^100) as a polynomial in A of degree at most n - 1.
2. Computing matrix inverse: If c_0 = (-1)^n * Det(A) != 0, then multiplying p(A) = 0 by A^{-1} gives:
A^{-1} = (-1 / c_0) [ A^{n-1} + c_{n-1}A^{n-2} + ... + c_1*I ]
For a 2x2 matrix: lambda^2 - Trace(A)lambda + Det(A) = 0. Always check the two invariant constraints first: lambda_1 + lambda_2 = a_11 + a_22 and lambda_1 lambda_2 = a_11a_22 - a_12a_21. This often eliminates 3 out of 4 multiple-choice options immediately.
2. Worked Problem: Matrix Inverse and Powers via Cayley-Hamilton
When applying Cayley-Hamilton to find A^{-1}, remember that the constant term c_0 becomes c_0 * A^{-1}, not c_0 * I.
Problem [Original Practice Problem · Modeled on GATE Pattern]:
Let A = [ [ 1, 2 ], [ 3, 2 ] ].
(a) Find the eigenvalues of A.
(b) Express A^4 as a linear combination of A and I using the Cayley-Hamilton theorem.
Step 1: Compute Characteristic Equation:
Trace(A) = 1 + 2 = 3
Det(A) = (1)(2) - (2)(3) = 2 - 6 = -4
Char Eq: lambda^2 - Trace(A)*lambda + Det(A) = 0
lambda^2 - 3*lambda - 4 = 0
(lambda - 4)(lambda + 1) = 0 => lambda_1 = 4, lambda_2 = -1.
Step 2: Apply Cayley-Hamilton Theorem:
A^2 - 3A - 4I = 0 => A^2 = 3A + 4I
Step 3: Compute higher powers recursively:
A^3 = A * A^2 = A(3A + 4I) = 3A^2 + 4A
Substitute A^2 = 3A + 4I:
A^3 = 3(3A + 4I) + 4A = 9A + 12I + 4A = 13A + 12I
A^4 = A * A^3 = A(13A + 12I) = 13A^2 + 12A
Substitute A^2 = 3A + 4I:
A^4 = 13(3A + 4I) + 12A = 39A + 52I + 12A = 51A + 52I
Verification via Division Algorithm:
Dividing lambda^4 by lambda^2 - 3*lambda - 4:
lambda^4 = (lambda^2 - 3lambda - 4)(lambda^2 + 3lambda + 13) + (51*lambda + 52).
Since p(A) = 0, A^4 = 51A + 52I. Verified perfectly.
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