Cayley-Hamilton Theorem & Minimal Polynomials
“p(A) = 0, polynomial reduction of high powers, matrix inverse derivation, and the minimal polynomial monic divisor.”
1.The Cayley-Hamilton Theorem & Polynomial Algebra
§ 3.1.The Cayley-Hamilton Theorem states that if p(lambda) = det(A - lambda I) is the characteristic polynomial of square matrix A, then p(A) = 0 (the zero matrix).
Application to Matrix Powers:
Any polynomial f(A) of degree k >= n can be divided by p(A):
f(A) = q(A) p(A) + r(A) = q(A) 0 + r(A) = r(A).
Here r(A) has degree at most n - 1. Thus, ANY high matrix power can be expressed as a linear combination of {I, A, A^2, ..., A^{n-1}}.
The Minimal Polynomial m(lambda):
The minimal polynomial m(lambda) is the monic polynomial of lowest degree satisfying m(A) = 0.
Key Theorem: A matrix A is diagonalizable if and only if m(lambda) = (lambda - lambda_1)(lambda - lambda_2)...(lambda - lambda_k) has NO repeated roots.
2. Worked Practice Problem: Evaluating Matrix Polynomial via Cayley-Hamilton
Use the characteristic polynomial identity A^2 = tr(A)*A - det(A)*I to reduce high powers iteratively.
Problem: Let A = [ [1, 2], [3, 4] ]. (a) Use the Cayley-Hamilton theorem to find A^{-1}. (b) Express A^4 as a linear combination of A and I.
Step 1: Compute characteristic polynomial p(lambda):
Trace(A) = 1 + 4 = 5.
det(A) = (1)(4) - (2)(3) = 4 - 6 = -2.
p(lambda) = lambda^2 - Trace(A)lambda + det(A) = lambda^2 - 5lambda - 2.
Step 2: Apply Cayley-Hamilton theorem p(A) = 0:
A^2 - 5A - 2I = 0 => A^2 = 5A + 2I.
Step 3: Derive A^{-1} by multiplying identity by A^{-1}:
A - 5I - 2A^{-1} = 0
2A^{-1} = A - 5I => A^{-1} = (1 / 2) (A - 5I).
A - 5*I = [ [1 - 5, 2], [3, 4 - 5] ] = [ [-4, 2], [3, -1] ].
A^{-1} = [ [-2, 1], [1.5, -0.5] ].
Verification via formula: (1 / -2) * [ [4, -2], [-3, 1] ] = [ [-2, 1], [1.5, -0.5] ]. (Matches exactly).
Step 4: Compute A^4 via iterative reduction:
A^2 = 5A + 2I.
A^3 = A A^2 = A (5A + 2I) = 5A^2 + 2A = 5(5A + 2I) + 2A = 27A + 10I.
A^4 = A A^3 = A (27A + 10I) = 27A^2 + 10A = 27(5A + 2I) + 10A = 135A + 54I + 10*A.
A^4 = 145A + 54I.
Conclusion: A^{-1} = 0.5(A - 5I) and A^4 = 145A + 54I.
Primary Academic References & Exam Sources
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