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Cayley-Hamilton: an Operator Satisfies Its Own Equation

The most quotable theorem in the subject: substitute T into its own characteristic polynomial and you get zero. We prove it honestly (no fake 'set x = T in det(xI - T) = 0'), then harvest a free formula for T inverse.

Stating the theorem and a tempting wrong proof

The Cayley-Hamilton theorem says: every operator annihilates its own characteristic polynomial. If p_T(x) = det(xI - T), then p_T(T) = 0. In one line: an operator satisfies its own characteristic equation.

An honest proof via the adjugate

Work over the ring of matrices whose entries are polynomials in x. Let B(x) = adj(xI - A), the adjugate (classical adjoint) of xI - A. The fundamental adjugate identity says B(x) (xI - A) = det(xI - A) I = p(x) I. Now B(x) is a matrix whose entries are polynomials of degree at most n-1, so we can write B(x) = B_{n-1} x^{n-1} + ... + B_1 x + B_0 with constant matrices B_i.

  1. Expand both sides of B(x)(xI - A) = p(x) I as polynomials in x with matrix coefficients, where p(x) = x^n + c_{n-1} x^{n-1} + ... + c_0.
  2. Match coefficients of each power x^k on both sides. This gives n+1 matrix equations relating the B_i to A and the c_k — pure algebra, no substitution of a matrix for x.
  3. Multiply the k-th equation on the left by A^k, then add all of them. The B_i terms telescope and cancel completely.
  4. What survives on the left is exactly A^n + c_{n-1} A^{n-1} + ... + c_0 I = p(A); the right collapses to 0. Hence p(A) = 0.
A = [1, 2; 3, 4]
p(x) = det(xI - A) = x^2 - (trace)x + det
     = x^2 - 5x - 2

Check p(A) = A^2 - 5A - 2I:
  A^2  = [ 7, 10; 15, 22]
 -5A   = [-5,-10;-15,-20]
 -2I   = [-2,  0;  0, -2]
  ----------------------------
  sum  = [ 0,  0;  0,  0]   <- p(A) = 0, as promised.
Cayley-Hamilton checked numerically on a 2x2: p(A) = 0.

Payoff: the inverse from the characteristic polynomial

Because p_T annihilates T, the minimal polynomial m_T DIVIDES p_T — the minimal one is the least-degree annihilator, and any other annihilator (including p_T) is a polynomial multiple of it. That single divisibility fact organizes everything in guides 3-5.

Cayley-Hamilton also hands you a cheap inverse. Write p_T(x) = x^n + ... + c_1 x + c_0. From p_T(T) = 0, isolate the constant term: T (T^{n-1} + ... + c_1 I) = -c_0 I. If c_0 != 0 (equivalently 0 is not an eigenvalue, so T is invertible), then T^{-1} = (-1/c_0)(T^{n-1} + c_{n-1} T^{n-2} + ... + c_1 I) — an inverse expressed as a polynomial in T.