Linear Algebra

392 Chapter Five. SimilarityWe refer to that result by saying that a matrix or map satisfies its characteristicpolynomial.1.10 Lemma Where f(x) is a polynomial, if f(T) is the zero matrix then f(x)is divisible by the minimal polynomial of T. That is, any polynomial that issatisfied by T is divisible by T’s minimal polynomial.Proof Let m(x) be minimal for T. The Division Theorem for Polynomials givesf(x) = q(x)m(x) + r(x) where the degree of r is strictly less than the degree ofm. Plugging in T shows that r(T) is the zero matrix, because T satisfies both fand m. That contradicts the minimality of m unless r is the zero polynomial.QEDCombining the prior two lemmas gives that the minimal polynomial dividesthe characteristic polynomial. Thus, any root of the minimal polynomial isalso a root of the characteristic polynomial. That is, so far we have that ifm(x) = (x−λ 1 ) q1 · · · (x−λ i ) q ithen c(x) has the form (x−λ 1 ) p1 · · · (x−λ i ) p i(x−λ i+1 ) p i+1· · · (x − λ z ) p zwhere each q j is less than or equal to p j . We finishthe proof of the Cayley-Hamilton Theorem by showing that the characteristicpolynomial has no additional roots, that is, there are no λ i+1 , λ i+2 , etc.1.11 Lemma Each linear factor of the characteristic polynomial of a square matrixis also a linear factor of the minimal polynomial.Proof Let T be a square matrix with minimal polynomial m(x) and assumethat x − λ is a factor of the characteristic polynomial of T, that λ is an eigenvalueof T. We must show that x − λ is a factor of m, i.e., that m(λ) = 0.Suppose that λ is an eigenvalue of T with associated eigenvector ⃗v. ThenT 2 has the eigenvalue λ 2 associated with ⃗v because T · T⃗v = T · λ⃗v = λT⃗v = λ 2 ⃗v.Similarly T k has the eigenvalue λ n associated with ⃗v.With that we have that for any polynomial function p(x), application of thematrix p(T) to ⃗v equals the result of multiplying ⃗v by the scalar p(λ).p(T) · ⃗v = (c k T k + · · · + c 1 T + c 0 I) · ⃗v = c k T k ⃗v + · · · + c 1 T⃗v + c 0 ⃗v= c k λ k ⃗v + · · · + c 1 λ⃗v + c 0 ⃗v = p(λ) · ⃗vSince m(T) is the zero matrix, ⃗0 = m(T)(⃗v) = m(λ) · ⃗v for all ⃗v, and hencem(λ) = 0.QEDThat concludes the proof of the Cayley-Hamilton Theorem.1.12 Example We can use the Cayley-Hamilton Theorem to find the minimalpolynomial of this matrix.⎛⎞2 0 0 11 2 0 2T = ⎜⎟⎝0 0 2 −1⎠0 0 0 1