From: Robin Chapman Subject: Re: Cayley-Hamilton Theorem Date: Mon, 19 Jul 1999 07:43:37 GMT Newsgroups: sci.math Keywords: a proof In article <19990718212715.02163.00002004@ng-cm1.aol.com>, prestupnik@aol.com (Prestupnik) wrote: > How does one prove the Cayley Hamilton Theorem? > The book i was reading proved it for 2*2 matrices(A) by actually > expanding the characteristic equation (det(A-kI)=0) (i use k for > lambda because it is easier) and than plugging > in A. Afterwards the author possed the following proof of the theorem > in general and asked what is wrong with this proof. What is wrong > with this proof? > > " The characterisit equation of A is det(A-kI)=0. Now, substitute k=A > in this equation. We obtain det(A-AI)=0. But A-AI=0, and det 0=0. > Hence the matrix A satisfies its own characteristic equation." > from "The Mathematics of Matrices" by Philip Davis. > > If this is not a valid proof, of the theorem, than can anyone please > point me in the direction of a general proof of the Cayley-Hamilton. > Thank You. > For a square matrix X one has det(X) I = X adj(X) where adj(X) is the matrix of cofactors. Let X = A - tI. This gives c(t) I = (A - tI)adj(A - tI) where c(t) is the characteristic polynomial of A. The entries in adj(A - tI) are polynomials in t of degree <= n - 1 (where A is n by n). Thus adj(A-tI) = B_0 + tB_1 + ... + t^{n-1}B_{n-1} where the B_j are (scalar) matrices. Write c(t) = a_0 + a_1 t + ... + a_n t^n. Comparing coefficients on both sides of c(t) I = (A - tI)adj(A - tI) gives a_0 I = A B_0 a_1 I = A B_1 - B_0 a_2 I = A B_2 - B_1 .... a_{n-1} I = A B_{n-1} - B_{n-2} a_n I = -B_{n-1}. Thus a_0 I = A B_0 a_1 A = A^2 B_1 - A B_0 a_2 A^2 = A^3 B_2 - A^2 B_1 .... a_{n-1} A^{n-1} = A^n B_{n-1} - A^{n-1} B_{n-2} a_n A^n = - A^n B_{n-1}. Adding gives a_0 I + a_1 A + a_2 A^2 + ... + a_n A^n = 0 or c(A) = 0: Cayley-Hamilton. -- Robin Chapman http://www.maths.ex.ac.uk/~rjc/rjc.html "They did not have proper palms at home in Exeter." Peter Carey, _Oscar and Lucinda_ Sent via Deja.com http://www.deja.com/ Share what you know. Learn what you don't. ============================================================================== From: djb@koobera.math.uic.edu (D. J. Bernstein) Subject: Re: Cayley-Hamilton Theorem Date: 22 Jul 1999 05:51:42 GMT Newsgroups: sci.math Achava Nakhash, the Loving Snake wrote: > The Cayley-Hamilton theorem is not a triviality, Take any matrix C = (c_11 ... c_1n,c_21 ... c_2n,...,c_n1 ... c_nn). Expand the characteristic polynomial p(x) = det(xI - C) by minors along the first row: there are cofactors p_1(x), ..., p_n(x) such that p(x) = p_1(x) (x-c_11) + ... + p_n(x) ( -c_1n) 0 = p_1(x) ( -c_21) + ... + p_n(x) ( -c_2n) ... 0 = p_1(x) ( -c_n1) + ... + p_n(x) (x-c_nn) and in particular p(C) = p_1(C) (C-c_11 I) + ... + p_n(C) ( -c_1n I) 0 = p_1(C) ( -c_21 I) + ... + p_n(C) ( -c_2n I) ... 0 = p_1(C) ( -c_n1 I) + ... + p_n(C) (C-c_nn I) Write e_1, ..., e_n for the usual unit vectors; then p(C) e_1 = p_1(C) (C e_1-c_11 e_1) + ... + p_n(C) ( -c_1n e_1) 0 e_2 = p_1(C) ( -c_21 e_2) + ... + p_n(C) ( -c_2n e_2) ... 0 e_n = p_1(C) ( -c_n1 e_n) + ... + p_n(C) (C e_n-c_nn e_n) which conveniently add up to p(C) e_1 = 0. Similarly p(C) e_2 = 0, ..., p(C) e_n = 0. Thus p(C) = 0. This proof is widely known as ``the determinant trick.'' It is a standard feature of textbooks on commutative algebra. Somehow it seems to have escaped the attention of authors of books on linear algebra. ---Dan ============================================================================== From: Ken.Pledger@vuw.ac.nz (Ken Pledger) Subject: Re: Cayley-Hamilton Theorem Date: Tue, 27 Jul 1999 10:48:56 +1200 Newsgroups: sci.math In article <379363E9.E95B2163@hotmail.com>, "Achava Nakhash, the Loving Snake" wrote: > .... I don't know the spiffiest proof > of the Cayley-Hamilton theorem. Surely someone out there knows a one or > two liner and we would all love to see it.... This takes more than two lines; but is spiffier than some, if you're happy to use determinants and adjugates. It also has the advantage of not needing a field of scalars containing the eigenvalues. I learned it long ago from Sam Perlis, "Theory of Matrices," Addison-Wesley, 1952, p.136. Every square matrix B satisfies (adj B).B = (det B).I. Let B = xI - A, and write Q = adj(xI - A), p(x) = det(xI - A). Then Q.(xI - A) = p(x).I The matrices Q, xI - A, and p(x).I can each be expressed as a polynomial in x with matrix coefficients. The last equation shows that the remainder on right division of p(x).I by xI - A is 0. But it is p(A) by the remainder theorem. Therefore p(A) = 0. O.K.: it needs some stuff about polynomials with matrix coefficients, but anyway I've always found this proof more intuitively helpful than others I've seen. Ken Pledger.