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_
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==============================================================================
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.