Re: Cayley-Hamilton theorem

mariano.suarezalvarez@xxxxxxxxx wrote:
Stephen Montgomery-Smith wrote:

mariano.suarezalvarez@xxxxxxxxx wrote:

Stephen Montgomery-Smith wrote:


Jim Rockford wrote:


I teach an advanced linear algebra class and often assign the homework
problem of proving the Cayley-Hamilton theorem. Just to refresh you,
this theorem states that every square matrix satisfies its own
characteristic equation:

det(A-lambda*I) = lambda^n + c(n-1)*lambda^(n-1) + ... + c1*lambda + c0
= 0 ------>

A^n + c(n-1)*A^(n-1) + ... + c1*A + c0*I = 0

where the c(i) are scalars.

You can prove this, for instance, simply by recognizing that every
matrix has a Jordan form, and then plugging in the Jordan decomposition
and doing some algebra.

However, some students ask me why they can't just begin with the
definition of the characteristic equation, det(A - lambda*I)=0, plug
in A for lambda, and then conclude det(0)=0. Voila, the proof is
finished.

Usually I dodge this issue and simply say that I don't want them to do
the proof "that way," but I'm wondering if I can say something more
authoritative than this! Is there actually any sort of logical flaw in
proving this theorem the easy way simply by substituting A for lambda
in the defining equation det(A-lambda*I) = 0 ???

This proof, as stated, is definitely wrong. Basically when you
substitute lambda=A, you are really putting matrices into the entries of
the matrix, in effect having a matrix of matrices. So what you really
have is

det(AxI-IxA) = 0

where x denotes tensor product, and the det only operates on the first
factor of the tensor product (and in general I suspect that a det that
operates on only one factor of a tensor product cannot be well defined,
but I don't really know).


Without a *lot* of work, you can only define the determinant of
a matrix with commuting coefficients. There are extensions to
the non-commutative case, like Dieudonné's determinant in the
local case (cf. any book on algebraic K-theory) (but this determinant
takes values not in the coefficient ring but in its abelianization...).

That is a seriously interesting answer. I think I'll read up about
this. But before I go to the library, could you give me a quick idea
what the abelianization is?


If G is a (not necessarily abelian) group, consider the subgroup
G' of G generated by all elements of the form

g h g^{-1} h^{-1}

for g and h in G.

It is easy to see that G' is a normal subgroup of G, called the
_derived_subgroup_, and that the quotient G/G' is abelian. One
usually calls G/G' the _abelianization_ of G, and writes it G_ab

It is an easy exercice that G_ab is the largest quotient of G
which is abelian. Precisely, this can be stated as follows: let
p:G->G_ab be the natural projection onto the quotient; then:

If f:G->H is a morphism of groups and H is abelian, there
exists exactly one morphism of groups g:G_ab -> H such that
one has the equality g p = f : G ->H.

Now let R be a not-necessarily commutative, local ring and
let GL(R,n) be the set of invertible n-by-n matrices with
entries in R. Consider the multiplicative group U of invertible
elements in R; this will in general be non-abelian. Then the
Dieudonné determinant is a group morphism

det : GL(R,n) -> U_ab

Its construction is explain in most introductions to algebraic
K-theory (Rosenberg's ``Algebraic K-theory and its applications''
and Srinivas' ``Algebraic K-theory'', for example).

Thank you very much for your detailed answer.

So presumably, if R=GL(C,m) (C=complex numbers) then U'=SL(C,m) and so U_ab=C and so the det function on GL(R,n) reduces to the usual det function on GL(C,mn), which means in effect that the det on the "first factor" really isn't feasible.

The reason your answer caught my interest is that recently I have considered a simpler problem, that is, how do you canonically define the geometric mean of a collection of elements from a non-commutative algebra (typically GL(C,m)). This is a special case of computing the determinant of diagonal elements from GL(R,n), and even that seems to be a very difficult problem which I could not satisfactorily solve, and indeed I was believing that there is no satisfactory solution, and to some extent you seem to be comfirming this.

And now I have some good places to read about this.

Thanks Stephen
.

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