Re: Cayley-Hamilton theorem

Stephen Montgomery-Smith wrote:
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.

Note that R is to be a *ring*. You then need R = M(C, m), the
algebra of all m-by-m matrices. Then you are correct that the
multiplicative group U of invertible elements in R is GL(C, m),
and that its derived group is U' = SL(C,m) so U_ab = C\{0}
(note that I removed the zero: this is *multiplicative* group
of nonzero complex numbers)

Now... The construction of Dieudonné does *not* work for the
ring R, because R is not local
<http://en.wikipedia.org/wiki/Local_ring>:
it is easy to see that is non-invertible elements do not form
an ideal, for example.

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.

The classical definition of determinats depends very heavily
on the fact that fields are commutative. It does extend to
arbitrary commutative rings very easily, but not to
non-commutative rings like M(R,m).

Part of K-theory is the study of the K_1-functor, which is an
assigment to each ring R (commutative or not) of an abelian
group K_1(R) which, among other marvelous properties,
comes with a map

GL(R,m) -> K_1(R)

which can be thought of as a determinant. For example, if
your ring R is a field F, then one can show that K_1(F) is
precisely the multiplicative group F\{0} of non-zero elements
in F, and that the above map, which in this case is

GL(F,m) -> F\{0} = K_1(F),

is precisely the determinant. This, in fact, is true even
when R is more generally a local ring---in proving this is
that one has to build the Dieudonné determinant.

Thus, in a way, given a ring R, one can think of K_1(R)
as "the set of values of determinants of matrices with
coefficients in R.

-- m

PS: One may wonder what the 1 means in K_1(R). Well,
there is a whole sequence of groups K_n(R) for each
integer n. Just as K_1(R) can be seen as the "values of
determinant function on matrices", K_0(R) can be seen
as (slightly massaged version of) the set of "dimensions"
of R-modules. For example, if F is a field,

K_0(F) = ZZ, the group on integers,

and this more or less corresponds to the fact that one can
attach a "dimension" to each finite dimensional vector
space over F.

The meaning of the K_n(R) for n not 0 nor 1 is quite more
involved, and their study is the so called "higher K-theory".
Note that this is quite deep stuff: Dan Quillen got awarded
a Fields medal (in 78?) because he showed that the K_n(R)
can be constructed, and Vladimir Voevodsky got his
medal (in part) because he computed the K_n groups for
a field.

.

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