Re: Cayley-Hamilton theorem

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...).

--m

.

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