Re: Confused by linear algebra (invariant factors and characteristic poly)

In article <1170012339.079417.231040@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<stonemcstone@xxxxxxxxx> wrote:

Hi,

My book *defines* the characteristic polynomial of the nxn matrix A
over field k as the product of the invariant factors of k^n, viewed as
a k[x]-module via the rule xv = Av.

Then later, when convenient, it pretends that it defined the character
polynomial in the usual way, as the determinant of (A-xI). I guess
these definitions are equivalent, otherwise why would they use the
first one at all-- but I have no idea why this is so. It's not at ALL
obvious (even though what little I could find on google basically said
"it's clear the characteristic polynomial is the product of the
invariant factors"... it's not clear to me at all!)

How does one PROVE that the characteristic polynomial (using the
normal definition, as det(A-x)) equals the product of the invariant
factors?

[[Of course, everyone is implicitly meaning "equality up to units"
since invariant factors are only determined up to units, and anyway
the two equally canonical definitions, det(A-xI) vs. det(-A+xI),
likewise]]

I have the feeling there is something obvious I'm overlooking, and for
that reason it's really frustrating! =( Thanks for helping.


In a nutshell:

It is a non-trivial Theorem that A - xI can be transformed, using
elementary row operations, into a matrix D = diag(f_1(x), ..., f_t(x)).
The non-constant f_i(x) are the invariant factors of A.

Clearly det(A - xI) = det(D) which is the result you want.

--
Paul Sperry
Columbia, SC (USA)
.

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