Re: Jacobian determinant and inverse function theorem
- From: Jannick Asmus <jannick.news@xxxxxx>
- Date: Sun, 03 Sep 2006 22:01:43 +0200
On 03.09.2006 21:39, frank wrote:
Yes, this is a very clear explanation indeed.
On the question why det is continuous I would answer that by definition
det is a sum of products of a sign function and matrix entries (which
in the case at hand are known to be continuous). Is this correct?
Yes, it is correct - and very clear, too.
This all leaves me wondering why my textbook (SA Douglass) glossed
completely over the issue you just clarified for me, while it usually
explains and proves everything into the smallest detail.
I don't know this textbook.
Thanks for your kind effort.
No problem. :-) Let me say that while reading your second posting I was
wondering where in the proof this result could be used for. But maybe I
will look it up one day.
Frank
Best wishes,
J.
.Ok. I was a bit hasty with my first reply. So here is what you might be
expecting:
Define on the n-fold product U^(n) of U, i.e. U x ... x U (n times), the
matrix-valued map F: U^(n) -> R^(nxn) where F(z1,...,zn) is the matrix
you mentioned above. F is continuous, since any of the component
functions of F is continuous by assumption. Now consider the composition
G of maps det o F: U^(n) -> R^(nxn) -> R where det denotes the
determinant function. Since det is continuous (why?), G is continuous.
It holds G(c^(n))=Jf(c)<>0 (with c^(n)=(c,...,c) (n times)).
Can you conclude from here what you need?
Best wishes,
J.
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