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From: Vachel on 23 May 2010 15:46
suppos matrix A = X-GSF' , and J = (||A||F ) 2 = tr(AA*) is the square
of A's Frobenius norm. now how to calculate the derivative of J
respect to S?

because J = tr(AA*), and d(J)/d(A) = 2A , can i calculate it like the
following?
d(J)/d(S) = [d(J)/d(A)] [d(A)/d(S)] ? it seems not right.
how to solve this problem?
is there any books or articles on this subject?
any suggestions can help!
thanks for anything useful!
From: Herman Rubin on 24 May 2010 13:49
On 2010-05-23, Vachel <liangjuntu(a)gmail.com> wrote:
> suppos matrix A = X-GSF' , and J = (||A||F ) 2 = tr(AA*) is the square
> of A's Frobenius norm. now how to calculate the derivative of J
> respect to S?

> because J = tr(AA*), and d(J)/d(A) = 2A , can i calculate it like the
> following?
> d(J)/d(S) = [d(J)/d(A)] [d(A)/d(S)] ? it seems not right.
> how to solve this problem?
> is there any books or articles on this subject?
> any suggestions can help!
> thanks for anything useful!

In general it is necessary to use differentials. The
differential of J is tr(dA*A')+tr(A*dA')=2tr(A'*dA).
Now dA=-G*dS*F', so we get

dJ=-2tr(A'*G*dS*F') = -2tr(F'*A'*G*dS),

which shows how to get the derivative of J with
respect to any element of S.

--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
From: Ronald Bruck on 24 May 2010 16:48
In article
<cc42f3a6-ecef-4f1d-a42c-d034bf5c56d1(a)23g2000pre.googlegroups.com>,
Vachel <liangjuntu(a)gmail.com> wrote:

> suppos matrix A = X-GSF' , and J = (||A||F ) 2 = tr(AA*) is the square
> of A's Frobenius norm. now how to calculate the derivative of J
> respect to S?
>
> because J = tr(AA*), and d(J)/d(A) = 2A , can i calculate it like the
> following?
> d(J)/d(S) = [d(J)/d(A)] [d(A)/d(S)] ? it seems not right.
> how to solve this problem?
> is there any books or articles on this subject?
> any suggestions can help!
> thanks for anything useful!

How about from the DEFINITION? Compute

tr ((A+h)(A*+h*)) = tr(AA*) + 2 tr (A*h) + tr (hh*)

which is the classical (base point) + (linear) + (superlinear).

Then use the chain rule. Herman Rudin does it for you in another post.

*****************

I remember Arnie Liulevicius, (too) many years ago in a global analysis
class at the U of Chicago, complaining how many students knew the
abstract definition of the derivative, knew all the theory, could toss
Hom(E, Hom(E, F)) around with the best of them, but couldn't COMPUTE a
derivative worth a damn. I guess it's still true.

-- Ron Bruck
From: Ronald Bruck on 24 May 2010 18:29
In article <240520101348089359%bruck(a)math.usc.edu>, Ronald Bruck
<bruck(a)math.usc.edu> wrote:

> In article
> <cc42f3a6-ecef-4f1d-a42c-d034bf5c56d1(a)23g2000pre.googlegroups.com>,
> Vachel <liangjuntu(a)gmail.com> wrote:
>
> > suppos matrix A = X-GSF' , and J = (||A||F ) 2 = tr(AA*) is the square
> > of A's Frobenius norm. now how to calculate the derivative of J
> > respect to S?
> >
> > because J = tr(AA*), and d(J)/d(A) = 2A , can i calculate it like the
> > following?
> > d(J)/d(S) = [d(J)/d(A)] [d(A)/d(S)] ? it seems not right.
> > how to solve this problem?
> > is there any books or articles on this subject?
> > any suggestions can help!
> > thanks for anything useful!
>
> How about from the DEFINITION? Compute
>
> tr ((A+h)(A*+h*)) = tr(AA*) + 2 tr (A*h) + tr (hh*)
>
> which is the classical (base point) + (linear) + (superlinear).
>
> Then use the chain rule. Herman Rudin does it for you in another post.

Oops--mea culpa! We have

tr ((A+h)(A*+h*)) = tr (AA* + Ah* + A*h + hh*)
= tr (AA*) + tr (hA* + A*h) + hh*

and h |--> tr (hA* + A*h) is linear.

-- Ron Bruck
From: Vachel on 25 May 2010 09:24
On May 25, 6:29 am, Ronald Bruck <br...(a)math.usc.edu> wrote:
> In article <240520101348089359%br...(a)math.usc.edu>, Ronald Bruck
>
>
>
>
>
> <br...(a)math.usc.edu> wrote:
> > In article
> > <cc42f3a6-ecef-4f1d-a42c-d034bf5c5...(a)23g2000pre.googlegroups.com>,
> > Vachel <liangju...(a)gmail.com> wrote:
>
> > > suppos matrix A = X-GSF' , and J = (||A||F ) 2 = tr(AA*) is the square
> > > of A's Frobenius norm. now how to calculate the derivative of J
> > > respect to S?
>
> > > because J = tr(AA*), and d(J)/d(A) = 2A , can i calculate it like the
> > > following?
> > > d(J)/d(S) = [d(J)/d(A)] [d(A)/d(S)] ? it seems not right.
> > > how to solve this problem?
> > > is there any books or articles on this subject?
> > > any suggestions can help!
> > > thanks for anything useful!
>
> > How about from the DEFINITION?  Compute
>
> >    tr ((A+h)(A*+h*)) = tr(AA*) + 2 tr (A*h) + tr (hh*)
>
> > which is the classical (base point) + (linear) + (superlinear).
>
> > Then use the chain rule.  Herman Rudin does it for you in another post.
>
> Oops--mea culpa!  We have
>
>     tr ((A+h)(A*+h*)) = tr (AA* + Ah* + A*h + hh*)
>       = tr (AA*) + tr (hA* + A*h) +hh*
>
> and h |--> tr (hA* + A*h) is linear.
>
> -- Ron Bruck

thanks Ron
where can i find some materials about the classical (base point) +
(linear) + (superlinear) ?
cause i don't know how to use the chain rule.
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