Path: news.mathworks.com!not-for-mail
From: <HIDDEN>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Question on the derivate /calculus of a 2-norm matrix . Thanks a lot
Date: Mon, 26 Jul 2010 03:01:04 +0000 (UTC)
Organization: Anhui University
Lines: 14
Message-ID: <i2itpg$h3j$1@fred.mathworks.com>
References: <i2he90$21a$1@fred.mathworks.com> <i2i2st$n4j$1@fred.mathworks.com> <i2irns$94i$1@fred.mathworks.com> <i2it4t$73b$1@fred.mathworks.com>
Reply-To: <HIDDEN>
NNTP-Posting-Host: webapp-05-blr.mathworks.com
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: 8bit
X-Trace: fred.mathworks.com 1280113264 17523 172.30.248.35 (26 Jul 2010 03:01:04 GMT)
X-Complaints-To: news@mathworks.com
NNTP-Posting-Date: Mon, 26 Jul 2010 03:01:04 +0000 (UTC)
X-Newsreader: MATLAB Central Newsreader 2419503
Xref: news.mathworks.com comp.soft-sys.matlab:656060

"Matt J " <mattjacREMOVE@THISieee.spam> wrote in message <i2it4t$73b$1@fred.mathworks.com>...
> "Antony " <mutang.bing@gmail.com> wrote in message <i2irns$94i$1@fred.mathworks.com>...
> 
> > I have another problem. Maybe we can not directly solve it and I think the result might be more complxe than my former problem. The problem is: 
> >   if g(x) = ||KX-B||^0.6 with all the other settings as the former problem, what is \partial{g}/\partial{x}? 
> ========
> 
> This is equivalent to  (||KX-B||^2)^0.3
> 
> So you can use your original result, with one more step of the chain rule leading to
> 
> Gradient = 0.3*(||KX-B||^2)^(-.7)  * 2*K'*(K*X-B)

Why not write it as Gradient = 0.3 *2*K'*(K*X-B)*(||KX-B||^2)^(-.7) according to the chain rule? It is because K'*(K*X-B) is a scalar and there is no difference between them? Thank you!