Path: news.mathworks.com!not-for-mail
From: "Titus" <titus.edelhofer@mathworks.de>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Eigen value and Eigen vector?
Date: Fri, 17 Aug 2007 17:18:49 +0200
Organization: The MathWorks, Inc.
Lines: 53
Message-ID: <fa4e8p$473$1@fred.mathworks.com>
References: <fa2bhu$5bp$1@fred.mathworks.com> <1187310068.805535.110740@19g2000hsx.googlegroups.com> <fa3hql$1so$1@fred.mathworks.com> <fa3kac$rlr$1@fred.mathworks.com> <fa4a5e$35c$1@fred.mathworks.com>
NNTP-Posting-Host: de-edelhoft-x.ac.mathworks.de
X-Trace: fred.mathworks.com 1187363930 4323 172.16.75.150 (17 Aug 2007 15:18:50 GMT)
X-Complaints-To: news@mathworks.com
NNTP-Posting-Date: Fri, 17 Aug 2007 15:18:50 +0000 (UTC)
X-Priority: 3
X-MSMail-Priority: Normal
X-Newsreader: Microsoft Outlook Express 6.00.2900.3028
X-RFC2646: Format=Flowed; Original
X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028
Xref: news.mathworks.com comp.soft-sys.matlab:424319




"stephen " <huangj5@nationwide.com> schrieb im Newsbeitrag 
news:fa4a5e$35c$1@fred.mathworks.com...
> Thanks for all the responses, guys.
>
> I am a little confused now as I look at the results. I
> have an Excel VBA program which could calculate the
> Eigenvalues in descending order and the associated
> Eigenvetors. If I use the suggested method to calculate
> the same things in Matlab, some values of Eigenvector are
> exactly the same while some values of Eigenvector are the
> same in absolute terms but have opposite signs (plus vs
> minus).  I use formula V*D*V&#8217; to check if they are getting
> the same correlation matrix, and they are.
>
> Does anyone have any idea what&#8217;s going on here?
>
> Thanks,
> Stephen
>
>
>
>
> "us " <us@neurol.unizh.ch> wrote in message
> <fa3kac$rlr$1@fred.mathworks.com>...
>> Titus:
>> <SNIP the big easy...
>>
>> > > D = D(length(D(1,:)):-1:1,length(D(:,1)):-1:1) ;
>>
>> > slightly easier: use "end" instead of length:
>> > D = D(end:-1:1,end:-1:1)
>>
>> ... or, yet another solution
>>
>>      D=rot90(D,2)
>>
>> us
>

Hi Stephen,
yes I have ;-)
Eigenvectors are not unique in the following sense: if x is an eigenvector, 
k*x for every k~=0 is an eigenvector as well, since an eigenvector is a non 
trivial vector that solves
A * x = lambda * x
so: for any scalar k~=0
A * (k*x) = k * (A * x) = k * (lambda * x) = lambda * (k * x),
so k*x is an eigenvector as well, especially for k=-1...

Titus