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A*x <= h*x ? x is random vector, A is square matrix, h is eigenvalue
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hhwolf76  
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 More options Nov 6, 4:58 am
Newsgroups: sci.math
From: hhwolf76 <james.zho...@gmail.com>
Date: Thu, 05 Nov 2009 15:58:02 EST
Local: Fri, Nov 6 2009 4:58 am
Subject: A*x <= h*x ? x is random vector, A is square matrix, h is eigenvalue
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I got a question. If A is a square matrix, h is one of
A's eigenvalues. If x is the corresponding eigenvector,
then A*x=h*x. But if x is not eigenvector, it is a random vector, is it true that A*x<=h*x ?
  Not familar with the property of eigenvector. Thanks!

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hhwolf76  
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 More options Nov 6, 5:02 am
Newsgroups: sci.math
From: hhwolf76 <james.zho...@gmail.com>
Date: Thu, 05 Nov 2009 16:02:41 EST
Local: Fri, Nov 6 2009 5:02 am
Subject: Re: A*x <= h*x ? x is random vector, A is square matrix, h is eigenvalue
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Or in other way,

Is it true that x'*A*x<=x'*h*x ?
Still, A is square matrix, h is eigenvalue, x is random vector.


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Ken Pledger  
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 More options Nov 6, 7:04 am
Newsgroups: sci.math
From: Ken Pledger <ken.pled...@mcs.vuw.ac.nz>
Date: Fri, 06 Nov 2009 12:04:08 +1300
Local: Fri, Nov 6 2009 7:04 am
Subject: Re: A*x <= h*x ? x is random vector, A is square matrix, h is eigenvalue
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In article
<854775835.17564.1257454992142.JavaMail.r...@gallium.mathforum.org>,

 hhwolf76 <james.zho...@gmail.com> wrote:
> Or in other way,

> Is it true that x'*A*x<=x'*h*x ?
> Still, A is square matrix, h is eigenvalue, x is random vector.

Try   A =

(1 0)
(0 2)

with  h = 1  and  x = (0 1)

      Ken Pledger.


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hhwolf76  
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 More options Nov 6, 10:23 am
Newsgroups: sci.math
From: hhwolf76 <james.zho...@gmail.com>
Date: Thu, 05 Nov 2009 21:23:45 EST
Local: Fri, Nov 6 2009 10:23 am
Subject: Re: A*x <= h*x ? x is random vector, A is square matrix, h is eigenvalue
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I assume A=R'*R (R is a square matrix),
h is the biggest eigenvalue.

Is my assumption right?


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Ray Vickson  
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 More options Nov 8, 1:23 am
Newsgroups: sci.math
From: Ray Vickson <RGVick...@shaw.ca>
Date: Sat, 7 Nov 2009 09:23:57 -0800 (PST)
Local: Sun, Nov 8 2009 1:23 am
Subject: Re: A*x <= h*x ? x is random vector, A is square matrix, h is eigenvalue
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On Nov 5, 6:23 pm, hhwolf76 <james.zho...@gmail.com> wrote:

> I assume A=R'*R (R is a square matrix),
> h is the biggest eigenvalue.

> Is my assumption right?

Yes. All you need is for A to be real and symmetric. See
http://en.wikipedia.org/wiki/Rayleigh_quotient and
http://www.umiacs.umd.edu/~shaohua/enee739q_cmsc858c/RayleighsQuotien...
.
This last link shows that a stationary point of the Rayleigh quotient
yields an eigenvalue, and since the max/min of the rayleigh quotient
corresponds to a stationary point, these max and min ratios are the
largest and smallest eigenvalues.

R.G. Vickson


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