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A*x <= h*x ? x is random vector, A is square matrix, h...

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hhwolf76...
Posted: Thu Nov 05, 2009 10:58 am
Guest
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!
 
hhwolf76...
Posted: Thu Nov 05, 2009 11:02 am
Guest
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.
 
hhwolf76...
Posted: Thu Nov 05, 2009 4:23 pm
Guest
I assume A=R'*R (R is a square matrix),
h is the biggest eigenvalue.

Is my assumption right?
 
Ken Pledger...
Posted: Thu Nov 05, 2009 6:04 pm
Guest
In article
<854775835.17564.1257454992142.JavaMail.root at (no spam) gallium.mathforum.org>,
hhwolf76 <james.zhou76 at (no spam) gmail.com> wrote:

[quote]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.
[/quote]

Try A =

(1 0)
(0 2)

with h = 1 and x = (0 1)

Ken Pledger.
 
Ray Vickson...
Posted: Sat Nov 07, 2009 7:23 am
Guest
On Nov 5, 6:23 pm, hhwolf76 <james.zho... at (no spam) gmail.com> wrote:
[quote]I assume A=R'*R (R is a square matrix),
h is the biggest eigenvalue.

Is my assumption right?
[/quote]
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/RayleighsQuotient.pdf
..
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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