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

[hhwolf76]

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]

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.

[Ken Pledger]

In article
<854775835.17564.12574...@gallium.mathforum.org>,
hhwolf76 <james....@gmail.com> wrote:

Try A =

(1 0)
(0 2)

with h = 1 and x = (0 1)

Ken Pledger.

[hhwolf76]

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

Is my assumption right?

[Ray Vickson]

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