Re: bound on eigenvalues...
- From: "comtech" <comtech.usa@xxxxxxxxx>
- Date: 7 Dec 2005 12:42:09 -0800
Robert Low wrote:
> José Carlos Santos wrote:
> > The eigenvalues are the roots of the characteristic polynomial, which is
> > P(x) = -x^3 + a x^2 + b x + c, with a = 0.501, b = -0.024089, and
> > c = 0.0000524046. Since it's a third degree polynomial, it can't have
> > more than three roots. Now, since P(0) = 0.0000524046, P(0.05) =
> > -0.0000245449, P(0.1) = 0.00165351, P(0.4) = 0.00657681, and P(0.5) =
> > -0.0117421, there must be a root between 0 and 0.1, a second one between
> > 0.1 and 0.4, and a third one between 0.4 and 0.5. This proves that all
> > roots are smaller than 0.5.
>
> But I think he *meant* to write that there must be a root
> between 0 and 0.05, a second one between 0.05 and 0.1,
> and a third one between 0.4 and 0.5, since that's how the
> sign changes go.
But these approaches look like numerical... it is perfectly OK for
practice ... but here we need analytical methods...
.
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