Re: bound on eigenvalues...
On 06-12-2005 8:08, comtech wrote:
Given a matrix
A =
0.1337 0.1115 0.1647
0.1115 0.1584 0.1362
0.1647 0.1362 0.2089
I know by numerical computation that its eigenvalues are:
ans =
0.0023
0.0513
0.4475
so any number > 0.4475 is the bound on the eigenvalues...
But how do I find bound analytically?
Can I prove that the eigenvalues are less than a number, say, 0.8, or
0.9, or x, where x<1 strictly analytically?
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.
Best regards,
Jose Carlos Santos
.
