Re: Relation of Matrix eigenvalues to product of that matrix and a diagonal matrix




In article <9fc7d44f-bb60-46e8-8f2d-53a08f5fe389@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Gregory C. Jones" <GregoryCJones@xxxxxxxxx> writes:
On Dec 6, 1:44 pm, michaelgrunew...@xxxxxxxx (Michaël Grünewald)
wrote:
"GregoryC.Jones" <GregoryCJo...@xxxxxxxxx> writes:

Are there any simple results connecting the eigenvalues of a matrix

A

to the eigenvalues of a matrix

AB

where B is diagonal?
[SNIP]
Thanks for any insights/references!

This makes me think to the Horn conjecture (now a theorem) that gives
strucutre information about eigenvalues of three hermitian matrices A,
B and C with A + B + C =3D 0. I was once told that analogous questions
with products were actively studied, so maybe the expressions ``Horn
conjecture'' and ``Weyl inequalities'' may be a start for your
search.
--=20
Best wishes,
Micha=EBl

Hmm... reading about these it seems my problem is related yet
different.

The multiplicative version of "sum of Hermitian matrices" seems to be
"product of unitary matrices." The product of Hermitian matrices is
not Hermitian so I should recast my problem to show its more natural
origin.

How do we relate the eigenvalues of Hermitian V, to its general linear
transform, G^T V G. My problem is simpler because we can assume
G is diagonal. Thus spectrum(V) vs. spectrum(Lambda V Lambda), and
multiplying by Lambda(-1) on the right and Lambda on the left, this is
the same as spectrum(Lambda^2 V) so this looks a little simpler in
this
case. (In general spectrum(G^T V G) = spectrum (G G^T V))

Gregory Jones


hence there is practically no relation between the two:
Lambda^2 V x = mu x <=> V x = Lambda^{-2} mu x
this is known as the generalized eigenvalue problem, occurs often in
applications and has as eigenvalues the stationary vales of the
generalized Rayleigh-Quotient

y^H V y / y^H Lambda^{-2} y
which might behave quite different from

y^H V y / y^H y
but the signature is the same as
the one of V (by Sylvesters theorem on inertia)
hth
peter

.



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