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From: HardySpicer on 2 Jun 2010 04:28
I have a partitioned matrix (order 2n square)

I A
B I

where I is the identity matrix of order n and all thesub matrices are
square of order n.

Is there a way of working out it's eigenvalues in terms of the sub-
matrices?

My thoughts are that I have to find the determinant of

eI-I -A
-B eI-I

in terms of e (I have no lambda on the keyboard!) and use a partician
determinant formula of some sort.


Hardy
From: Alois Steindl on 2 Jun 2010 04:43
HardySpicer <gyansorova(a)gmail.com> writes:

> I have a partitioned matrix (order 2n square)
>
> I A
> B I
>
> where I is the identity matrix of order n and all thesub matrices are
> square of order n.
>
> Is there a way of working out it's eigenvalues in terms of the sub-
> matrices?
>
> My thoughts are that I have to find the determinant of
>
> eI-I -A
> -B eI-I
>
> in terms of e (I have no lambda on the keyboard!) and use a partician
> determinant formula of some sort.
>
>
> Hardy
Hello,
first two remarks:
i) Adding a constant diagonal matrix just shifts the eigenvalues, so you
could equally well search the eigenvalues and -vectors of the matrix

0 A
B 0

and then add 1 to the eigenvalues to get those of your matrix

ii) working with the characteristic polynomial makes only sense for extremely
small matrices.

By playing around with the equations

A y = lambda x
B x = lambda y

you will find, that you can reduce the problem to

B A y = lambda^2 y

or equivalently

A B x = lambda^2 x

Good luck
Alois
From: HardySpicer on 3 Jun 2010 03:48
On Jun 2, 8:43 pm, Alois Steindl <Alois.Stei...(a)tuwien.ac.at> wrote:
> HardySpicer <gyansor...(a)gmail.com> writes:
> > I have a partitioned matrix (order 2n square)
>
> > I A
> > B I
>
> > where I is the identity matrix of order n and all thesub  matrices are
> > square of order n.
>
> > Is there a way of working out it's eigenvalues in terms of the sub-
> > matrices?
>
> > My thoughts are that I have to find the determinant of
>
> > eI-I  -A
> > -B   eI-I
>
> > in terms of e (I have no lambda on the keyboard!) and use a partician
> > determinant formula of some sort.
>
> > Hardy
>
> Hello,
> first two remarks:
> i) Adding a constant diagonal matrix just shifts the eigenvalues, so you
> could equally well search the eigenvalues and -vectors of the matrix
>
> 0  A
> B  0
>
> and then add 1 to the eigenvalues to get those of your matrix
>
> ii) working with the characteristic polynomial makes only sense for extremely
> small matrices.
>
> By playing around with the equations
>
> A y = lambda x
> B x = lambda y
>
> you will find, that you can reduce the problem to
>
> B A y = lambda^2 y
>
> or equivalently
>
> A B x = lambda^2 x
>
> Good luck
> Alois

Thanks for that. Just one question, when I find lambda^2 and hence
lambda is the square root of these values - do I take both positive
and negative roots?


Hardy
From: Alois Steindl on 7 Jun 2010 04:35
HardySpicer <gyansorova(a)gmail.com> writes:

>
> Thanks for that. Just one question, when I find lambda^2 and hence
> lambda is the square root of these values - do I take both positive
> and negative roots?
>
>
> Hardy
Hello,
you should really be able to answer this question yourself.
Alois
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