Please check logic of a simple linear algebra proof

Hi Folks,

Could someone please verify that my answer to part b in section 2.1 of
the following document is correct?

  http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf

In particular, could you verify that the result of part a is not
sufficient for the proof of part b?

This is a homework that has already been submitted so you will not
be violating any academic integrity.

Thanks in advance.
--
%  Randy Yates                  % "She's sweet on Wagner-I think she'd die for Beethoven.
%% Fuquay-Varina, NC            %  She love the way Puccini lays down a tune, and
%%% 919-577-9882                %  Verdi's always creepin' from her room." 
%%%% <yates@ieee.org>           % "Rockaria", *A New World Record*, ELO   
http://home.earthlink.net/~yatescr

Randy Yates wrote:
> Hi Folks,
>
> Could someone please verify that my answer to part b in section 2.1 of
> the following document is correct?
>
>   http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf
>
> In particular, could you verify that the result of part a is not
> sufficient for the proof of part b?

Logically, I think you're correct. Your proof in part (a) only states
that (lambda, x) being an eigenpair of A implies that (lambda^k, x) is
an eigenpair of A^k; it doesn't state that the implication goes the
other way also. Whether the implication in the other direction could
have been shown more easily, I don't know.

Jason

Jason wrote:

> Randy Yates wrote:
> > Hi Folks,
> >
> > Could someone please verify that my answer to part b in section 2.1 of
> > the following document is correct?
> >
> >   http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf
> >
> > In particular, could you verify that the result of part a is not
> > sufficient for the proof of part b?
>
> Logically, I think you're correct. Your proof in part (a) only states
> that (lambda, x) being an eigenpair of A implies that (lambda^k, x) is
> an eigenpair of A^k; it doesn't state that the implication goes the
> other way also.

Good points: indeed, in general the other implication is wrong (think
of how to construct a counter example!). However, for real symmetric
matrices, the inverse implication is also true (this follows
immediately from the eigendecomposition property of symmetric
matrices).

Regards,
Andor

Randy Yates skrev:
> Hi Folks,
>
> Could someone please verify that my answer to part b in section 2.1 of
> the following document is correct?
>
>   http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf
>
> In particular, could you verify that the result of part a is not
> sufficient for the proof of part b?

There is no definition of matrix norm in your paper. If I remember
correctly,

||A|| = max_x in R^N  {||Ax||/||x||}

The norm of a matrix is the maximum elongation of any vector.
Once this is established, one can proceed in two steps:

- The eigenvectors of symmetric matrix form an orthogonal basis
- The elongation corresponding to the maximum norm only
  occurs for vectors in the subspace spanned by the eigenvector(s)
  that correspond to the maximum eigenvalue.

Rune

Randy,

What degree are you working on?

Dirk

Randy Yates wrote:
> Hi Folks,
>
> Could someone please verify that my answer to part b in section 2.1 of
> the following document is correct?
>
>   http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf
>
> In particular, could you verify that the result of part a is not
> sufficient for the proof of part b?
>
> This is a homework that has already been submitted so you will not
> be violating any academic integrity.
>
> Thanks in advance.
> --
> %  Randy Yates                  % "She's sweet on Wagner-I think she'd die for Beethoven.
> %% Fuquay-Varina, NC            %  She love the way Puccini lays down a tune, and
> %%% 919-577-9882                %  Verdi's always creepin' from her room."
> %%%% <yates@ieee.org>           % "Rockaria", *A New World Record*, ELO
> http://home.earthlink.net/~yatescr

Hi Jason,

Thanks for your feedback - it is very useful. A couple of
comments below, but basically we understand each other and
agree.

cincydsp@gmail.com writes:

> Randy Yates wrote:
>> Hi Folks,
>>
>> Could someone please verify that my answer to part b in section 2.1 of
>> the following document is correct?
>>
>>   http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf
>>
>> In particular, could you verify that the result of part a is not
>> sufficient for the proof of part b?
>
> Logically, I think you're correct. Your proof in part (a) only states
> that (lambda, x) being an eigenpair of A implies that (lambda^k, x) is
> an eigenpair of A^k; it doesn't state that the implication goes the
> other way also. 

That's exactly what I thought. Thanks so much for confirming that I didn't
do something stupid.

> Whether the implication in the other direction could
> have been shown more easily, I don't know.

In general, as Andor said, the implication doesn't go the other
way. When I say "in general," I mean that, even for the specific case
of k=2, an *eigenpair* doesn't exist in which the eigenvectors are the
same. 

Here's a simple counterexample. Let A = [1 0; 0 -1]. The eigenvalues
of A are +1 and -1. An eigenpair of A is (-1,[0 1]'). 
A^2 = [1 0;0 1], which means that (1, [0 1]') is an eigenpair of
A^2. However, (1, [0 1]') is not an eigenpair of A. 
-- 
%  Randy Yates                  % "Though you ride on the wheels of tomorrow,
%% Fuquay-Varina, NC            %  you still wander the fields of your
%%% 919-577-9882                %  sorrow."
%%%% <yates@ieee.org>           % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

"dbell" <bellda2005@cox.net> writes:

> Randy,
>
> What degree are you working on?

MSEE. Hopefully I can complete it before I get Alzheimer's....

--Randy


>
> Dirk
>
> Randy Yates wrote:
>> Hi Folks,
>>
>> Could someone please verify that my answer to part b in section 2.1 of
>> the following document is correct?
>>
>>   http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf
>>
>> In particular, could you verify that the result of part a is not
>> sufficient for the proof of part b?
>>
>> This is a homework that has already been submitted so you will not
>> be violating any academic integrity.
>>
>> Thanks in advance.
>> --
>> %  Randy Yates                  % "She's sweet on Wagner-I think she'd die for Beethoven.
>> %% Fuquay-Varina, NC            %  She love the way Puccini lays down a tune, and
>> %%% 919-577-9882                %  Verdi's always creepin' from her room."
>> %%%% <yates@ieee.org>           % "Rockaria", *A New World Record*, ELO
>> http://home.earthlink.net/~yatescr
>

-- 
%  Randy Yates                  % "Bird, on the wing,
%% Fuquay-Varina, NC            %   goes floating by
%%% 919-577-9882                %   but there's a teardrop in his eye..."
%%%% <yates@ieee.org>           % 'One Summer Dream', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

"Andor" <andor.bariska@gmail.com> writes:

> Jason wrote:
>
>> Randy Yates wrote:
>> > Hi Folks,
>> >
>> > Could someone please verify that my answer to part b in section 2.1 of
>> > the following document is correct?
>> >
>> >   http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf
>> >
>> > In particular, could you verify that the result of part a is not
>> > sufficient for the proof of part b?
>>
>> Logically, I think you're correct. Your proof in part (a) only states
>> that (lambda, x) being an eigenpair of A implies that (lambda^k, x) is
>> an eigenpair of A^k; it doesn't state that the implication goes the
>> other way also.
>
> Good points: indeed, in general the other implication is wrong (think
> of how to construct a counter example!). However, for real symmetric
> matrices, the inverse implication is also true (this follows
> immediately from the eigendecomposition property of symmetric
> matrices).

Hey Andor,

Thanks for taking a look. I think that even for symmetric matrices
the converse of part a is not true. I gave a counterexample to Jason.

What do you mean by "the eigendecomposition property of symmetric
matrices?"
-- 
%  Randy Yates                  % "She tells me that she likes me very much,
%% Fuquay-Varina, NC            %     but when I try to touch, she makes it
%%% 919-577-9882                %                            all too clear."
%%%% <yates@ieee.org>           %        'Yours Truly, 2095', *Time*, ELO  
http://home.earthlink.net/~yatescr

On 11 Dez., 20:23, Randy Yates <y...@ieee.org> wrote:
> "Andor" <andor.bari...@gmail.com> writes:
> > Jason wrote:
>
> >> Randy Yates wrote:
> >> > Hi Folks,
>
> >> > Could someone please verify that my answer to part b in section 2.1 of
> >> > the following document is correct?
>
> >> >  http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf
>
> >> > In particular, could you verify that the result of part a is not
> >> > sufficient for the proof of part b?
>
> >> Logically, I think you're correct. Your proof in part (a) only states
> >> that (lambda, x) being an eigenpair of A implies that (lambda^k, x) is
> >> an eigenpair of A^k; it doesn't state that the implication goes the
> >> other way also.
>
> > Good points: indeed, in general the other implication is wrong (think
> > of how to construct a counter example!). However, for real symmetric
> > matrices, the inverse implication is also true (this follows
> > immediately from the eigendecomposition property of symmetric
> > matrices).
> Hey Andor,
>
> Thanks for taking a look. I think that even for symmetric matrices
> the converse of part a is not true. I gave a counterexample to Jason.

That isn't a counter-example to the correct inverse implication, which
states:

Let A be a (real) symmetric matrix, k a positve integer and x an
eigenvector of the matrix A^k. Then x is also an eigenvector of A. Let
lambda_x be the corresponding eigenvalue of A, then lambda_x^k is the
eigenvalue of x for A^k.

The same statement for general (non-symmetric) matrices isn't true.

>
> What do you mean by "the eigendecomposition property of symmetric
> matrices?"

Perhaps I should have said the diagonalisation property. It states that
a (real) symmetric matrix A is diagonalisable, ie. there exists a basis
of eigenvectors of A.


Regards,
Andor

Randy Yates wrote:
> "dbell" <bellda2005@cox.net> writes:
> 
>> Randy,
>>
>> What degree are you working on?
> 
> MSEE. Hopefully I can complete it before I get Alzheimer's....
> 
> --Randy
> 
> 
>> Dirk
>>
>> Randy Yates wrote:
>>> Hi Folks,
>>>
>>> Could someone please verify that my answer to part b in section 2.1 of
>>> the following document is correct?
>>>
>>>   http://www.digitalsignallabs.com/ma580/hw/hw5partII/response/hw.pdf
>>>
>>> In particular, could you verify that the result of part a is not
>>> sufficient for the proof of part b?
>>>
>>> This is a homework that has already been submitted so you will not
>>> be violating any academic integrity.
>>>
>>> Thanks in advance.
>>> --
>>> %  Randy Yates                  % "She's sweet on Wagner-I think she'd die for Beethoven.
>>> %% Fuquay-Varina, NC            %  She love the way Puccini lays down a tune, and
>>> %%% 919-577-9882                %  Verdi's always creepin' from her room."
>>> %%%% <yates@ieee.org>           % "Rockaria", *A New World Record*, ELO
>>> http://home.earthlink.net/~yatescr
> 

Good luck to you! [although I hope you don't need luck :)]

Cheers

PeteS