FPGA Central

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/ma5...esponse/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."
%%%% <[email protected]> % "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/ma5...esponse/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/ma5...esponse/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/ma5...esponse/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/ma5...esponse/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."
> %%%% <[email protected]> % "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.

[email protected] 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/ma5...esponse/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."
%%%% <[email protected]> % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr

"dbell" <[email protected]> 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/ma5...esponse/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."
>> %%%% <[email protected]> % "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..."
%%%% <[email protected]> % 'One Summer Dream', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

"Andor" <[email protected]> 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/ma5...esponse/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."
%%%% <[email protected]> % '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/ma5...esponse/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" <[email protected]> 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/ma5...esponse/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."
>>> %%%% <[email protected]> % "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

PeteS <[email protected]> writes:

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

Thanks, Peter! Well-wishing is good.
--
% 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."
%%%% <[email protected]> % "Rockaria", *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Randy Yates wrote:
> PeteS <[email protected]> writes:
>
>> Good luck to you! [although I hope you don't need luck ]
>
> Thanks, Peter! Well-wishing is good.

I can only hope we are up to the other questions I know you'll want to
ask later!

I remember a very good friend I worked with in 98-00 who was doing a
MSEE at Rutgers, and he used to drop by my cube regularly; not so much
just for clarifications, but sometimes merely for a little reaffirmation.

He's of sicilian extraction and somewhat emotional, and he just liked to
sit and ask me what my answers were to feel better about his own
abilities (which were perfectly up to the task), I think

Cheers

PeteS

"Rune Allnor" <[email protected]> writes:

> 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/ma5...esponse/hw.pdf
>>
>> In particular, could you verify that the result of part a is not
>> sufficient for the proof of part b?

Hi Rune,

Thanks for responding. I'm sorry it took me all day to get back to
your post - I got preoccupied with the other posts and forgot you had
posted.

> There is no definition of matrix norm in your paper. If I remember
> correctly,
>
> ||A|| = max_x in R^N {||Ax||/||x||}

Then the question becomes, "What vector norm (||x||) do you use?"
In any case, this is called a "matrix norm subordinate to a vector
norm."

The matrix 2-norm is not a subordinate matrix norm. The matrix
2-norm of A is defined as the "spectral radius" of A^T*A. The
spectral radius of A is defined to be the maximum eigenvalue
magnitude of A.

> 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.

While this is a very nice insight on symmetric matrices (I didn't
realize a plain, symmetric matrix was orthogonally similar to a
diagonal matrix), I don't think it helps in this problem since we had
to use the 2-norm.
--
% Randy Yates % "Midnight, on the water...
%% Fuquay-Varina, NC % I saw... the ocean's daughter."
%%% 919-577-9882 % 'Can't Get It Out Of My Head'
%%%% <[email protected]> % *El Dorado*, Electric Light Orchestra
http://home.earthlink.net/~yatescr

Randy Yates wrote:
> "Rune Allnor" <all...@tele.ntnu.no> writes:
> > 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/ma5...esponse/hw.pdf
>
> >> In particular, could you verify that the result of part a is not
> >> sufficient for the proof of part b?Hi Rune,
>
> Thanks for responding. I'm sorry it took me all day to get back to
> your post - I got preoccupied with the other posts and forgot you had
> posted.
>
> > There is no definition of matrix norm in your paper. If I remember
> > correctly,
>
> > ||A|| = max_x in R^N {||Ax||/||x||}
>
>Then the question becomes, "What vector norm (||x||) do you use?"
> In any case, this is called a "matrix norm subordinate to a vector
> norm."
>
> The matrix 2-norm is not a subordinate matrix norm.

Randy, why do you think it is called the 2-norm?

> The matrix
> 2-norm of A is defined as the "spectral radius" of A^T*A. The
> spectral radius of A is defined to be the maximum eigenvalue
> magnitude of A.

Sometimes, there is more than one equivalent definition:

http://mathworld.wolfram.com/SpectralNorm.html

Regards,
Andor

"Andor" <[email protected]> writes:

> Randy Yates wrote:
>> "Rune Allnor" <all...@tele.ntnu.no> writes:
>> > 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/ma5...esponse/hw.pdf
>>
>> >> In particular, could you verify that the result of part a is not
>> >> sufficient for the proof of part b?Hi Rune,
>>
>> Thanks for responding. I'm sorry it took me all day to get back to
>> your post - I got preoccupied with the other posts and forgot you had
>> posted.
>>
>> > There is no definition of matrix norm in your paper. If I remember
>> > correctly,
>>
>> > ||A|| = max_x in R^N {||Ax||/||x||}
>>
>>Then the question becomes, "What vector norm (||x||) do you use?"
>> In any case, this is called a "matrix norm subordinate to a vector
>> norm."
>>
>> The matrix 2-norm is not a subordinate matrix norm.
>
> Randy, why do you think it is called the 2-norm?

So you define things by their labels?

Seriously, it wasn't defined to me as an induced norm, even
though I see now from a couple of reliable sources that it
is. My mistake.
--
% Randy Yates % "Watching all the days go by...
%% Fuquay-Varina, NC % Who are you and who am I?"
%%% 919-577-9882 % 'Mission (A World Record)',
%%%% <[email protected]> % *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Randy Yates <[email protected]> writes:
> [...]
> The matrix 2-norm is not a subordinate matrix norm.

As Andor pointed out, this is wrong. I apologize - it
was not defined to me as an induced norm.

> The matrix 2-norm of A is defined as the "spectral radius" of A^T*A.

Again I made an error here. The corected statement is

The matrix 2-norm of A is defined as the square root of the
"spectral radius" of A^T*A.

Sorry for the confusion.
--
% Randy Yates % "...the answer lies within your soul
%% Fuquay-Varina, NC % 'cause no one knows which side
%%% 919-577-9882 % the coin will fall."
%%%% <[email protected]> % 'Big Wheels', *Out of the Blue*, ELO
http://home.earthlink.net/~yatescr

Randy Yates skrev:
> Randy Yates <[email protected]> writes:
> > [...]
> > The matrix 2-norm is not a subordinate matrix norm.
>
> As Andor pointed out, this is wrong. I apologize - it
> was not defined to me as an induced norm.
>
> > The matrix 2-norm of A is defined as the "spectral radius" of A^T*A.
>
> Again I made an error here. The corected statement is
>
> The matrix 2-norm of A is defined as the square root of the
> "spectral radius" of A^T*A.
>
> Sorry for the confusion.

I think these are consistent with the definition I used.
If we start with A being symmetric, we fine that the
eigenvectors are urthogonal and the matrix of egienvectors
is unitary,

A = ELE' = A'

where E'E = EE' = I and L is diagonal, so

A'A = ELE'ELE' = ELILE' = E'L^2E

And so the largest eigenvalue of A equals the square rooth
of the largest eigenvalue of A'A, which is consistent with my
definition, provided A is symmetric.

I can see the point of goint the roundabout way via A'A
in the case when A is *not* symmetric. In that case,
forming the product A'A is a way to obtain a symmetric
matrix from A, and thus to squeeze the definition of a
matrix norm into a familiar framework.

Rune

On 2006-12-12 14:33:44 -0400, "Rune Allnor" <[email protected]> said:

>
> Randy Yates skrev:
>> Randy Yates <[email protected]> writes:
>>> [...]
>>> The matrix 2-norm is not a subordinate matrix norm.
>>
>> As Andor pointed out, this is wrong. I apologize - it
>> was not defined to me as an induced norm.
>>
>>> The matrix 2-norm of A is defined as the "spectral radius" of A^T*A.
>>
>> Again I made an error here. The corected statement is
>>
>> The matrix 2-norm of A is defined as the square root of the
>> "spectral radius" of A^T*A.
>>
>> Sorry for the confusion.
>
> I think these are consistent with the definition I used.
> If we start with A being symmetric, we fine that the
> eigenvectors are urthogonal and the matrix of egienvectors
> is unitary,
>
> A = ELE' = A'
>
> where E'E = EE' = I and L is diagonal, so
>
> A'A = ELE'ELE' = ELILE' = E'L^2E
>
> And so the largest eigenvalue of A equals the square rooth
> of the largest eigenvalue of A'A, which is consistent with my
> definition, provided A is symmetric.
>
> I can see the point of goint the roundabout way via A'A
> in the case when A is *not* symmetric. In that case,
> forming the product A'A is a way to obtain a symmetric
> matrix from A, and thus to squeeze the definition of a
> matrix norm into a familiar framework.
>
> Rune

If your first definition is of the matrix norm induced by the L_2
vector norm then you have theorem that it has an explicit formulae
which is the same as the spectral radius.

If your first definition is of the spectral radius then you need to
prove the theorem that you have indeed defined an induced norm. And
that the vector norm is L_2.

If you become too familiar with the results it is easy to forget
which are the definitions and which are the theorems. Sometimes
they are interchangeable but it is never the case that they are
all only definitions. It is common to see careful mathematical texts
with a theorm that says "the following many characteriztions are
equivalent" with a proof that 1 implies 2, 2 implies 3 and so on
until the last implies the first in a big circle. Often there will
only be two or three of the steps that take any effort. The fact
that some of the steps take some work is the old "no free lunch"
principle at work.

Gordon Sande skrev:
> On 2006-12-12 14:33:44 -0400, "Rune Allnor" <[email protected]> said:
>
> >
> > Randy Yates skrev:
> >> Randy Yates <[email protected]> writes:
> >>> [...]
> >>> The matrix 2-norm is not a subordinate matrix norm.
> >>
> >> As Andor pointed out, this is wrong. I apologize - it
> >> was not defined to me as an induced norm.
> >>
> >>> The matrix 2-norm of A is defined as the "spectral radius" of A^T*A.
> >>
> >> Again I made an error here. The corected statement is
> >>
> >> The matrix 2-norm of A is defined as the square root of the
> >> "spectral radius" of A^T*A.
> >>
> >> Sorry for the confusion.
> >
> > I think these are consistent with the definition I used.
> > If we start with A being symmetric, we fine that the
> > eigenvectors are urthogonal and the matrix of egienvectors
> > is unitary,
> >
> > A = ELE' = A'
> >
> > where E'E = EE' = I and L is diagonal, so
> >
> > A'A = ELE'ELE' = ELILE' = E'L^2E
> >
> > And so the largest eigenvalue of A equals the square rooth
> > of the largest eigenvalue of A'A, which is consistent with my
> > definition, provided A is symmetric.
> >
> > I can see the point of goint the roundabout way via A'A
> > in the case when A is *not* symmetric. In that case,
> > forming the product A'A is a way to obtain a symmetric
> > matrix from A, and thus to squeeze the definition of a
> > matrix norm into a familiar framework.
> >
> > Rune
>
> If your first definition is of the matrix norm induced by the L_2
> vector norm then you have theorem that it has an explicit formulae
> which is the same as the spectral radius.

I can't see why the norm

||A|| = max_{x in R^N} ||Ax||/||x|| (1)

should be constrained to the L2 norm. Restricting the
discussion to symmetric A, we find some Ax = vx where
(v,x) is the maximum eigenvalue v with eigenvector x for A.

Regardless of what norm one uses,

||Ax|| = ||vx|| = v||x|| (2)

the latter equality following directly from the usual axioms
of norms. Since (2) is independent of the norm used,
(1) is also independent of the norms used, and is
thus not restricted to the L2 norm.

> If your first definition is of the spectral radius then you need to
> prove the theorem that you have indeed defined an induced norm. And
> that the vector norm is L_2.

The proof consists of checking whether the expression
satisfies the axioms of the norm. There aren't too many
to check; I think there are 4 (it's been a few years since I
last read about these things).

Rune

On 2006-12-12 15:39:21 -0400, "Rune Allnor" <[email protected]> said:

>
> Gordon Sande skrev:
>> On 2006-12-12 14:33:44 -0400, "Rune Allnor" <[email protected]> said:
>>
>>>
>>> Randy Yates skrev:
>>>> Randy Yates <[email protected]> writes:
>>>>> [...]
>>>>> The matrix 2-norm is not a subordinate matrix norm.
>>>>
>>>> As Andor pointed out, this is wrong. I apologize - it
>>>> was not defined to me as an induced norm.
>>>>
>>>>> The matrix 2-norm of A is defined as the "spectral radius" of A^T*A.
>>>>
>>>> Again I made an error here. The corected statement is
>>>>
>>>> The matrix 2-norm of A is defined as the square root of the
>>>> "spectral radius" of A^T*A.
>>>>
>>>> Sorry for the confusion.
>>>
>>> I think these are consistent with the definition I used.
>>> If we start with A being symmetric, we fine that the
>>> eigenvectors are urthogonal and the matrix of egienvectors
>>> is unitary,
>>>
>>> A = ELE' = A'
>>>
>>> where E'E = EE' = I and L is diagonal, so
>>>
>>> A'A = ELE'ELE' = ELILE' = E'L^2E
>>>
>>> And so the largest eigenvalue of A equals the square rooth
>>> of the largest eigenvalue of A'A, which is consistent with my
>>> definition, provided A is symmetric.
>>>
>>> I can see the point of goint the roundabout way via A'A
>>> in the case when A is *not* symmetric. In that case,
>>> forming the product A'A is a way to obtain a symmetric
>>> matrix from A, and thus to squeeze the definition of a
>>> matrix norm into a familiar framework.
>>>
>>> Rune
>>
>> If your first definition is of the matrix norm induced by the L_2
>> vector norm then you have theorem that it has an explicit formulae
>> which is the same as the spectral radius.
>
> I can't see why the norm
>
> ||A|| = max_{x in R^N} ||Ax||/||x|| (1)
>
> should be constrained to the L2 norm. Restricting the
> discussion to symmetric A, we find some Ax = vx where
> (v,x) is the maximum eigenvalue v with eigenvector x for A.

Each vector norm yields its own induced matrix norm. There are
more matrix norms than there are vector norms. Only some matrix
norms are induced by vector norms.

> Regardless of what norm one uses,
>
> ||Ax|| = ||vx|| = v||x|| (2)
>
> the latter equality following directly from the usual axioms
> of norms. Since (2) is independent of the norm used,
> (1) is also independent of the norms used, and is
> thus not restricted to the L2 norm.

The max may not occur on an eigenvector under some of the other
norms. That (2) implies (1) does not hold because of this technical
issue.

Eigenvectors are interesting under L_2 but less interesting
under other vector norms.

If you give a definition of an induced matrix norm starting from L_2
thne you can get to the spectral radius.

If you start from L_1, for example, then you will not get there as that
induced norm has a different form. If you want to get to the spectral
radius then it matters which vector norm you chose as the starting point.

>> If your first definition is of the spectral radius then you need to
>> prove the theorem that you have indeed defined an induced norm. And
>> that the vector norm is L_2.
>
> The proof consists of checking whether the expression
> satisfies the axioms of the norm. There aren't too many
> to check; I think there are 4 (it's been a few years since I
> last read about these things).

Exactly. But if you claim to be able to have both as definitions of
being matrix norms then you get into logic problems as the original
poster did. He tried to prove a result by assuming it. It was a
true result that he had learned in the past. But asserting it as a
definition did not help him with the logic of the course he was trying
to study.

> Rune

I wrote:

> 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.

Randy,

have you come up with a counter-example that you can use to demonstrate
that a) is not sufficient to prove b) in your homework? The
counter-example must be a (necessarily non-symmetric) matrix A, some
positive integer k and a vector x such that x is an eigenvector of A^k
but not of A ...

Also, do you see why the above statement has to be true for symmetric
matrices?

Regards,
Andor

"Andor" <[email protected]> writes:

> I wrote:
>
>> 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.
>
> Randy,
>
> have you come up with a counter-example that you can use to demonstrate
> that a) is not sufficient to prove b) in your homework? The
> counter-example must be a (necessarily non-symmetric) matrix A, some
> positive integer k and a vector x such that x is an eigenvector of A^k
> but not of A ...
>
> Also, do you see why the above statement has to be true for symmetric
> matrices?

Hi Andor,

By above do you mean the text you quoted from yourself? Presuming that
is the case, then I don't see that it's relevent.

What we need to show in part b is that if l is the maximum eigenvalue
magnitude of A^T*A = A^2, then +sqrt(l) is the maximum eigenvalue
magnitude of A.

What you have shown is that if k is an eigenvalue of A, then k^2 is an
eigenvalue of A^2. What if there is another eigenvalue of A^2 that
isn't an eigenvalue of A and which has magnitude that is greater
than or equal to the magnitude of k^2?

I see this as two different directions of a logic statement. One is
"if p then q"; the other is "if q then p".

Therefore I don't see that these two statements are equivalent. I also
don't see that the eigenvector is directly relevent. What we need for
the problem is the eigenvalues to be related, regardless of their
eigenvectors.

Am I wrong? If so, please show me where my error is.
--
% 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."
%%%% <[email protected]> % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

Randy Yates <[email protected]> writes:

> What if there is another eigenvalue of A^2 that
> isn't an eigenvalue of A

I should have written, "what if there is another eigenvalue of
A^2 that isn't k^2...".
--
% Randy Yates % "And all that I can do
%% Fuquay-Varina, NC % is say I'm sorry,
%%% 919-577-9882 % that's the way it goes..."
%%%% <[email protected]> % Getting To The Point', *Balance of Power*, ELO
http://home.earthlink.net/~yatescr

Randy Yates wrote:
> "Andor" <andor.bari...@gmail.com> writes:
> > I wrote:
>
> >> 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.
>
> > Randy,
>
> > have you come up with a counter-example that you can use to demonstrate
> > that a) is not sufficient to prove b) in your homework? The
> > counter-example must be a (necessarily non-symmetric) matrix A, some
> > positive integer k and a vector x such that x is an eigenvector of A^k
> > but not of A ...
>
> > Also, do you see why the above statement has to be true for symmetric
> > matrices?
>
> Hi Andor,
>
> By above do you mean the text you quoted from yourself?

Yes.

> Presuming that is the case, then I don't see that it's relevent.

>From the extended statement of part a (extended to include my remark),
the statements of part b.) follow immediately. I thought that was the
route you wanted to take for your proofs. If that isn't the case, I
must have misunderstood you. Apologies.

Regards,
Andor

"Andor" <[email protected]> writes:

> Randy Yates wrote:
>> "Andor" <andor.bari...@gmail.com> writes:
>> > I wrote:
>>
>> >> 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.
>>
>> > Randy,
>>
>> > have you come up with a counter-example that you can use to demonstrate
>> > that a) is not sufficient to prove b) in your homework? The
>> > counter-example must be a (necessarily non-symmetric) matrix A, some
>> > positive integer k and a vector x such that x is an eigenvector of A^k
>> > but not of A ...
>>
>> > Also, do you see why the above statement has to be true for symmetric
>> > matrices?
>>
>> Hi Andor,
>>
>> By above do you mean the text you quoted from yourself?
>
> Yes.
>
>> Presuming that is the case, then I don't see that it's relevent.
>
>>From the extended statement of part a (extended to include my remark),
> the statements of part b.) follow immediately.

I just listed the reasons (in the material from my previous post,
which you have omitted) why I think it DOES NOT follow immediately from
the first paragraph above in the quoted text.

Am I not making sense? Am I not following you?

Maybe I'm having a problem in knowing which pieces and parts you are
referring to among various posts. If you don't mind, Andor, can you
write a new post which makes the proof of part b) in an explicit and
orderly way? In other words, can you please spell it out for me?

Sorry if I'm being dense.
--
% Randy Yates % "Though you ride on the wheels of tomorrow,
%% Fuquay-Varina, NC % you still wander the fields of your
%%% 919-577-9882 % sorrow."
%%%% <[email protected]> % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr