FPGA Central

Stated in the form of a homework problem, but it's something I'm working
on for real:

Consider the similarity transform

Q * A * Q' = B (where Q' = transpose of Q)

where A is any real square matrix, and Q is unitary.

This is linear in A, so in general we know there must be a matrix R that
satisfies

R * A = B.

So -- is there a way to find R given Q? A general formula, perhaps?
Even better, one that's expressed in matrix form? Does anyone have a
suggestion for search terms that I might Google? Something more specific
than "linear algebra", which is going to snow me under with stuff I
already know?

In this case the matrix A is orthogonal, although not orthonormal, if
that restriction helps my cause.

(I'd submit this to the applied math group, but when I checked it
recently it had turned into a venue for flame wars -- any sensible
questions would just go up in smoke).

--
www.wescottdesign.com

On Nov 10, 12:12 pm, Tim Wescott <t...@seemywebsite.com> wrote:
> Stated in the form of a homework problem, but it's something I'm working
> on for real:
>
> Consider the similarity transform
>
> Q * A * Q' = B (where Q' = transpose of Q)
>
> where A is any real square matrix, and Q is unitary.
>
> This is linear in A, so in general we know there must be a matrix R that
> satisfies
>
> R * A = B.
>
> So -- is there a way to find R given Q? A general formula, perhaps?
> Even better, one that's expressed in matrix form? Does anyone have a
> suggestion for search terms that I might Google? Something more specific
> than "linear algebra", which is going to snow me under with stuff I
> already know?
>
> In this case the matrix A is orthogonal, although not orthonormal, if
> that restriction helps my cause.
>
> (I'd submit this to the applied math group, but when I checked it
> recently it had turned into a venue for flame wars -- any sensible
> questions would just go up in smoke).
>
> --www.wescottdesign.com

Tim, maybe I am missing something but...

If A is orthogonal, it must be full rank and hence invertible.

Thus, R = B*inv(A) satisfies your requirement.

In terms of just Q and A, that would be:
R = Q*A*conj(Q)*inv(A)

Are you trying to get a closed form without inverting A?

On Nov 10, 8:01*pm, Dilip Warrier <dili...@yahoo.com> wrote:
> On Nov 10, 12:12 pm, Tim Wescott <t...@seemywebsite.com> wrote:
> > Consider the similarity transform
>
> > Q * A * Q' = B (where Q' = transpose of Q)
>
> > where A is any real square matrix, and Q is unitary.
>
> > This is linear in A, so in general we know there must be a matrix R that
> > satisfies
>
> > R * A = B.
>
...
> Tim, maybe I am missing something but...
>
> If A is orthogonal, it must be full rank and hence invertible.
>
> Thus, R = B*inv(A) satisfies your requirement.
>
> In terms of just Q and A, that would be:
> R = Q*A*conj(Q)*inv(A)

If Q * A * Q' = B is "linear in A" I would take this as B is a
function of A, B(A) = QAQ', and this function is supposed to be
representable in the form of B(A) = RA for some R depending only on Q.

I don't see why this should be possible (not saying it isn't, I just
can't see why it should). Actually I would put some effort to find a
counter example first. That could bring some light in.

On Tue, 10 Nov 2009 10:01:30 -0800, Dilip Warrier wrote:

> On Nov 10, 12:12 pm, Tim Wescott <t...@seemywebsite.com> wrote:
>> Stated in the form of a homework problem, but it's something I'm
>> working on for real:
>>
>> Consider the similarity transform
>>
>> Q * A * Q' = B (where Q' = transpose of Q)
>>
>> where A is any real square matrix, and Q is unitary.
>>
>> This is linear in A, so in general we know there must be a matrix R
>> that satisfies
>>
>> R * A = B.
>>
>> So -- is there a way to find R given Q? A general formula, perhaps?
>> Even better, one that's expressed in matrix form? Does anyone have a
>> suggestion for search terms that I might Google? Something more
>> specific than "linear algebra", which is going to snow me under with
>> stuff I already know?
>>
>> In this case the matrix A is orthogonal, although not orthonormal, if
>> that restriction helps my cause.
>>
>> (I'd submit this to the applied math group, but when I checked it
>> recently it had turned into a venue for flame wars -- any sensible
>> questions would just go up in smoke).
>>
>> --www.wescottdesign.com
>
> Tim, maybe I am missing something but...
>
> If A is orthogonal, it must be full rank and hence invertible.
>
> Thus, R = B*inv(A) satisfies your requirement.
>
> In terms of just Q and A, that would be: R = Q*A*conj(Q)*inv(A)
>
> Are you trying to get a closed form without inverting A?

I want to know R (a) without having to compute B for a given A, and (b)
in a general, symbolic form.

--
www.wescottdesign.com

On Nov 10, 8:25*pm, "Mr.Capsicum" <mr.capsi...@gmail.com> wrote:
> If Q * A * Q' = B is "linear in A" I would take this as B is a
> function of A, B(A) = QAQ', and this function is supposed to be
> representable in the form of B(A) = RA for some R depending only on Q.
>
> I don't see why this should be possible (not saying it isn't, I just
> can't see why it should). Actually I would put some effort to find a
> counter example first. That could bring some light in.

I poked around a bit and it seems that if A = [a,b;c,d] and Q in R^
{2x2} then every element of product QAQ' depends on all of a, b, c and
d thus its impossible to find R which would give the same result and
would depend only on Q.

Is there a reason to think that QAQ' is linear?

On Nov 10, 1:51 pm, Tim Wescott <t...@seemywebsite.com> wrote:
> On Tue, 10 Nov 2009 10:01:30 -0800, Dilip Warrier wrote:
> > On Nov 10, 12:12 pm, Tim Wescott <t...@seemywebsite.com> wrote:
> >> Stated in the form of a homework problem, but it's something I'm
> >> working on for real:
>
> >> Consider the similarity transform
>
> >> Q * A * Q' = B (where Q' = transpose of Q)
>
> >> where A is any real square matrix, and Q is unitary.
>
> >> This is linear in A, so in general we know there must be a matrix R
> >> that satisfies
>
> >> R * A = B.
>
> >> So -- is there a way to find R given Q? A general formula, perhaps?
> >> Even better, one that's expressed in matrix form? Does anyone have a
> >> suggestion for search terms that I might Google? Something more
> >> specific than "linear algebra", which is going to snow me under with
> >> stuff I already know?
>
> >> In this case the matrix A is orthogonal, although not orthonormal, if
> >> that restriction helps my cause.
>
> >> (I'd submit this to the applied math group, but when I checked it
> >> recently it had turned into a venue for flame wars -- any sensible
> >> questions would just go up in smoke).
>
> >> --www.wescottdesign.com
>
> > Tim, maybe I am missing something but...
>
> > If A is orthogonal, it must be full rank and hence invertible.
>
> > Thus, R = B*inv(A) satisfies your requirement.
>
> > In terms of just Q and A, that would be: R = Q*A*conj(Q)*inv(A)
>
> > Are you trying to get a closed form without inverting A?
>
> I want to know R (a) without having to compute B for a given A, and (b)
> in a general, symbolic form.
>
> --www.wescottdesign.com

OK, I think I understand now. I believe the flaw, as pointed out by
the other poster, is in assuming that if a function f(A) is linear in
the matrix A, then it can be represented as the multiplication of
another matrix R with A.
Thus, f(A) = R*A does not hold for all linear f.

As a counterexample, consider g(A) = transpose(A). This is linear in A
since g(alpha*A) = alpha*g(A) where alpha is a scalar and g(A+B) = g
(A) + g(B).

If there existed an R for g as assumed above, then using the special
case of the identity matrix,
g(I) = R*I = transpose(I) where I is an identity matrix.

Solving, you have R = I. i.e. g(A) = A. But, A = transpose(A) is
clearly incorrect for any non-symmetric matrix.

On Tue, 10 Nov 2009 11:01:02 -0800, Mr.Capsicum wrote:

> On Nov 10, 8:25Â*pm, "Mr.Capsicum" <mr.capsi...@gmail.com> wrote:
>> If Q * A * Q' = B is "linear in A" I would take this as B is a function
>> of A, B(A) = QAQ', and this function is supposed to be representable in
>> the form of B(A) = RA for some R depending only on Q.
>>
>> I don't see why this should be possible (not saying it isn't, I just
>> can't see why it should). Actually I would put some effort to find a
>> counter example first. That could bring some light in.
>
> I poked around a bit and it seems that if A = [a,b;c,d] and Q in R^
> {2x2} then every element of product QAQ' depends on all of a, b, c and d
> thus its impossible to find R which would give the same result and would
> depend only on Q.
>
> Is there a reason to think that QAQ' is linear?

Apparently not in general!

I was thinking specifically in terms of matrix representations of
quaternions, and the direction cosine matrix that you can make from
them. In this case the expression _is_ linear, but there are some pretty
severe restrictions on A, which appear to be necessary.

Sigh. It looks like I'm going to have to crank through this element-by-
element. I was hoping to avoid that.

--
www.wescottdesign.com

>On Tue, 10 Nov 2009 11:01:02 -0800, Mr.Capsicum wrote:
>
>> On Nov 10, 8:25Â*pm, "Mr.Capsicum" <mr.capsi...@gmail.com> wrote:
>>> If Q * A * Q' = B is "linear in A" I would take this as B is
function
>>> of A, B(A) = QAQ', and this function is supposed to be representabl
in
>>> the form of B(A) = RA for some R depending only on Q.
>>>
>>> I don't see why this should be possible (not saying it isn't, I just
>>> can't see why it should). Actually I would put some effort to find a
>>> counter example first. That could bring some light in.
>>
>> I poked around a bit and it seems that if A = [a,b;c,d] and Q in R^
>> {2x2} then every element of product QAQ' depends on all of a, b, c an
d
>> thus its impossible to find R which would give the same result an
would
>> depend only on Q.
>>
>> Is there a reason to think that QAQ' is linear?
>
>Apparently not in general!
>
>I was thinking specifically in terms of matrix representations of
>quaternions, and the direction cosine matrix that you can make from
>them. In this case the expression _is_ linear, but there are some prett

>severe restrictions on A, which appear to be necessary.

Could you give more details on your Q and A? It does seem some of you
questions are phrased with the intent to have broader appeal, but tha
makes them more abstract than they need to be, and therefore more difficul
to understand what you're really after. As pointed out in Dilip'
counterexample, the fact that Q*A*Q' is linear in A doesn't mean you ca
represent it as R*A, with R a function only of Q, in the general case.
Specific details seem crucial here.

On Tue, 10 Nov 2009 15:37:56 -0600, Michael Plante wrote:

>>On Tue, 10 Nov 2009 11:01:02 -0800, Mr.Capsicum wrote:
>>
>>> On Nov 10, 8:25Â*pm, "Mr.Capsicum" <mr.capsi...@gmail.com> wrote:
>>>> If Q * A * Q' = B is "linear in A" I would take this as B is a
> function
>>>> of A, B(A) = QAQ', and this function is supposed to be representable
> in
>>>> the form of B(A) = RA for some R depending only on Q.
>>>>
>>>> I don't see why this should be possible (not saying it isn't, I just
>>>> can't see why it should). Actually I would put some effort to find a
>>>> counter example first. That could bring some light in.
>>>
>>> I poked around a bit and it seems that if A = [a,b;c,d] and Q in R^
>>> {2x2} then every element of product QAQ' depends on all of a, b, c and
> d
>>> thus its impossible to find R which would give the same result and
> would
>>> depend only on Q.
>>>
>>> Is there a reason to think that QAQ' is linear?
>>
>>Apparently not in general!
>>
>>I was thinking specifically in terms of matrix representations of
>>quaternions, and the direction cosine matrix that you can make from
>>them. In this case the expression _is_ linear, but there are some
>>pretty
>
>>severe restrictions on A, which appear to be necessary.
>
> Could you give more details on your Q and A? It does seem some of your
> questions are phrased with the intent to have broader appeal, but that
> makes them more abstract than they need to be, and therefore more
> difficult to understand what you're really after. As pointed out in
> Dilip's counterexample, the fact that Q*A*Q' is linear in A doesn't mean
> you can represent it as R*A, with R a function only of Q, in the general
> case. Specific details seem crucial here.


I was hoping that there _was_ a broader solution, but it appears that is
not the case. In fact, it appears that the whole thing doesn't fly.

In quaternion math, you can rotate a real 3-D vector arbitrarily by using
it as the imaginary part of a quaternion, then multiplying it fore and
aft by a unit-length rotation quaternion and it's conjugate. The result
is guaranteed to be rotated in space, not rendered into 4-D, and not have
it's length changed.

Since quaternions can be expressed as matrices, this is where the QAQ'
came from.

You can also multiply that 3-D vector by a direction cosine matrix, b =
Ra.

Where I erred was in thinking that after the real 3-D vector is made into
a quaternion and then a matrix, that the matrix operation RA (i.e. some
analog of the DCM times the matrixified, quaternionified, 3-D vector)
would hold -- it doesn't, so I didn't have to waste any more time
figuring out how to find a nonexistent operation.

Now I'm grinding through this stuff element by element, making progress,
and being grateful that there are only 3 spatial dimensions (in common
usage at least), which holds down the number of individual terms I need
to juggle.

--
www.wescottdesign.com

In article <[email protected]>,
Tim Wescott <[email protected]> wrote:

> I was hoping that there _was_ a broader solution, but it appears that is
> not the case. In fact, it appears that the whole thing doesn't fly.

Hi Tim,

I'm very curious about the original problem. You seem to be trying to
avoid computing something -- and that's a laudable goal... but you seem
to be asking for help with attempted solutions.

I'd really like to know what the starting point was. I may bring a
different point of view to the problem.

(Of course, if you're planning to patent the solution, you can't very
well tell us.)

vale,
rip

--
email address is r i p 1 AT c o m c a s t DOT n e t

On Tue, 10 Nov 2009 16:36:23 -0800, richard i pelletier wrote:

> In article <[email protected]>,
> Tim Wescott <[email protected]> wrote:
>
>> I was hoping that there _was_ a broader solution, but it appears that
>> is not the case. In fact, it appears that the whole thing doesn't fly.
>
> Hi Tim,
>
> I'm very curious about the original problem. You seem to be trying to
> avoid computing something -- and that's a laudable goal... but you seem
> to be asking for help with attempted solutions.
>
> I'd really like to know what the starting point was. I may bring a
> different point of view to the problem.
>
> (Of course, if you're planning to patent the solution, you can't very
> well tell us.)
>
> vale,
> rip

I'm working on an unscented Kalman filter. It works great, but it's
sloooooooow. The textbook unscented Kalman filter has you calculating
the difference equation over and over again; I'm trying to get all the
matrices in symbolic form so I can (hopefully) speed up the calculation.

--
www.wescottdesign.com

In article <[email protected]>,
Tim Wescott <[email protected]> wrote:

> I'm working on an unscented Kalman filter. It works great, but it's
> sloooooooow. The textbook unscented Kalman filter has you calculating
> the difference equation over and over again; I'm trying to get all the
> matrices in symbolic form so I can (hopefully) speed up the calculation.

Thanks. Alas, it's not at my fingertips, but let me think about it.

vale,
rip

--
email address is r i p 1 AT c o m c a s t DOT n e t