On Aug 9, 11:36�pm, Rune Allnor <all...@tele.ntnu.no> wrote:> On 10 Aug, 03:13, HardySpicer <gyansor...@gmail.com> wrote: > > > On Aug 10, 12:47�pm, Gordon Sande <g.sa...@worldnet.att.net> wrote: > > > All you can say is that given B and a nonrandom algorithm you will > > > get the same A every time. > > > > > Hardy > > > Yes exactly - that was my original question. So I expect there is no > > solution that is unique. > > Rememnber, B = A'A is the matrix equivalent of an autocorrelation > sequence. There is no way to recover the original sequence x[n] > from its autocorrelation rxx[k]. > > The best you can do is to impose additional constraints like > 'zero phase', 'minimum phase' or something like that. > > RuneThat;s right for sure. You can however factor a matrix Laurent series into a unique spectral-factor polynomial matrix ie A(z^-1)A'(z) = Matrix Laurent series in + and - powers of z. This too is an autocorrelation. Hardy
Matrix Problem
Started by ●August 9, 2009
Reply by ●August 10, 20092009-08-10
Reply by ●August 10, 20092009-08-10
On 10 Aug, 09:44, HardySpicer <gyansor...@gmail.com> wrote:> On Aug 9, 11:36�pm, Rune Allnor <all...@tele.ntnu.no> wrote: > > > > > > > On 10 Aug, 03:13, HardySpicer <gyansor...@gmail.com> wrote: > > > > On Aug 10, 12:47�pm, Gordon Sande <g.sa...@worldnet.att.net> wrote: > > > > All you can say is that given B and a nonrandom algorithm you will > > > > get the same A every time. > > > > > > Hardy > > > > Yes exactly - that was my original question. So I expect there is no > > > solution that is unique. > > > Rememnber, B = A'A is the matrix equivalent of an autocorrelation > > sequence. There is no way to recover the original sequence x[n] > > from its autocorrelation rxx[k]. > > > The best you can do is to impose additional constraints like > > 'zero phase', 'minimum phase' or something like that. > > > Rune > > That;s right for sure. You can however factor a matrix Laurent series > into a unique spectral-factor polynomial matrix ie > > A(z^-1)A'(z) = Matrix Laurent series in + and - powers of z. This too > is an autocorrelation.I don't know or remember the technicalities of Laurent series, so I will not comment on the details. However, don't be surprised if the same argument can be used all over, that one forms a matrix C = TA and the z and z^-1 contributions from T then cancel to produce the same Laurent series as for A. Then consider this: If there was any way *whatsoever* to recover the data from the autocorrelation, all the textbooks would at least mention it. No matter how technically involved the procedure itself might be. Remember, this might be the first time *you* face this sort of problem, but the problem is as old as the field of mathemathical data analysis itself. People started searching for such kinds of methods maybe 100 years ago. No such method has been found, which indicates one of two things: - No such method exists. - Such a method exists but can not be found with the mathemathical tools people have tried so far. Rune
Reply by ●August 10, 20092009-08-10
Rune Allnor <allnor@tele.ntnu.no> writes:> On 10 Aug, 03:13, HardySpicer <gyansor...@gmail.com> wrote: >> On Aug 10, 12:47 pm, Gordon Sande <g.sa...@worldnet.att.net> wrote: > >> > All you can say is that given B and a nonrandom algorithm you will >> > get the same A every time. >> >> > > Hardy >> >> Yes exactly - that was my original question. So I expect there is no >> solution that is unique. > > Rememnber, B = A'A is the matrix equivalent of an autocorrelation > sequence. There is no way to recover the original sequence x[n] > from its autocorrelation rxx[k].Right. In particular, if you don't know how many rows are in A, you can't even recover that. Scott -- Scott Hemphill hemphill@alumni.caltech.edu "This isn't flying. This is falling, with style." -- Buzz Lightyear
Reply by ●August 10, 20092009-08-10
HardySpicer <gyansorova@gmail.com> writes:> On Aug 10, 12:47 pm, Gordon Sande <g.sa...@worldnet.att.net> wrote: >> On 2009-08-09 20:13:50 -0300, HardySpicer <gyansor...@gmail.com> said: >> >> >> >> > On Aug 10, 10:50 am, Gordon Sande <g.sa...@worldnet.att.net> wrote: >> >> On 2009-08-09 17:38:35 -0300, HardySpicer <gyansor...@gmail.com> said: >> >> >>> On Aug 10, 8:16 am, Gordon Sande <g.sa...@worldnet.att.net> wrote: >> >>>> On 2009-08-09 16:19:42 -0300, HardySpicer <gyansor...@gmail.com> said: >> >> >>>>> I have a 2X2 matrix >> >> >>>>> A=[1 2 >> >>>>> 3 7] >> >> >>>>> and I form B=AA' where ' is transpose. This B is symmetric of cours >> > e >> >>>>> though A is not. >> >>>>> I now decompose B into >> >> >>>>> B=USV where S is the diagonal eigenvalue matrix and U and V are >> >>>>> orthogonal matrices. >> >> >>>>> Now I find U.(S^0.5). since the square root of a diagonal matrix S is >> >>>>> a diagonal matrix. I do not get back to A though. I know the >> >>>>> decomposition is right since when I work out >> >>>>> USV I get back to B. >> >> >>>>> How to get back to A given B? Cholesky factorization doesn't work >> >>>>> either since it gives a symmetric answer. >> >>>>> You can multiply the result from a Cholesky factorization via a >> >>>>> Unitary matrix but how do you find such a matrix? >> >> >>>>> Hardy >> >> >>>> Since this is an elementary homework style question all you can expect >> >>>> are hints. So here are two. >> >> >>>> 1. How does B change when A is multiplied by a 2x2 orthogonal matrix? >> >> >>>> 2. Why would one think the first question is an interesting question? >> >> >>> 1. B stays the same. >> >>> Are you saying the answer is not unique? >> >> >> Yes. A and AQ give the same B so either is a solution. >> >> And you can use any Q you might care to. >> >> >>> Actually I am a bit old for >> >>> homework! >> >>> There is only one unique A that gives B but thereare many other >> >>> symmetric matrices say C that give B ie CC'=B where C=AX and X is >> >>> unitary. >> >> >>> Am I closer? >> >> > Yes - trouble is I need A. >> >> Then you need to find some way of deciding which Q in AQ will be >> the one which meets you needs. Have you forgotten to specify some >> additional property that will choose Q? Or will any Q actually do >> the job and you have not realized it? >> >> All you can say is that given B and a nonrandom algorithm you will >> get the same A every time. >> >> > Hardy > > Yes exactly - that was my original question. So I expect there is no > solution that is unique.There is no unique solution unless you impose additional constraints. For example, since B is a symmetric, positive definite matrix, there is a unique symmetric, positive definite square root: [ 6 17 17 59 ] / sqrt(65) Scott -- Scott Hemphill hemphill@alumni.caltech.edu "This isn't flying. This is falling, with style." -- Buzz Lightyear
Reply by ●August 10, 20092009-08-10
On 2009-08-10 07:14:49 -0300, Rune Allnor <allnor@tele.ntnu.no> said:> On 10 Aug, 09:44, HardySpicer <gyansor...@gmail.com> wrote: >> On Aug 9, 11:36�pm, Rune Allnor <all...@tele.ntnu.no> wrote: >> >> >> >> >> >>> On 10 Aug, 03:13, HardySpicer <gyansor...@gmail.com> wrote: >> >>>> On Aug 10, 12:47�pm, Gordon Sande <g.sa...@worldnet.att.net> wrote: >>>>> All you can say is that given B and a nonrandom algorithm you will >>>>> get the same A every time. >> >>>>>> Hardy >> >>>> Yes exactly - that was my original question. So I expect there is no >>>> solution that is unique. >> >>> Rememnber, B = A'A is the matrix equivalent of an autocorrelation >>> sequence. There is no way to recover the original sequence x[n] >>> from its autocorrelation rxx[k]. >> >>> The best you can do is to impose additional constraints like >>> 'zero phase', 'minimum phase' or something like that. >> >>> Rune >> >> That;s right for sure. You can however factor a matrix Laurent series >> into a unique spectral-factor polynomial matrix ie >> >> A(z^-1)A'(z) = Matrix Laurent series in + and - powers of z. This too >> is an autocorrelation. > > I don't know or remember the technicalities of Laurent > series, so I will not comment on the details. However, > don't be surprised if the same argument can be used > all over, that one forms a matrix C = TA and the z > and z^-1 contributions from T then cancel to produce > the same Laurent series as for A. > > Then consider this: If there was any way *whatsoever* > to recover the data from the autocorrelation, all the > textbooks would at least mention it. No matter how > technically involved the procedure itself might be.In one dimension the problem is often called the zero flipping problem and is tied to the fundemental theoren of algebra. In two and hihger dimensions the problem goes away as there is no corresponding fundemental theorem although there are occassional (sets of measure zero) degeneracies that lead to limited ambiguities.> Remember, this might be the first time *you* face this > sort of problem, but the problem is as old as the field > of mathemathical data analysis itself. People started > searching for such kinds of methods maybe 100 years ago. > > No such method has been found, which indicates one of > two things: > > - No such method exists. > - Such a method exists but can not be found with the > mathemathical tools people have tried so far. > > Rune
Reply by ●August 10, 20092009-08-10
On 10 Aug, 14:39, Gordon Sande <g.sa...@worldnet.att.net> wrote:> On 2009-08-10 07:14:49 -0300, Rune Allnor <all...@tele.ntnu.no> said:> > Then consider this: If there was any way *whatsoever* > > to recover the data from the autocorrelation, all the > > textbooks would at least mention it. No matter how > > technically involved the procedure itself might be. > > In one dimension the problem is often called the zero > flipping problem and is tied to the fundemental theoren of > algebra. In two and hihger dimensions the problem goes away > as there is no corresponding fundemental theorem although there > are occassional (sets of measure zero) degeneracies that lead > to limited ambiguities.If this is what I think it is, the algorithm requires that the sequence complies to an ARMA(p,q) model where bot p and q are known. Not all sequences comply to an ARMA model. Even if the noiseless data do, the parameters might not be accurately estimated in the presence of additive noise. Rune
Reply by ●August 10, 20092009-08-10
On Aug 10, 3:14�am, Rune Allnor <all...@tele.ntnu.no> wrote:> On 10 Aug, 09:44, HardySpicer <gyansor...@gmail.com> wrote: > > > > > On Aug 9, 11:36�pm, Rune Allnor <all...@tele.ntnu.no> wrote: > > > > On 10 Aug, 03:13, HardySpicer <gyansor...@gmail.com> wrote: > > > > > On Aug 10, 12:47�pm, Gordon Sande <g.sa...@worldnet.att.net> wrote: > > > > > All you can say is that given B and a nonrandom algorithm you will > > > > > get the same A every time. > > > > > > > Hardy > > > > > Yes exactly - that was my original question. So I expect there is no > > > > solution that is unique. > > > > Rememnber, B = A'A is the matrix equivalent of an autocorrelation > > > sequence. There is no way to recover the original sequence x[n] > > > from its autocorrelation rxx[k]. > > > > The best you can do is to impose additional constraints like > > > 'zero phase', 'minimum phase' or something like that. > > > > Rune > > > That;s right for sure. You can however factor a matrix Laurent series > > into a unique spectral-factor polynomial matrix ie > > > A(z^-1)A'(z) = Matrix Laurent series in + and - powers of z. This too > > is an autocorrelation. > > I don't know or remember the technicalities of Laurent > series, so I will not comment on the details. However, > don't be surprised if the same argument can be used > all over, that one forms a matrix C = TA and the z > and z^-1 contributions from T then cancel to produce > the same Laurent series as for A. > > Then consider this: If there was any way *whatsoever* > to recover the data from the autocorrelation, all the > textbooks would at least mention it. No matter how > technically involved the procedure itself might be. > > Remember, this might be the first time *you* face this > sort of problem, but the problem is as old as the field > of mathemathical data analysis itself. People started > searching for such kinds of methods maybe 100 years ago. > > No such method has been found, which indicates one of > two things: > > - No such method exists. > - Such a method exists but can not be found with the > � mathemathical tools people have tried so far. > > RuneYou can make the spectral factors unique by taking out the A0 matrix term and defining A(z^-1)RA(z) where R is a positive-deinite symmetric matrix and A(0) =I. (eye). This has been known for some time. Doesn't work for a zeroth order though since there is nothing to take out!
