I have a 2X2 matrix
A=[1 2
3 7]
and I form B=AA' where ' is transpose. This B is symmetric of course
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
Matrix Problem
Started by ●August 9, 2009
Reply by ●August 9, 20092009-08-09
On 2009-08-09 16:19:42 -0300, HardySpicer <gyansorova@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 course > 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? > > HardySince 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?
Reply by ●August 9, 20092009-08-09
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 course > > 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? 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?
Reply by ●August 9, 20092009-08-09
On 9 Aug, 22:38, HardySpicer <gyansor...@gmail.com> wrote:> 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 course > > > though A is not....> There is only one unique A that givesWrong. If B = A'A where A is arbitrary, then there exist infinitely many unitary transforms T T'T = TT' = I such that (TA)'TA = A'T'TA = A'IA = A'A = B. Rune
Reply by ●August 9, 20092009-08-09
On Sun, 09 Aug 2009 13:38:35 -0700, HardySpicer wrote:> 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 course >> > 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? 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?I'm not sure what you meant to say, but you just disproved what appears to be your own statement: Given a matrix B the decomposition into A where B = AA' is _not_ unique. (Note that this is the case even when "matrix" B is 1x1 -- A can be either sqrt(B) or -sqrt(B)). -- http://www.wescottdesign.com
Reply by ●August 9, 20092009-08-09
On 2009-08-09 17:38:35 -0300, HardySpicer <gyansorova@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 course >>> 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?
Reply by ●August 9, 20092009-08-09
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 course > >>> 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. Hardy
Reply by ●August 9, 20092009-08-09
On 2009-08-09 20:13:50 -0300, HardySpicer <gyansorova@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
Reply by ●August 9, 20092009-08-09
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. > > > HardyYes exactly - that was my original question. So I expect there is no solution that is unique. Hardy
Reply by ●August 10, 20092009-08-10
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
