[SciPy-User] qr decompostion gives negative q, r ?

[Virgil Stokes]

On 2012-11-21 01:36, Virgil Stokes wrote:
> On 2012-11-21 00:29, Robert Kern wrote:
>> On Tue, Nov 20, 2012 at 11:21 PM, Virgil Stokes <vs at it.uu.se> wrote:
>>> On 2012-11-20 23:59, Robert Kern wrote:
>>>> On Tue, Nov 20, 2012 at 10:49 PM, Virgil Stokes <vs at it.uu.se> wrote:
>>>>
>>>>> Ok Skipper,
>>>>> Unfortunately, things are worse than I had hoped, numpy sometimes
>>>>> returns the negative of the q,r and other times the same as MATLAB!
>>>>> Thus, as someone has already mentioned in this discussion, the "sign"
>>>>> seems to depend on the matrix being decomposed. This could be a
>>>>> nightmare to track down.
>>>>>
>>>>> I hope that I can return to some older versions of numpy/scipy to 
>>>>> work
>>>>> around this problem until this problem is fixed. Any suggestions 
>>>>> on how
>>>>> to recover earlier versions would be appreciated.
>>>> That's not going to help you. The only thing that we guarantee (or
>>>> have *ever* guaranteed) is that the result is a valid QR
>>>> decomposition. If you need to swap signs to normalize things to your
>>>> desired convention, you will need to do that as a postprocessing step.
>>> But why do I need to normalize with numpy (at least with latest
>>> release); but not with MATLAB.
>> Because MATLAB decided to do the normalization step for you. That's a
>> valid decision. And so is ours.
>>
>>> A simple question for you.
>>>
>>> In my application MATLAB generates a sequence of QR factorizations for
>>> covariance matrices in which R is always PD --- which is corect!
>> That is not part of the definition of a QR decomposition. Failing to
>> meet that property does not make the QR decomposition incorrect.
>>
>> The only thing that is incorrect is passing an arbitrary, but valid,
>> QR decomposition to something that is expecting a strict *subset* of
>> valid QR decompositions.
> Sorry but I do not understand this...
> Let me give you an example that I believe illustrates the problem in 
> numpy
>
> I have the following matrix, A:
>
> array([[  7.07106781e+02,   5.49702852e-04,   1.66675481e-19],
>        [ -3.53553391e+01,   7.07104659e+01,   1.66675481e-19],
>        [  0.00000000e+00,  -3.97555166e+00,   7.07106781e-03],
>        [ -7.07106781e+02,  -6.48214647e-04,   1.66675481e-19],
>        [  3.53553391e+01,  -7.07104226e+01,   1.66675481e-19],
>        [  0.00000000e+00,   3.97560687e+00,  -7.07106781e-03],
>        [  0.00000000e+00,   0.00000000e+00,   0.00000000e+00],
>        [  0.00000000e+00,   0.00000000e+00,   0.00000000e+00],
>        [  0.00000000e+00,   0.00000000e+00,   0.00000000e+00]])
>
> Note, this is clearly not a covariance matrix, but, it does contain a 
> covariance matrix (3x3). I refer you to the paper for how this matrix 
> was generated.
>
> Using np.linalg.qr(A) I get the following for R (3x3) which is 
> "square-root" of the covariance matrix:
>
> array([[ -1.00124922e+03,   4.99289918e+00,   0.00000000e+00],
>        [  0.00000000e+00,  -1.00033071e+02,   5.62045938e-04],
>        [  0.00000000e+00,   0.00000000e+00,  -9.98419272e-03]])
>
> which is clearly not PD, since the it's 3 eigenvalues (diagonal 
> elements) are all negative.
>
> Now, if I use qr(A,0) in MATLAB:
>
> I get the following for R (3x3)
>
>  1001.24922,    -4.99290,      0.00000
>      0.00000,    100.03307,     -0.00056
>     -0.00000,       0.00000,       0.00998
>
> This is obviously PD, as it should be, and gives the correct results. 
> Note, it is the negative of the R obtained with numpy.
>
> I can provide other examples in which both R's obtained are the same 
> and they both lead to correct results. That is, when the R's are 
> different, the R obtained with MATLAB is always PD and always gives 
> the correct end result, while the R with numpy is not PD and does not  
> give the correct end result.
>
> I hope that this helps you to understand my problem better. If there 
> are more details that you need then let me know what, please.
Sorry, I forgot the paper (it is attached).
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