[R] Matrix inversion-different answers from LAPACK and LINPACK

Avraham.Adler at guycarp.com Avraham.Adler at guycarp.com
Wed Jun 17 19:10:13 CEST 2009


Thank you VERY much, that was fantastic. I wish I understood WHY your
suggestion works.

To extend that to an n-by-n square matrix, the proper procedure would be
(example, n=5 - most I would usually use (mixture of lognormals)):

Hinv <- matrix(NA, 5, 5)
Hinv[, 1] <- solve(qr(PLLH, LAPACK=TRUE), c(1,0,0,0,0))
Hinv[, 2] <- solve(qr(PLLH, LAPACK=TRUE), c(0,1,0,0,0))
Hinv[, 3] <- solve(qr(PLLH, LAPACK=TRUE), c(0,0,1,0,0))
Hinv[, 4] <- solve(qr(PLLH, LAPACK=TRUE), c(0,0,0,1,0))
Hinv[, 5] <- solve(qr(PLLH, LAPACK=TRUE), c(0,0,0,0,1))

I'm glad that I know the size of the matrix a priori

Also, is this something of which the development list shou;d be made aware?

Once again, thank you very much!

--Avraham



                                                                           
             "Ravi Varadhan"                                               
             <RVaradhan at jhmi.e                                             
             du>                                                        To 
                                       <Avraham.Adler at guycarp.com>,        
             06/17/2009 12:56          <r-help at r-project.org>              
             PM                                                         cc 
                                                                           
                                                                   Subject 
                                       RE: [R] Matrix inversion-different  
                                       answers from LAPACK and LINPACK     
                                                                           
                                                                           
                                                                           
                                                                           
                                                                           
                                                                           




Avraham,

You can make LAPACK work by doing the following:

Hinv[, 1] <- solve(qr(PLLH, LAPACK=TRUE), c(1,0))
Hinv[, 2] <- solve(qr(PLLH, LAPACK=TRUE), c(0,1))

Here is an example:

H <- matrix(runif(4), 2, 2)
H <- H + t(H)
Hinv <- solve(qr(H))  # this is the correct inverse from LINPACK

Hinv1 <- matrix(NA, 2, 2)
Hinv1[, 1] <- solve(qr(H, LAPACK=TRUE), c(1,0))
Hinv1[, 2] <- solve(qr(H, LAPACK=TRUE), c(0,1))
Hinv2 <- solve(qr(H, LAPACK=TRUE))  # this won't work, as you found out!

             all.equal(Hinv, Hinv1)
             all.equal(Hinv, Hinv2)


Hope this helps,
Ravi.

----------------------------------------------------------------------------

-------

Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: rvaradhan at jhmi.edu

Webpage:
http://www.jhsph.edu/agingandhealth/People/Faculty_personal_pages/Varadhan.h

tml



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--------


-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On
Behalf Of Avraham.Adler at guycarp.com
Sent: Wednesday, June 17, 2009 11:38 AM
To: r-help at r-project.org
Subject: [R] Matrix inversion-different answers from LAPACK and LINPACK


Hello.

I am trying to invert a matrix, and I am finding that I can get different
answers depending on whether I set LAPACK true or false using "qr". I had
understood that LAPACK is, in general more robust and faster than LINPACK,
so I am confused as to why I am getting what seems to be invalid answers.
The matrix is ostensibly the Hessian for a function I am optimizing. I want
to get the parameter correlations, so I need to invert the matrix. There
are
times where the standard "solve(X)" returns an error, but "solve(qr(X,
LAPACK=TRUE))" returns values. However, there are times, where the latter
returns what seems to be bizarre results.

For example, an example matrix is PLLH (Pareto LogLikelihood Hessian)

                      alpha                    theta
alpha 1144.6262175141619082 -0.012907777205604828788
theta   -0.0129077772056048  0.000000155437688485563

Running plain "solve (PLLH)" or "solve (qr(PLLH))" returns:

                       [,1]                  [,2]
alpha    0.0137466171688024      1141.53956787721
theta 1141.5395678772131305 101228592.41439932585

However, running "solve(qr(PLLH, LAPACK=TRUE)) returns:

                      [,1]                  [,2]
[1,]    0.0137466171688024    0.0137466171688024
[2,] 1141.5395678772131305 1141.5395678772131305

which seems wrong as the original matrix had identical entries on the off
diagonals.

I am neither a programmer nor an expert in matrix calculus, so I do not
understand why I should be getting different answers using different
libraries to perform the ostensibly same function. I understand the
extremely small value of d²X/d(theta)² (PLLH[2,2]) may be contributing to
the error, but I would appreciate confirmation, or correction, from the
experts on this list.

Thank you very much,

--Avraham Adler



PS: For completeness, the QR decompositions per "R" under LINPACK and
LAPACK
are shown below:

> qr(PLLH)
$qr
                          alpha                     theta
alpha -1144.6262175869414932095 0.01290777720653695122277
theta    -0.0000112768491646264 0.00000000987863187747112

$rank
[1] 2

$qraux
[1] 1.99999999993641619511209 0.00000000987863187747112

$pivot
[1] 1 2

attr(,"class")
[1] "qr"
>

> qr(PLLH, LAPACK=TRUE)
$qr
                           alpha                     theta
alpha -1144.62621758694149320945 0.01290777720653695122277
theta    -0.00000563842458249248 0.00000000987863187747112

$rank
[1] 2

$qraux
[1] 1.99999999993642 0.00000000000000

$pivot
[1] 1 2

attr(,"useLAPACK")
[1] TRUE
attr(,"class")
[1] "qr"
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