Positive Definite Matrix

Thank you for all your input but I'm afraid I dont know what the  
final conclusion is. I will have to check the the eigenvalues if any  
are negative. Why would setting them to zero make a difference?  
Sorry to drag this on.
The discussion is proceeding on the assumption that the "true"  
matrix is PD and that only because of numerical imprecision has a  
negative eigenvalue been reported. You would only decide to set the  
negative eigenvalues to zero if you had prior knowledge that the  
matrix _should_ be PD and that you needed to so something further with  
the matrix on that basis. Usually the  matrices  in question are the  
result of many calculations that may have introduced sufficient  
numerical round-off error to distort the result.
David.

>
> Thanks
>
> On Sat, Jan 29, 2011 at 9:00 PM, Prof Brian Ripley <ripley at stats.ox.ac.uk 
> > wrote:
> On Sat, 29 Jan 2011, David Winsemius wrote:
>
>
> On Jan 29, 2011, at 12:17 PM, Prof Brian Ripley wrote:
>
> On Sat, 29 Jan 2011, David Winsemius wrote:
>
>
> On Jan 29, 2011, at 10:11 AM, David Winsemius wrote:
>
> On Jan 29, 2011, at 9:59 AM, John Fox wrote:
> Dear David and Alex,
> I'd be a little careful about testing exact equality as in all(M ==  
> t(M) and
> careful as well about a test such as all(eigen(M)$values > 0) since  
> real
> arithmetic on a computer can't be counted on to be exact.
> Which was why I pointed to that thread from 2005 and the existing  
> work that had been put into packages. If you want to substitute  
> all.equal for all, there might be fewer numerical false alarms, but  
> I would think there could be other potential problems that might  
> deserve warnings.
>
> In addition to the two "is." functions cited earlier there is also a  
> "posdefify" function by Maechler in the sfsmisc package: "  
> Description : From a matrix m, construct a "close" positive definite  
> one."
>
> But again, that is not usually what you want.  There is no guarantee  
> that the result is positive-definite enough that the Cholesky  
> decomposition will work.
>
> I don't see a Cholesky decomposition method being used in that  
> function. It appears to my reading to be following what would be  
> called an eigendecomposition.
>
> Correct, but my point is that one does not usually want a
>
> "close" positive definite one
>
> but a 'square root'.
>
>
>
> -- 
> David.
>
>
> Give up on Cholesky factors unless you have a matrix you know must  
> be symmetric and strictly positive definite, and use the  
> eigendecomposition instead (setting negative eigenvalues to zero).   
> You can then work with the factorization to ensure that (for  
> example) variances are always non-negative because they are always  
> computed as sums of squares.
>
> This sort of thing is done in many of the multivariate analysis  
> calculations in R (e.g. cmdscale) and in well-designed packages.
>
>
> -- 
> David.
> On Jan 29, 2011, at 7:58 AM, David Winsemius wrote:
> On Jan 29, 2011, at 7:22 AM, Alex Smith wrote:
> Hello I am trying to determine wether a given matrix is symmetric and
> positive matrix. The matrix has real valued elements.
> I have been reading about the cholesky method and another method is
> to find the eigenvalues. I cant understand how to implement either of
> the two. Can someone point me to the right direction. I have used
> ?chol to see the help but if the matrix is not positive definite it
> comes up as error. I know how to the get the eigenvalues but how can
> I then put this into a program to check them as the just come up with
> $values.
> Is checking that the eigenvalues are positive enough to determine
> wether the matrix is positive definite?
> That is a fairly simple linear algebra fact that googling or pulling
> out a standard reference should have confirmed.
> Just to be clear (since on the basis of some off-line communications  
> it
> did not seem to be clear):  A real, symmetric matrix is Hermitian (and
> therefore all of its eigenvalues are real). Further, it is positive-
> definite if and only if its eigenvalues are all positive.
> qwe<-c(2,-1,0,-1,2,-1,0,1,2)
> q<-matrix(qwe,nrow=3)
> isPosDef <- function(M) { if ( all(M == t(M) ) ) {  # first test
> symmetric-ity
>                            if (  all(eigen(M)$values > 0) ) {TRUE}
>                               else {FALSE} } #
>                            else {FALSE}  # not symmetric
>
>                      }
> isPosDef(q)
> [1] FALSE
> m
> [,1] [,2] [,3] [,4] [,5]
> [1,]  1.0  0.0  0.5 -0.3  0.2
> [2,]  0.0  1.0  0.1  0.0  0.0
> [3,]  0.5  0.1  1.0  0.3  0.7
> [4,] -0.3  0.0  0.3  1.0  0.4
> [5,]  0.2  0.0  0.7  0.4  1.0
> isPosDef(m)
> [1] TRUE
> You might want to look at prior postings by people more  
> knowledgeable than
> me:
> http://finzi.psych.upenn.edu/R/Rhelp02/archive/57794.html
> Or look at what are probably better solutions in available packages:
> http://finzi.psych.upenn.edu/R/library/corpcor/html/ 
> rank.condition.html
> http://finzi.psych.upenn.edu/R/library/matrixcalc/html/is.positive.definit
> e.html
> --
> David.
> this is the matrix that I know is positive definite.
> eigen(m)
> $values
> [1] 2.0654025 1.3391291 1.0027378 0.3956079 0.1971228
> $vectors
>     [,1]        [,2]         [,3]        [,4]        [,5]
> [1,] -0.32843233  0.69840166  0.080549876  0.44379474  0.44824689
> [2,] -0.06080335  0.03564769 -0.993062427 -0.01474690  0.09296096
> [3,] -0.64780034  0.12089168 -0.027187620  0.08912912 -0.74636235
> [4,] -0.31765040 -0.68827876  0.007856812  0.60775962  0.23651023
> [5,] -0.60653780 -0.15040584  0.080856897 -0.65231358  0.42123526
> and this are the eigenvalues and eigenvectors.
> I thought of using
> eigen(m,only.values=T)
> $values
> [1] 2.0654025 1.3391291 1.0027378 0.3956079 0.1971228
> $vectors
> NULL
> m <- matrix(scan(textConnection("
> 1.0  0.0  0.5 -0.3  0.2
> 0.0  1.0  0.1  0.0  0.0
> 0.5  0.1  1.0  0.3  0.7
> -0.3  0.0  0.3  1.0  0.4
> 0.2  0.0  0.7  0.4  1.0
> ")), 5, byrow=TRUE)
> #Read 25 items
> m
> [,1] [,2] [,3] [,4] [,5]
> [1,]  1.0  0.0  0.5 -0.3  0.2
> [2,]  0.0  1.0  0.1  0.0  0.0
> [3,]  0.5  0.1  1.0  0.3  0.7
> [4,] -0.3  0.0  0.3  1.0  0.4
> [5,]  0.2  0.0  0.7  0.4  1.0
> all( eigen(m)$values >0 )
> #[1] TRUE
> Then i thought of using logical expression to determine if there are
> negative eigenvalues but couldnt work. I dont know what error this is
> b<-(a<0)
> Error: (list) object cannot be coerced to type 'double'
> ??? where did "a" and "b" come from?
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> -- 
> Brian D. Ripley,                  ripley at stats.ox.ac.uk
> Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> University of Oxford,             Tel:  +44 1865 272861 (self)
> 1 South Parks Road,                     +44 1865 272866 (PA)
> Oxford OX1 3TG, UK                Fax:  +44 1865 272595
>
> David Winsemius, MD
> West Hartford, CT
>
> -- 
> Brian D. Ripley,                  ripley at stats.ox.ac.uk
> Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> University of Oxford,             Tel:  +44 1865 272861 (self)
> 1 South Parks Road,                     +44 1865 272866 (PA)
> Oxford OX1 3TG, UK                Fax:  +44 1865 272595
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

David Winsemius, MD
West Hartford, CT

Positive Definite Matrix

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