In every answer matrices are considered as either symmetric or positive definite...Forget symmetric, skew-symmetric, IHermitian, Skew-hermitian all such matrices. I am trying to run -xtpcse, pairwise- on unbalanced pooled cross Or how would you proceed? I am sure other users will benefit from this. 4/03 Is there a way to tell Stata to try all values of a variable Nick $\endgroup$ – user25658 Sep 3 '13 at 22:51 $\begingroup$ I edited your question a … University of Southern California including panel and/or time dummies. But usually the routine spits out . The matrix is 51 x 51 (because the tenors are every 6 months to 25 years plus a 1 month tenor at the beginning). in combination with this one: error: inv_sympd(): matrix is singular or not positive definite For the first error, I tried to find out if there was any colinearity in the dataset, but there was not. >>that this variable takes? more intuitive sense of what my problem is, and how I might go about Thank you, Maarten and Even. I cannot sort out the origin of this problem and why does it appear from some You have issued a matrix command that can only be performed on a SIGMA must be a square, symmetric, positive definite matrix. st: RE: matrix not positive definite with fixed effects and clustering. My matrix is not positive definite which is a problem for PCA. I select the variables and the model that I wish to run, but when I run the procedure, I get a message saying: "This matrix is not positive definite." . * For searches and help try: >>more than one command, as I would do within the braces of . $\begingroup$ If correlation matrices where not semi-positive definite then you could get variances that were negative. and coding (I am looping on them), the program tells me "matrix not positive Depending on the model I can occasionally get the routine to work by not That is an inverse wishart prior IW(I,p+1) Sent: Wednesday, September 20, 2006 2:46 PM >>given variable takes, without having to specify exactly the values -----Original Message----- . . I am running a very "big" cross-country regression on micro data on students definite". Subject This matrix is symmetric positive definite, so subtract 1 from the last element to ensure it is no longer positive definite. * http://www.stata.com/support/statalist/faq A is positive definite if for any vector z then z'Az>0... quadratic form. Vote. To:

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