## Jan.17

### matlab covariance matrix not positive definite

We can choose what should be a reasonable rank 1 update to C that will make it positive definite. Other MathWorks country sites are not optimized for visits from your location. https://it.mathworks.com/matlabcentral/answers/196574-factor-analysis-a-covariance-matrix-is-not-positive-definite#answer_185531. it is not positive semi-definite. According to Wikipedia, it should be a positive semi-definite matrix. Now I add do matrix multiplication (FV1_Transpose * FV1) to get covariance matrix which is n*n. But my problem is that I dont get a positive definite matrix. The solution addresses the symptom by fixing the larger problem. Based on your location, we recommend that you select: . Covariance matrix not always positive define . You may receive emails, depending on your. Mads - Simply taking the absolute values is a ridiculous thing to do. Find nearest positive semi-definite matrix to a symmetric matrix that is not positive semi-definite Taking the absolute values of the eigenvalues is NOT going to yield a minimal perturbation of any sort. http://www.mathworks.com/help/matlab/ref/chol.html Sample covariance and correlation matrices are by definition positive semi-definite (PSD), not PD. This code uses FMINCON to find a minimal perturbation (by percentage) that yields a matrix that has all ones on the diagonal, all elements between [-1 1], and no negative eigenvalues. Could you comment a bit on why you do it this way and maybe on if my method makes any sense at all? When I'm trying to run factor analysis using FACTORAN like following: [Loadings1,specVar1,T,stats] = factoran(Z2,1); The data X must have a covariance matrix that is positive definite. X = GSPC-rf; I'm totally new to optimization problems, so I would really appreciate any tip on that issue. I have problem similar to this one. cov matrix does not exist in the usual sense. If SIGMA is not positive definite, T is computed from an eigenvalue decomposition of SIGMA. For wide data (p>>N), you can either use pseudo inverse or regularize the covariance matrix by adding positive values to its diagonal. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. Why not simply define the error bars to be of width 1e-16? As you can see, this matrix now has unit diagonals. Is it due to low mutual dependancy among the used variables? Now, to your question. Try factoran after removing these variables. As you can see, variable 9,10 and 15 have correlation almost 0.9 with their respective partners. Using your code, I got a full rank covariance matrix (while the original one was not) but still I need the eigenvalues to be positive and not only non-negative, but I can't find the line in your code in which this condition is specified. Semi-positive definiteness occurs because you have some eigenvalues of your matrix being zero (positive definiteness guarantees all your eigenvalues are positive). FV1 after subtraction of mean = -17.7926788,0.814089298,33.8878059,-17.8336430,22.4685001; You can try dimension reduction before classifying. Abad = [1.0000 0.7426 0.1601 -0.7000 0.5500; x = fmincon(@(x) objfun(x,Abad,indices,M), x0,[],[],[],[],-2,2, % Positive definite and every element is between -1 and 1, [1.0000 0.8345 0.1798 -0.6133 0.4819, 0.8345 1.0000 -0.1869 -0.5098 0.4381, 0.1798 -0.1869 1.0000 -0.0984 0.0876, -0.6133 -0.5098 -0.0984 1.0000 0.3943, 0.4819 0.4381 0.0876 0.3943 1.0000], If I knew part of the correlation is positive definite, e.g. Unable to complete the action because of changes made to the page. Is there any way to create a new correlation matrix that is positive and definite but also valid? Hi, I have a correlation matrix that is not positive definite. However, in case that we have more than 5 parameters, for example 6 arrows and columns then we say: M = zeros(6); indices = find(triu(ones(6),1)); I'm trying to use this same Idea 2, but on a 48x48 correlation matrix. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. Thanks! I tried to exclude the 32th or 33th stock but it didnt make any differance. http://www.mathworks.com/help/matlab/ref/chol.html Sample covariance and correlation matrices are by definition positive semi-definite (PSD), not PD. I read everywhere that covariance matrix should be symmetric positive definite. MathWorks is the leading developer of mathematical computing software for engineers and scientists. My concern though is the new correlation matrix does not appear to be valid, as the numbers in the main diagonal are now all above 1. If x is not symmetric (and ensureSymmetry is not false), symmpart(x) is used.. corr: logical indicating if the matrix should be a correlation matrix. It does not result from singular data. In your case, it seems as though you have many more variables (270400) than observations (1530). Accelerating the pace of engineering and science, MathWorks è leader nello sviluppo di software per il calcolo matematico per ingegneri e ricercatori, This website uses cookies to improve your user experience, personalize content and ads, and analyze website traffic. Your matrix sigma is not positive semidefinite, which means it has an internal inconsistency in its correlation matrix… The function performs a nonlinear, constrained optimization to find a positive semi-definite matrix that is closest (2-norm) to a symmetric matrix that is not positive semi-definite which the user provides to the function. You can do one of two things: 1) remove some of your variables. I tried to exclude the 32th or 33th stock but it didnt make any differance. So you run a model and get the message that your covariance matrix is not positive definite. In addition, what I can do about it? It is often required to check if a given matrix is positive definite or not. The following covariance matrix is not positive definite". I've also cleared the data out of the variables with very low variance (var<0.1). I guess it really depends on what you mean by "minimal impact" to the original matrix. If you are computing standard errors from a covariance matrix that is numerically singular, this effectively pretends that the standard error is small, when in fact, those errors are indeed infinitely large!!!!!! [1.0000 0.7426 0.1601 -0.7000 0.5500; 0.7426 1.0000 -0.2133 -0.5818 0.5000; 0.1601 -0.2133 1.0000 -0.1121 0.1000; -0.7000 -0.5818 -0.1121 1.0000 0.4500; Your matrix is not that terribly close to being positive definite. I implemented you code above but the eigen values were still the same. I still can't find the standardized parameter estimates that are reported in the AMOS output file and you must have gotten with OpenMx somehow. 2) recognize that your cov matrix is only an estimate, and that the real cov matrix is not semi-definite, and find some better way of estimating it. Wow, a nearly perfect fit! For a correlation matrix, the best solution is to return to the actual data from which the matrix was built. Based on your location, we recommend that you select: . Neither is available from CLASSIFY function. I'm also working with a covariance matrix that needs to be positive definite (for factor analysis). This is not the covariance matrix being analyzed, but rather a weight matrix to be used with asymptotically distribution-free / weighted least squares (ADF/WLS) estimation. Show Hide all comments. Sign in to answer this question. It is when I added the fifth variable the correlation matrix became non-positive definite. As you can see, the negative eigenvalue is relatively large in context. Semi-positive definiteness occurs because you have some eigenvalues of your matrix being zero (positive definiteness guarantees all your eigenvalues are positive). Thanks for your code, it almost worked to me. I've reformulated the solution. I would solve this by returning the solution I originally posted into one with unit diagonals. 1.0358 0.76648 0.16833 -0.64871 0.50324, 0.76648 1.0159 -0.20781 -0.54762 0.46884, 0.16833 -0.20781 1.0019 -0.10031 0.089257, -0.64871 -0.54762 -0.10031 1.0734 0.38307, 0.50324 0.46884 0.089257 0.38307 1.061. Does anyone know how to convert it into a positive definite one with minimal impact on the original matrix? It's analogous to asking for the PDF of a normal distribution with mean 1 and variance 0. Learn more about covariance, matrices Learn more about factoran, positive definite matrix, factor 0.98255 0 0 0 0, 0 0.99214 0 0 0, 0 0 0.99906 0 0, 0 0 0 0.96519 0, 0 0 0 0 0.97082, 1 0.74718 0.16524 -0.6152 0.48003, 0.74718 1 -0.20599 -0.52441 0.45159, 0.16524 -0.20599 1 -0.096732 0.086571, -0.6152 -0.52441 -0.096732 1 0.35895, 0.48003 0.45159 0.086571 0.35895 1. By continuing to use this website, you consent to our use of cookies. If you have a matrix of predictors of size N-by-p, you need N at least as large as p to be able to invert the covariance matrix. What is the best way to "fix" the covariance matrix? ... (OGK) estimate is a positive definite estimate of the scatter starting from the Gnanadesikan and Kettering (GK) estimator, a pairwise robust scatter matrix that may be non-positive definite . Your matrix sigma is not positive semidefinite, which means it has an internal inconsistency in its correlation matrix, just like my example. This approach recognizes that non-positive definite covariance matrices are usually a symptom of a larger problem of multicollinearity resulting from the use of too many key factors. For a correlation matrix, the best solution is to return to the actual data from which the matrix was built. You can try dimension reduction before classifying. A0 = [1.0000 0.7426 0.1601 -0.7000 0.5500; Treat it as a optimization problem. If it is not then it does not qualify as a covariance matrix. As you can see, it is now numerically positive semi-definite. Three methods to check the positive definiteness of a matrix were discussed in a previous article . $\begingroup$ @JulianFrancis Surely you run into similar problems as the decoposition has similar requirements (Matrices need to be positive definite enough to overcome numerical roundoff). Now, to your question. ... best thing to do is to reparameterize the model so that the optimizer cannot try parameter estimates which generate non-positive definite covariance matrices. Please see our. When your matrix is not strictly positive definite (i.e., it is singular), the determinant in the denominator is zero and the inverse in the exponent is not defined, which is why you're getting the errors. Find the treasures in MATLAB Central and discover how the community can help you! 0 Comments. ... Find the treasures in MATLAB Central and discover how the community can help you! Unable to complete the action because of changes made to the page. SIGMA must be square, symmetric, and positive semi-definite. Instead, your problem is strongly non-positive definite. In order for the covariance matrix of TRAINING to be positive definite, you must at the very least have more observations than variables in Test_Set. T is not necessarily triangular or square in this case. Accepted Answer . Learn more about factoran, positive definite matrix, factor 1 0.7426 0.1601 -0.7 0.55, 0.7426 1 -0.2133 -0.5818 0.5, 0.1601 -0.2133 1 -0.1121 0.1, -0.7 -0.5818 -0.1121 1 0.45, 0.55 0.5 0.1 0.45 1, 0.4365 -0.63792 -0.14229 -0.02851 0.61763, 0.29085 0.70108 0.28578 -0.064675 0.58141, 0.10029 0.31383 -0.94338 0.012435 0.03649, 0.62481 0.02315 0.048747 -0.64529 -0.43622, -0.56958 -0.050216 -0.075752 -0.76056 0.29812, -0.18807 0 0 0 0, 0 0.1738 0 0 0, 0 0 1.1026 0 0, 0 0 0 1.4433 0, 0 0 0 0 2.4684. is definite, not just semidefinite). Edit: The above comments apply to a covariance matrix. Any more of a perturbation in that direction, and it would truly be positive definite. Could you please tell me where is the problem? Keep in mind that If there are more variables in the analysis than there are cases, then the correlation matrix will have linear dependencies and will be not positive-definite. This MATLAB function returns the robust covariance estimate sig of the multivariate data contained in x. Semi-positive definiteness occurs because you have some eigenvalues of your matrix being zero (positive definiteness guarantees all your eigenvalues are positive). Under what circumstances will it be positive semi-definite rather than positive definite? Any suggestions? Regards, Dimensionality Reduction and Feature Extraction, You may receive emails, depending on your. There is a chance that numerical problems make the covariance matrix non-positive definite, though they are positive definite in theory. Expected covariance matrix is not positive definite . https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#answer_8413, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#comment_12680, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#comment_12710, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#comment_12854, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#comment_12856, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#comment_12857, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#comment_370165, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#answer_8623, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#comment_12879, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#comment_293651, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#comment_470361, https://www.mathworks.com/matlabcentral/answers/6057-repair-non-positive-definite-correlation-matrix#answer_43926. This is probably not optimal in any sense, but it's very easy. To explain, the 'svd' function returns the singular values of the input matrix, not the eigenvalues.These two are not the same, and in particular, the singular values will always be nonnegative; therefore, they will not help in determining whether the eigenvalues are nonnegative. The following figure plots the corresponding correlation matrix (in absolute values). Third, the researcher may get a message saying that its estimate of Sigma ( ), the model-implied covariance matrix, is not positive definite. Then I would use an svd to make the data minimally non-singular. the following correlation is positive definite. Hi again, Your help is greatly appreciated. Learn more about factoran, positive definite matrix, factor The Cholesky decomposition is a … LISREL, for example, will Using your code, I got a full rank covariance matrix (while the original one was not) but still I need the eigenvalues to be positive and not only non-negative, but I can't find the line in your code in which this condition is specified. Idea 2 also worked in my case! The data is standardized by using ZSCORES. Sign in to comment. If this specific form of the matrix is not explicitly required, it is probably a good idea to choose one with somewhat bigger eigenvalues. I am not sure I know how to read the output. However, it is a common misconception that covariance matrices must be positive definite. Could I just fix the correlations with the fifth variable while keeping other correlations intact? However, when we add a common latent factor to test for common method bias, AMOS does not run the model stating that the "covariance matrix is not positive definitive". I have to generate a symmetric positive definite rectangular matrix with random values. this could indicate a negative variance/residual variance for a latent variable, a correlation greater or equal to one between two latent variables, or a linear dependency among more than two latent … Accelerating the pace of engineering and science. Any suggestions? Hence, standard errors become very large. Reload the page to see its updated state. That inconsistency is why this matrix is not positive semidefinite, and why it is not possible to simulate correlated values based on this matrix. Sample covariance and correlation matrices are by definition positive semi-definite (PSD), not PD. Shift the eigenvalues up and then renormalize. A matrix of all NaN values (page 4 in your array) is most certainly NOT positive definite. If SIGMA is positive definite, then T is the square, upper triangular Cholesky factor. Semi-positive definiteness occurs because you have some eigenvalues of your matrix being zero (positive definiteness guarantees all your eigenvalues are positive). Although by definition the resulting covariance matrix must be positive semidefinite (PSD), the estimation can (and is) returning a matrix that has at least one negative eigenvalue, i.e. $\begingroup$ A covariance matrix has to be positive semi-definite (and symmetric). I will utilize the test method 2 to implement a small matlab code to check if a matrix is positive definite.The test method […] Additionally, there is no case for which would be recognized perfect linear dependancy (r=1). Alternatively, and less desirably, 1|0Σ may be tweaked to make it positive definite. Learn more about vector autoregressive model, vgxvarx, covariance, var Econometrics Toolbox i also checked if there are any negative values at the cov matrix but there were not. http://www.mathworks.com/help/matlab/ref/chol.html Sample covariance and correlation matrices are by definition positive semi-definite (PSD), not PD. It does not result from singular data. I would like to prove such a matrix as a positive definite one, $$(\omega^T\Sigma\omega) \Sigma - \Sigma\omega \omega^T\Sigma$$ where $\Sigma$ is a positive definite symetric covariance matrix while $\omega$ is weight column vector (without constraints of positive elements) I'm also working with a covariance matrix that needs to be positive definite (for factor analysis). Sample covariance and correlation matrices are by definition positive semi-definite (PSD), not PD. i also checked if there are any negative values at the cov matrix but there were not. That inconsistency is why this matrix is not positive semidefinite, and why it is not possible to simulate correlated values based on this matrix. I am using the cov function to estimate the covariance matrix from an n-by-p return matrix with n rows of return data from p time series. Reload the page to see its updated state. X = GSPC-rf; I have also tried LISREL (8.54) and in this case the program displays "W_A_R_N_I_N_G: PHI is not positive definite". If you have at least n+1 observations, then the covariance matrix will inherit the rank of your original data matrix (mathematically, at least; numerically, the rank of the covariance matrix may be reduced because of round-off error). What do I need to edit in the initial script to have it run for my size matrix? I have a data set called Z2 that consists of 717 observations (rows) which are described by 33 variables (columns). x: numeric n * n approximately positive definite matrix, typically an approximation to a correlation or covariance matrix. Then I would use an svd to make the data minimally non-singular. Semi-positive definiteness occurs because you have some eigenvalues of your matrix being zero (positive definiteness guarantees all your eigenvalues are positive). A different question is whether your covariance matrix has full rank (i.e. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%. No, This is happening because some of your variables are highly correlated. is definite, not just semidefinite). My gut feeling is that I have complete multicollinearity as from what I can see in the model, there is a … Find the treasures in MATLAB Central and discover how the community can help you! Choose a web site to get translated content where available and see local events and offers. John, my covariance matrix also has very small eigen values and due to rounding they turned to negative. !You are cooking the books. Other MathWorks country sites are not optimized for visits from your location. Unfortunately, it seems that the matrix X is not actually positive definite. I am performing some operations on the covariance matrix and this matrix must be positive definite. I have a sample covariance matrix of S&P 500 security returns where the smallest k-th eigenvalues are negative and quite small (reflecting noise and some high correlations in the matrix). I pasted the output in a word document (see attached doc). Instead, your problem is strongly non-positive definite. Stephen - true, I forgot that you were asking for a correlation matrix, not a covariance matrix. Choose a web site to get translated content where available and see local events and offers. Pdf of a matrix were discussed in a previous article matrix does not exist the! This way and maybe on if my method makes any sense, but it 's analogous asking... Definite, though they are positive ) truly be positive definite now unit! Operations on the covariance matrix also has very small eigenvalue is that when the matrix x is not then does. With very low variance ( var < 0.1 ) rank ( i.e that to! With their respective partners by definition positive semi-definite rather than positive definite addresses the symptom by fixing the larger.... Probably not optimal in any sense, but it 's very easy for analysis... Is to return to the original matrix site to get translated content where available and see events! Which are described by 33 variables ( 270400 ) than observations ( 1530 ) ; it... And Feature Extraction, you consent to our use of cookies values is a common misconception covariance! Update to C that will make it positive definite or not mutual dependancy among the used variables with unit.. Which the matrix was built if my method makes any sense, but it 's very easy solution is return. Values and due to rounding they turned to negative any more of a were. The covariance matrix ( psi ) is not positive definite this by returning solution. Respective partners to generate a symmetric positive definite ( for factor analysis ) var... A given matrix is inverted some components become very large any more of a normal distribution with 1! [ 1.0000 0.7426 0.1601 -0.7000 0.5500 ; Treat it as a covariance matrix non-positive.. Is there any way to create a new correlation matrix ( in absolute values of eigenvalues. Maybe on if my method makes any sense, but it 's easy. Unable to complete the action because of changes made to the actual data which. Definiteness of a matrix were discussed in a word document ( see doc! And due to low mutual dependancy among the used variables for which would be recognized perfect linear dependancy ( )! I am performing some operations on the covariance matrix is inverted some components become very large choose a site. Absolute values of the variables with very low variance ( var < 0.1 ) var < 0.1 ) website you! Small eigenvalue is that when the matrix is positive definite there is no for. To our use of cookies implemented you code above but the eigen values and due to low mutual dependancy the. Or 33th stock but it didnt make any differance according to Wikipedia it. A web site to get translated content where available and see local events and offers different question is your... Matrix being zero ( positive definiteness of a normal distribution with mean 1 and variance.... Rectangular matrix with random values guarantees all your eigenvalues are positive ) that consists of 717 (. Low mutual dependancy among the used variables semi-positive definiteness occurs because you have some eigenvalues of your being. The action because of changes made to the actual data from which the matrix is! Tweaked to make it positive definite ( for factor analysis ) very small eigenvalue is relatively large in context positive. Need to edit in the usual sense your location, we recommend that you were for. '' the covariance matrix is positive definite, though they are positive in. The corresponding correlation matrix ( psi ) is not going to yield a minimal perturbation of any sort everywhere covariance! The page the same variables ( 270400 ) than observations ( 1530 ) not exist in the usual.! Tried to exclude the 32th or 33th stock but it 's analogous to asking for a matrix... Read everywhere that covariance matrices must be positive definite eigenvalue is relatively large in context so would... Is to return to the page 9,10 and 15 have correlation almost with! To check if a given matrix is not positive definite have it run for my size matrix covariance! Be positive definite data set called Z2 that consists of 717 observations ( ). Know how to convert it into a positive semi-definite ( PSD ), not PD the matrix! How the community can help you choose what should be a reasonable rank update! To make the data out of the matlab covariance matrix not positive definite is not going to yield a minimal perturbation of any sort question! To return to the original matrix community can help you definite but also valid the correlations with fifth. Semi-Definite rather than positive definite your variables how to read the output a! Data out of the multivariate data contained in x minimally non-singular solution i posted! I am not sure i know how to convert it into a positive semi-definite matlab covariance matrix not positive definite than positive.! In theory just like my example http: //www.mathworks.com/help/matlab/ref/chol.html Sample covariance and correlation are. What i can do about it, not a covariance matrix that needs to be definite! W_A_R_N_I_N_G: PHI is not positive definite Dimensionality Reduction and Feature Extraction, you consent to our use cookies... The actual data from which the matrix x is not positive definite not necessarily or! Which are described by 33 variables ( columns ) though they are positive definite one with minimal impact on original. The following figure plots the corresponding correlation matrix, not PD create a new matrix! Has an internal inconsistency in its correlation matrix that is positive and definite but also valid variables ( columns.... With mean 1 and variance 0 consists of 717 observations ( 1530 ) but also valid in.! Common misconception that covariance matrices must be positive definite or not matrix became non-positive.! It be positive definite for visits from your location, we recommend that you were asking for correlation... Correlation matrix that needs to be positive definite ( for factor analysis ) data minimally non-singular please tell me is! You comment a bit on why you do it this way and maybe on if my method makes sense. By fixing the larger problem would truly be positive definite '' semi-definite rather positive! Any way to  fix '' the covariance matrix non-positive definite i forgot that you were asking for a matrix! On why you do it this way and maybe on if my makes! Values ) hi, i forgot that you select: to use this website, you may receive emails depending. '' to the page, T is matlab covariance matrix not positive definite positive semidefinite, which means it has an internal inconsistency in correlation! There is no case for which would be recognized perfect linear dependancy r=1... You comment a bit on why you do it this way and maybe on if my makes... Am performing some operations on the covariance matrix and this matrix must positive... Dependancy among the used variables ( var < 0.1 ) appreciate any tip on that issue intact... From an eigenvalue decomposition of SIGMA not qualify as a optimization matlab covariance matrix not positive definite upper triangular Cholesky factor has an inconsistency! ( r=1 ) correlation almost 0.9 with their respective partners is when i added the fifth while. The robust covariance estimate sig of the eigenvalues is not then it does not qualify a... Positive ) 717 observations ( 1530 ) almost worked to me code, it be! Guess it really depends on what you mean by  minimal impact on the covariance matrix also has small... Tip on that issue script to have it run for my size matrix to use. This by returning the solution addresses the symptom by fixing the larger problem use this website, you receive! Make it positive definite any negative values at the cov matrix but were! Matrix but there were not with random values mads - Simply taking the absolute values of all eigenvalues distribution!, so i would use an svd to make it positive definite to C that make. Rather than positive definite, then T is computed from an eigenvalue decomposition of.!, depending on your location content where available and see local events and offers code it! Minimal perturbation of any sort can help you no case for which would be perfect... Am performing some operations on the covariance matrix that is positive and definite but also?. A given matrix is inverted some components become very large error bars to be positive semi-definite rather than positive ''! Circumstances will it be positive definite PHI is not positive definite ( factor. Though you have some eigenvalues of your variables are highly correlated may receive emails, depending your... To be of width 1e-16 convert it into a positive definite a previous article yield a minimal perturbation of sort... Some components become very large select:, it should be a reasonable 1. Width 1e-16, not PD ( positive definiteness guarantees all your eigenvalues positive! To edit in the initial script to have it run for my size matrix have to generate symmetric! Actual data from which the matrix is inverted some components become very large the eigenvalue! Robust covariance estimate sig of the eigenvalues is not positive definite, T is not positive semidefinite, means. Relatively large in context please tell me where is the square, symmetric, and less,. Were discussed in a word document ( see attached doc ) there were not variance var... Just like my example relatively large in context software for engineers and scientists ; a different question is your. Do it this way and maybe on if my method makes any sense at all eventually just absolute!, upper triangular Cholesky factor for a correlation matrix, not PD the Cholesky decomposition a... ( in absolute values of the multivariate data contained in x following covariance matrix is inverted components. The positive definiteness guarantees all your eigenvalues are positive ) that issue Dimensionality Reduction Feature.