Warning: Matrix is singular to working precision

i am experiencing the above warning while running this code

sigma = eye(10);
mu = repelem(zeros,10);
xTrain = mvnrnd (mu,sigma,1000,10);
t = transpose (xTrain);
xTest = transpose (mvnrnd (mu,sigma,1000,10));
yTrain = randn(1000,1);
x = inv(t*xTrain);

Answer 1

[Walter Roberson]

You are doing inv(xTrain.'*xTrain) but rank(xTrain) is only 1

This is related to your passing in the undocumented fourth parameter to mvnrnd, which you give as 10. The third and fourth parameter together do not give an output size: the output size is determined by the third parameter (a size) and the length of the diagonal of the second parameter.

The undocumented fourth parameter is:

    %   R = MVNRND(MU,SIGMA,N,T) supplies the Cholesky factor T of
    %   SIGMA, so that SIGMA(:,:,J) == T(:,:,J)'*T(:,:,J) if SIGMA is a 3D array or SIGMA
    %   == T'*T if SIGMA is a matrix.  No error checking is done on T.

You encounter the line

          r = randn(n,size(T,1),'like',outtype) * T + mu;

where n is 1000 (your third parameter) and T is 10 so size(T,1) is 1 and mu is the vector of zeros you passed in as your first parameter, replicated to have n rows, so 1000 x 10 of zeros. This is then a 1000 x 1 vector plus a 1000 x 10 array of zeros.

If you were running R2016a or earlier you would have received an error, but in R2016b or later the operation is equivalent to

bsxfun(@plus, randn(n,size(T,1),'like',outtype) * T, mu)

so the 1000 x 1 vector is replicated to 1000 x 10 and the 1000 x 10 array of zeros is added. This gives you a 1000 x 10 matrix in which all of the rows are identical, and the rank of that is going to be 1. Multiply it by its transpose and the rank of the result cannot be greater than 1 so you cannot inv() it.

Answer 2

[Aaina]

I also have the warning

Warning: Matrix is singular to working precision. 
> In NormalizePopulation>FindHyperplaneIntercepts at 41
  In NormalizePopulation at 31

the matlab code goes like this:

function [pop, params] = NormalizePopulation(pop, params)

      params.zmin = UpdateIdealPoint(pop, params.zmin);

      fp = [pop.Cost] - repmat(params.zmin, 1, numel(pop));

      params = PerformScalarizing(fp, params);

      a = FindHyperplaneIntercepts(params.zmax);

      for i = 1:numel(pop)
          pop(i).NormalizedCost = fp(:,i)./a;
      end

end

function a = FindHyperplaneIntercepts(zmax)

      w = ones(1, size(zmax,2))/zmax;

      a = (1./w)';

end

What is the problem?

[Walter Roberson]

Your params.zmax is not a scalar, so ones(1, size(zmax,2))/zmax uses the mrdivide operator / and wants to do approximately

ones(1, size(zmax,2)) * inv(zmax)

but your zmax matrix is singular.

The code does seem to be expecting zmax to be an array -- otherwise it probably would not use size(zmax, 2)

In short: your matrix zmax is not linearly independent so the matrix division is not working properly.

[Aaina]

Thank you so much.... u solved my problem