Determinant is showing infinity instead of zero! Why? - matlab

Question

This is my matlab code I wrote for a problem I got as homework. after multiplication of A and its transpose the resulting square matrix should have determinant zero according all classmates as their codes (different one) gave them so. Why is my code not giving the determinant of c and d to be infinity
A = rand(500,1500);
b = rand(500,1);
c = (A.')*A;
detc = det(c);
cinv = inv((A.')*A);
d = A*(A.');
detd = det(d);
dinv = inv(A*(A.'));
x1 = (inv((A.')*A))*((A.')*b);
x2 = A.'*((inv(A*(A.')))*b);

Answer 1

This behavior is explained in the Limitations section of the det's documentation and exemplified in the Find Determinant of Singular Matrix subsection where it is stated:
The determinant of A is quite large despite the fact that A is singular. In fact, the determinant of A should be exactly zero! The inaccuracy of d is due to an aggregation of round-off errors in the MATLAB® implementation of the LU decomposition, which det uses to calculate the determinant.
That said, in this instance, you can produce your desired result by using the m-code implementation given on that same page but sorting the diagonal elements of U in an ascending matter. Consider the sample script:
clc();
clear();
A = rand(500,1500);
b = rand(500,1);
c = (A.')*A;
[L,U] = lu(c);
% Since det(L) is always (+/-)1, it doesn't impact anything
diagU = diag(U);
detU1 = prod(diagU);
detU2 = prod(sort(diagU,'descend'));
detU3 = prod(sort(diagU,'ascend'));
fprintf('Minimum: %+9.5e\n',min(abs(diagU)));
fprintf('Maximum: %+9.5e\n',max(abs(diagU)));
fprintf('Determinant:\n');
fprintf('\tNo Sort: %g\n' ,detU1);
fprintf('\tDescending Sort: %g\n' ,detU2);
fprintf('\tAscending Sort: %g\n\n',detU3);
This produces the output:
Minimum: +1.53111e-13
Maximum: +1.72592e+02
Determinant:
No Sort: Inf
Descending Sort: Inf
Ascending Sort: 0
Notice that the direction of the sort matters, and that no-sorting gives Inf since a true 0 doesn't exist on the diagonal. The descending sort sees the largest values multiplied first, and apparently, they exceed realmax and are never multiplied by a true 0, which would generate a NaN. The ascending sort clumps together all of the near-zero diagonal values with very few large negative values (in truth, a more robust method would sort based on magnitude, but that was not done here), and their multiplication generates a true 0 (meaning that the value falls below the smallest denormalized number available in IEEE-754 arithmetic) that produces the "correct" result.
All that written, and as others have implied, I'll quote original Matlab developer and Mathworks co-founder Cleve Moler:
[The determinant] is useful in theoretical considerations and hand calculations, but does not provide a sound basis for robust numerical software.

Answer 2

Ok. So the fact that det(A'*A) is not zero is not a good indication of the (non-)singularity of A'*A.
The determinant depends on the scaling, and matrix clearly non-singular can have very small determinant. For instance, the matrix
1/2 * I_n
where I_n is the nxn identity has a determinant of (1/2)^n which is converging (quickly) to 0 as n goes to infinity. But 1/2 * I_n is not, at all, singular.
For this reason, a best idea to check the singularity of a matrix is the condition number.
In you case, after doing some tests
>> A = rand(500, 1500) ;
>> det(A'*A)
ans =
Inf
You can see that the (computed) determinant is clearly non-zero. But this is actually not surprising, and it should not really bother you. The determinant is fairly hard to compute, so yes, it is just rounding errors. If you want a better approximation, you can do the following
>> s = eig(A'*A) ;
>> prod(s)
ans =
0
There, you see it is closer to zero.
The condition number, on the other hand, is a much better estimator of the (non-)singularity of a matrix. Here, it is
>> cond(A'*A)
ans =
1.4853e+20
And, since it is much larger than 1e+16, the matrix is clearly singular. The reason for 1e+16 is a bit tedious, but is mostly due to the computer precision when doing floating point computations.

Answer 3

I think this is pretty much just a rounding problem, the Inf does not mean you are getting Infinity as an answer, it's just that your determinant is really big and exceeded realmax. As Adiel said, A*A.' generates a symmetric matrix, and should have a numerical value for its determinant. for example, set:
A=rand(5,15)
and you should find that the det of A*A.' is just a numerical value.
SO how did your friends get a ZERO, well it's easy to get 0 or inf for det of large matrices (why are you doing this in the first place I have no clue). So I think they are just getting the same/similar rounding issue.