I am newer to MATLAB. I followed getting started, and calculate the det of 4*4 magic square, it gives finite value instead of 0. Why? >> A=magic(4); >> det(A) ans = -1.4495e-012

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4/26/2010 8:27:05 AM

"Wei " <weitong@ustc.edu> wrote in message <hr3iop$mtm$1@fred.mathworks.com>... > I am newer to MATLAB. > I followed getting started, and calculate the det of 4*4 magic square, it gives finite value instead of 0. Why? > > >> A=magic(4); > >> det(A) > > ans = > > -1.4495e-012 Considering the magnitude of the magic square components this is numerically equivalent to zero. Its due to a rounding error somewhere.

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4/26/2010 8:59:05 AM

Yes, I know it (-1.4495e-012) is somehow "zero". But, I tried some previous version, e.g., 2009a, 7.1, and 6.1, and they all gave "0" answer as the Help says. So, I wish to know why it happens to 2010a, and where can I change it. Thanks. "Thomas Britton" <benjamin.britton@materials.ox.ac.remove.uk> wrote in message <hr3kkp$nt4$1@fred.mathworks.com>... > "Wei " <weitong@ustc.edu> wrote in message <hr3iop$mtm$1@fred.mathworks.com>... > > I am newer to MATLAB. > > I followed getting started, and calculate the det of 4*4 magic square, it gives finite value instead of 0. Why? > > > > >> A=magic(4); > > >> det(A) > > > > ans = > > > > -1.4495e-012 > > Considering the magnitude of the magic square components this is numerically equivalent to zero. Its due to a rounding error somewhere.

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4/26/2010 9:18:04 AM

"Wei " <weitong@ustc.edu> wrote in message <hr3loc$5sa$1@fred.mathworks.com>... > Yes, I know it (-1.4495e-012) is somehow "zero". > But, I tried some previous version, e.g., 2009a, 7.1, and 6.1, and they all gave "0" answer as the Help says. > So, I wish to know why it happens to 2010a, and where can I change it. Sigh. The determinant did not change, nor can you make it magically become zero again. This matrix is numerically singular. In fact, it is truly singular. Of course, testing for singularity using the determinant is a fool's game, best reserved for textbooks written by authors who live in a fool's world. The determinant is a terrible way to do that test. Here I will define a fool's world as someplace where the people understand (or care) nothing about floating point arithmetic. Computation of the determinant of a matrix is a nasty thing for large order matrices, at least if you do it using the textbook way. This will get impossibly inefficient for very reasonable size matrices. So MATLAB uses a better scheme. One such scheme is to recognize that the determinant is simply the product of the eigenvalues of the matrix. Thus M = magic(4); ans = -5.5968e-12 Note that this is a slightly different "zero" than det gives. This is because det actually uses a better scheme. det uses the product of the diagonal elements of U, taken from the LU factorization (with pivoting) of your matrix. [L,U,P] = lu(M); U U = 16 2 3 13 0 13.5 14.25 -2.25 0 0 -1.8889 5.6667 0 0 0 3.5527e-15 prod(diag(U)) ans = -1.4495e-12 Note that the result is not a true "zero". Get used to living in the numerical world of floating point arithmetic. John

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4/26/2010 9:56:07 AM

Thank you very much for exhaustive explanation of numerical algorithm. Why do the previous versions give exact "zero"? Is the newest 2010a using a different algorithm for calculating determinant? "John D'Errico" <woodchips@rochester.rr.com> wrote in message <hr3nvn$44$1@fred.mathworks.com>... > "Wei " <weitong@ustc.edu> wrote in message <hr3loc$5sa$1@fred.mathworks.com>... > > Yes, I know it (-1.4495e-012) is somehow "zero". > > But, I tried some previous version, e.g., 2009a, 7.1, and 6.1, and they all gave "0" answer as the Help says. > > So, I wish to know why it happens to 2010a, and where can I change it. > > Sigh. The determinant did not change, nor can you > make it magically become zero again. > > This matrix is numerically singular. In fact, it is truly > singular. Of course, testing for singularity using the > determinant is a fool's game, best reserved for > textbooks written by authors who live in a fool's > world. The determinant is a terrible way to do that > test. Here I will define a fool's world as someplace > where the people understand (or care) nothing > about floating point arithmetic. > > Computation of the determinant of a matrix is a > nasty thing for large order matrices, at least if you > do it using the textbook way. This will get impossibly > inefficient for very reasonable size matrices. So > MATLAB uses a better scheme. One such scheme is > to recognize that the determinant is simply the > product of the eigenvalues of the matrix. Thus > > M = magic(4); > ans = > -5.5968e-12 > > Note that this is a slightly different "zero" than > det gives. This is because det actually uses a better > scheme. det uses the product of the diagonal > elements of U, taken from the LU factorization > (with pivoting) of your matrix. > > [L,U,P] = lu(M); > U > U = > 16 2 3 13 > 0 13.5 14.25 -2.25 > 0 0 -1.8889 5.6667 > 0 0 0 3.5527e-15 > > prod(diag(U)) > ans = > -1.4495e-12 > > Note that the result is not a true "zero". Get > used to living in the numerical world of floating > point arithmetic. > > John

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4/27/2010 12:18:04 AM

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