Inversion of ill Contioned Matrices

In my code I have to take inversion of matrices at a certain point. When I am trying to take the inverse of ill conditioned matrices, I am getting a weird result.

I am not trying to solve equations. I have to take inverse of a matrix, as part of a similarity measure process. I was trying to take the inverse directly, but it was being scaled badly and when I tried to multiply the inverse with the matrix itself, I was nowhere near an Identity matrix. I even tried Singular Value Decomposition (SVD) but no luck. To get a good answer I need to be as close as possible to the inverse as possible.

[U,S,V] = svd(C1);

S1 = zeros(16,16);
for j = 1:16
    if S(j,j) ~= 0
        S1(j,j) = 1/S(j,j);
    end
end

C1_1 = V*S1*transpose(U);

C1*C1_1 was nowhere equal to identity

Can anyone suggest me a method about how to take inversion of ill conditioned matrices other than SVD. Thanks in advance.

[John D'Errico]

I don't even know from this example if C1 is a square matrix!!!!!!! Note that if C1 is not square, then it would NEVER result in an identity.

Answer 1

[Jan]

Have you tried Rump's INTLab?

Alain Barraud has publsihed some methods for accurate linear algebra also: http://www.mathworks.com/matlabcentral/fileexchange/10668-a-forward-stable-linear-solver

Answer 2

[Andrew Newell]

If you have the Symbolic Toolbox, you could try variable precision arithmetic:

C1 = vpa(C1);
C1_1 = inv(C1);

EDIT: If you don't have it, you could try downloading Multiple Precision Toolbox for MATLAB from the File Exchange and do the same thing with mp in place of vpa. I haven't used this toolbox myself, but judging by the list of functions you'll probably want to do this:

C1 = mp(C1);
C1_1 = inv(C1);

[Harsha]

I'm sorry but can you be a bit clearer.

[Andrew Newell]

I have added explicit commands above, but they're just guesses. You'll have to click on the link, download the package, and read the documentation.

Answer 3

[Harsha]

I don't think I have the symbolic tool box, do you have any other suggestions by chance.

[Andrew Newell]

See above. This Answer should really be a Comment.