Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.202823e-18.

hi guys I need a help I'm using LDA but when I run I got the message aforementioned

S1=cov(d1');
S2=cov(d2');
S3=cov(d3');
%% within-class scatter matrix
Sw=S1+S2+S3;
%% between -class scatter matrix
SB1=n1.* (mu1-mu)*(mu1-mu)';
SB2=n2.* (mu2-mu)*(mu2-mu)';
SB3=n3.* (mu3-mu)*(mu3-mu)';
SB=SB1+SB2+SB3;
%% Computing the LDA projection
W=inv(Sw)*SB;
%% getting the projection vectors
[V,D]=eig(W);

Answer 1

[Jan]

The error message is clear. I guess it occurs for this line:

W=inv(Sw)*SB;

This means, that Sw is ill-conditioned, such that there is no inverse of it. This is equivalent to dividing by zero. There is no "help" for this case, but the inputs do not allow this operation.

By the way, use

W=Sw \ SB

instead of the explicit inversion for numerical reasons. See doc inv .

[Rui Mauaie]

thanks Jan, I got it, I changed from inv(Sw)*SB to Sw/SB and I got good results. by the way what is the difference of using Sw/SB and pinv(Sw)*SB?

[Torsten]

Sw\SB, not Sw/SB !

[Bruno Luong]

Sw\SB 

is based on QR with column permutation, so if you system is ill-conditions, it let some of the components of the unknown W to be 0.

pinv(Sw)*SB

is based on SVD, so it forces the projection of the unknown W to the span of the degenerated singular vectors to be 0.

Generally SVD solution is "better", in the sense that is the smallest vector (in l2 norm) that meets the solution (or minimizes the residual l2 norm), therefore more stable with respect to noise.

Answer 2

[Tony Castillo]

Hello Guys,

I received this warning in a model made in Simescape Power Systems of Simulink,

Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.051867e-16.

Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.051867e-16.

May you help me to solve it ?

Thanks in advance.

[Marcel Johannesmann]

I have encountered a similar warning. I could narrow it down to the Load flow settings of the power gui block. The warning most likely occurs because your Base Power is too high so that your rated values (in pu) in the admittance matrix are getting too small. Adjusting the base power fixed this issue for me. Hope this helps.

[Mojtaba Raheli]

How did you do it? I cannot adjust base power. please write the direction of base power setting.

Answer 3

[Shaunak Bagade]

clc;
clear;
close all;
A = [2 1 -5; 0 1 -2; 0 0 2];
[P, D] = eig(A);
disp('P Matrix');

P Matrix

disp(P);

    1.0000   -0.7071    1.0000
         0    0.7071   -0.0000
         0         0    0.0000

disp('Diagonal form of A is');

Diagonal form of A is

disp(P\D*P);

Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 6.344132e-17.

    2.0000   -0.7071    0.0000
         0    1.0000    0.0000
         0         0    2.0000

[Shaunak Bagade]

How should I resolve this ?

[Jan]

This is a new question, although is concerns the same topic. Please open a new thread. Thanks.

[John D'Errico]

This question is completely different in nature form the original one posted. The reason it happens is only related in that you got a similar message. Yes. Thre is a singular matrix in the problem. But the reason why is completely different.

Regardless, you resolve it by posting a question, NOT an answer. I will not explain what happened until you post this as a question, nor should anyone else.

[Bruno Luong]

I find this question is interesting.

I believe the matrix A is actually not diagonalizable, it's has the Jordan form however.

Therefore P returned by EIG is singular (rank 2).

[John D'Errico]

Yes. It is interesting, in the sense that A is defective. But it is not a question, so I do not believe it should be answered. I will answer it in full, when the @Shaunak Bagade chooses to make it one. If this does not happen, then the person is not interested in learning the answer.

[Shaunak Bagade]

I didn't get you !

[Jan]

@Shaunak Bagade: Pleas stop the discussion here and open a new question. Thanks.

[John D'Errico]

@Shaunak Bagade - what did you not get? I will not answer your question, not UNTIL you bother to post this as a question. Nor should anyone else. So while your question is of technical interest, posting it as an answer to a different question is a problem, because that means nobody else will ever be able to find your question, and nobody else will be able to learn from the answer. Effectively, if you are interested in learning the answer to this problem, then you should be interested enough to learn to use the site properly.

Note that as quasi-moderators, we should have the capability to move this into a question by the OP. But that is beyond our abilities.

[John D'Errico]

@Shaunak Bagade - Since you seem not to want to follow my guidance, see my question, AND answer, here:

https://www.mathworks.com/matlabcentral/answers/1569563-why-do-i-see-a-singular-matrix-warning-here?s_tid=srchtitle

From now on, please learn to post a question as a question, rather than hijacking a separate only vaguely related question.

Answer 4

[Tugcan]

Hello guys i had a same error with different matlabs code. I tried this \ method but still i have this error. I get this error at values 80 120 and 320 for N. Best Regards.

Enter T:

1

Enter N:

80

Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 5.765040e-19. > In question4s (line 18)

error =

1.8689e+13

function mixedboundary()
T=input("Enter T: ");
N=input("Enter N: ");
h=T/N;
A=zeros(N+1,N+1);
F=zeros(N+1,1);
A(1,1)=1;
A(N+1,1:3)=[-3, 4, -1];
F(1,1)=1;
F(N+1,1)=6*h;
for i=2:N
    t(i)=(i-1)*h;
    A(i,i-1)=1/(h^2)-5/(h);
    A(i,i)=-2/(h^2)-3/(2*h);
    A(i,i+1)=1/(h^2)-5/(h);
    F(i,1)=14*exp(2*t(i))-3*exp(4*t(i));
end
Y=inv(A)*F;
exact=zeros(N+1,1);
for i=1:N+1
    t(i)=(i-1)*h;
    exact(i,1)=exp(2*t(i));
end
error=max(abs(exact-Y))
end

[Torsten]

Which 2nd order differential equation with which boundary conditions do you try to solve ?

[Tugcan]

I use kunte rotta method for this question and these are equations:

[Torsten]

The code has nothing to do with your assignment.

The method used in the code is suited for boundary value problems - thus problems where conditions on the solution are imposed at both ends of the interval of integration (in your case: t = 0 and t = T). In the assignment, both conditions are imposed at t = 0.

Further, you are told to first reduce the 2nd order differential equation to a system of two first order differential equations and then solve the system by a Runge-Kutta-method. Both does not happen in the code.

[Tugcan]

I see. I have reduced formst but i dont know how to evolve it to Runge-Kutta-Method but thank you for attention