Solution for a homogeneous equation using null space

 I have a homogeneous equation group as follows

A*b=0;where A is a 8 by 8 matrix, b is a 8 by 1 vector.

I have calculated that det(A)=5.141078303798737e-010, close to zero, so i use null(A) to find out the solution b, however i got a result like "Empty matrix: 8-by-0", I have checked the rank of the matrix A and got Rank(A)=8, since the determinant of A is close to zero, i thought that rank(A) should be smaller than 8 and I should be able to use null space to solve b. Any one can suggest a way to solve b?

Thanks.

Mike

[Wayne King]

Can you please post your matrix since it is only 8x8

[harry wang]

No problem, if i cut and paste the matrix, it seems terrible, so i post it as follows

(1,1) -7.257023337091996

(2,1) -2.440329565883115

(3,1) 0.000000000000001

(4,1) -0.000000000000000

(5,1) -0.000000000000000

(6,1) 0.000000000000003

(7,1) 3.658635913629607

(8,1) 0.000000000170258

(1,2) 1.282933883035595

(2,2) 1.201628321295855

(3,2) -0.000000000000000

(4,2) 0.000000000000000

(5,2) -0.000000000000000

(6,2) -0.000000000000000

(7,2) -0.852367515277986

(8,2) 0.000000000156291

(1,3) -10.998441425187863

(2,3) 5.209283612003834

(3,3) -6.338289276250228

(4,3) -0.000000000035585

(5,3) -11.027957706237926

(6,3) -2.473178531845574

(7,3) 8.881879896588558

(8,3) -0.000000000236824

(1,4) -10.998441425188025

(2,4) 5.209283612003911

(3,4) 6.338289276250308

(4,4) 0.000000000035585

(5,4) 11.027957706238094

(6,4) 2.473178531845598

(7,4) 8.881879896588693

(8,4) -0.000000000236824

(1,5) 0.000000000000000

(2,5) -0.000000000000003

(3,5) -3.658635913629607

(4,5) -0.000000000170258

(5,5) -7.257023337091996

(6,5) -2.440329565883115

(7,5) 0.000000000000001

(8,5) -0.000000000000000

(1,6) 0.000000000000000

(2,6) 0.000000000000000

(3,6) 0.852367515277986

(4,6) -0.000000000156291

(5,6) 1.282933883035595

(6,6) 1.201628321295855

(7,6) -0.000000000000000

(8,6) 0.000000000000000

(1,7) 11.027957706237926

(2,7) 2.473178531845574

(3,7) -8.881879896588558

(4,7) 0.000000000236824

(5,7) -10.998441425187863

(6,7) 5.209283612003834

(7,7) -6.338289276250228

(8,7) -0.000000000035585

(1,8) -11.027957706238094

(2,8) -2.473178531845598

(3,8) -8.881879896588693

(4,8) 0.000000000236824

(5,8) -10.998441425188025

(6,8) 5.209283612003911

(7,8) 6.338289276250308

(8,8) 0.000000000035585

[Wayne King]

Why can't you post this as MATLAB code??? Just use format short.

[Sean de Wolski]

post it as a full matrix so Wayne can copy and paste it:

full(A)

[Wayne King]

Oh, thank you Sean!! sorry I don't why it didn't occur to me that was sparse(A)

Answer 1

[John D'Errico]

Sigh. The determinant is a TERRRRRRRRRRRIBLE way to check for singularity. Godawful. A miserable computational tool, good only for homework problems. Did I say it was bad? Three times I've said it, so this must be true.

Do you need an example? Compute the determinant of

A = eye(8);
det(A)
ans =
   1

Yes, it is clearly nonsingular, and det predicted that. Since you used 1e-10 as a measure that a matrix must be singular, can we modify A just so subtly and get a determinant that small? How about this?

det(A*.05)
ans =
   3.9063e-11

Oops. I could have sworn that this matrix is also diagonal, just with 0.05 elements down the diagonal. Surely this is just as nonsingular as the original identity matrix was.

Gosh. I wonder why I should think it is singular just because det says so? I suppose if I go to extremes I can get a true underflow zero.

det(A*1e-50)
ans =
     0

Is that matrix any more singular than the others?

Why do people insist on using det?

[harry wang]

John,

I not only use det to check the sigularity but also use "rank" to check it, are there any other way to check the sinularity?

Harry

[Walter Roberson]

John, I guess nobody ever expects the det() to be a boojum.

[John D'Errico]

Don't use det. Period. PERIOD. PERIOD! Need I repeat myself? Under NO circumstances would I EVER advise that someone use det to test for singularity.

Rank is a good choice. An svd is a good choice, but of course rank is based on the svd.

Answer 2

[harry wang]

OK, i find the solution, if i use the similar solution as the "null" function, it works! if i use "null" directly, i will still get "empty matrix 8 by 0". Is it a bug of Matlab "null" function? the following is the code to simulate "null" function, Z is the null space for my matrix A.

[m,n] = size(A);
[U,S,V] = svd(A,0);
s = diag(S);
  tol = max(m,n) * max(s) * eps;
 r = sum(s > tol);
 Z = V(:,r+1:n);

[Dipendra Subedi]

I have the same issue. The null(A) returns empty matrix (specially when used inside a loop which is very strange) but the above implementation works fine (in and out of the loop).

Answer 3

[harry wang]

sorry about that, the following is the matrix in format short
 -7.2570    1.2829  -10.9984  -10.9984    0.0000    0.0000   11.0280  -11.0280
 -2.4403    1.2016    5.2093    5.2093   -0.0000    0.0000    2.4732   -2.4732
  0.0000   -0.0000   -6.3383    6.3383   -3.6586    0.8524   -8.8819   -8.8819
 -0.0000    0.0000   -0.0000    0.0000   -0.0000   -0.0000    0.0000    0.0000
 -0.0000   -0.0000  -11.0280   11.0280   -7.2570    1.2829  -10.9984  -10.9984
  0.0000   -0.0000   -2.4732    2.4732   -2.4403    1.2016    5.2093    5.2093
  3.6586   -0.8524    8.8819    8.8819    0.0000   -0.0000   -6.3383    6.3383
  0.0000    0.0000   -0.0000   -0.0000   -0.0000    0.0000   -0.0000    0.0000

Answer 4

[Wayne King]

Hi Harry, I get a rank of 6 for this matrix and therefore expect the nullspace to have dimension 2. I get the following ONB for the nullspace:

   -0.000000000000020  -0.756171511092082
     0.000000000000013   0.434311342718816
     0.339775797713826  -0.065890507565062
    -0.339775797713829  -0.065890507565044
    -0.756171511092085   0.000000000000021
     0.434311342718811  -0.000000000000010
    -0.065890507565062  -0.339775797713825
    -0.065890507565044   0.339775797713829

[harry wang]

Yeah, but if i recover the orginal long format, i would not be able to use "null" to solve it. Are there any way i can post the long format matrix?

Answer 5

[harry wang]

This is the only way i can think to post the original long format matrix, each row of the matrix is seperated by space. you can copy the whole matrix into matlab editor and delete the space. thanks

-7.257023337091996 1.282933883035595 -10.998441425187863 -10.998441425188025 0.000000000000000 0.000000000000000 11.027957706237926 -11.027957706238094

-2.440329565883115 1.201628321295855 5.209283612003834 5.209283612003911 -0.000000000000003 0.000000000000000 2.473178531845574 -2.473178531845598

0.000000000000001 -0.000000000000000 -6.338289276250228 6.338289276250308 -3.658635913629607 0.852367515277986 -8.881879896588558 -8.881879896588693

-0.000000000000000 0.000000000000000 -0.000000000035585 0.000000000035585 -0.000000000170258 -0.000000000156291 0.000000000236824 0.000000000236824

-0.000000000000000 -0.000000000000000 -11.027957706237926 11.027957706238094 -7.257023337091996 1.282933883035595 -10.998441425187863 -10.998441425188025

0.000000000000003 -0.000000000000000 -2.473178531845574 2.473178531845598 -2.440329565883115 1.201628321295855 5.209283612003834 5.209283612003911

3.658635913629607 -0.852367515277986 8.881879896588558 8.881879896588693 0.000000000000001 -0.000000000000000 -6.338289276250228 6.338289276250308

0.000000000170258 0.000000000156291 -0.000000000236824 -0.000000000236824 -0.000000000000000 0.000000000000000 -0.000000000035585 0.000000000035585

[Wayne King]

I see what you mean, but do you really need this kind of precision in your matrix?

Answer 6

[harry wang]

I have calculate the condition number of this matrix, it is 5.144376827909181e+012, quite large, so the matrix is close to a singular matrix, but how can i tell MATLAB that this is a singular matrix and i can use "null" to solve "b"?