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Making a magic square matrix singular

Asked by Feng Cheng Chang on 1 Apr 2013

We know that any magic square matrix of odd order is not singular. When each element of the matrix is subtracted by the sum-average of the total elements, then this perturbed matrix becomes singular, and the determinant of the resulted matrix is zero. That is,

     det(magic(n)-ones(n)*((1+n*n)/2)) = 0,    for any odd n.

Can anyone help me the proof or find literture in this subject?

8 Comments

Matt J on 1 Apr 2013

The matrix in det([... ]) is what I expected for the 9x9 perturbed matrix. Its determinant shold be equal to zero.

Feng, the determinant of the perturbed matrix is zero. Nobody disputes that and in fact, you've been given analytical proofs of why the matrix is singular.

The point, though, is that determinant computations are sensitive to floating point errors. I don't know how OCTAVE contends with that. What happens in OCTAVE when you compute the determinant as prod(eig(P))? In MATLAB, you get a very bad result,

>> P=magic(9)-41; prod(eig(P))

ans =

   -2.2875

However, MATLAB's rank() command clearly knows that P is singular

>> rank(P)
ans =
       8 
Feng Cheng Chang on 2 Apr 2013

Dear Matt J,

I checked the results using OCTAVE, and the results give prod(eig(P))= -3.7867 (but not -2.2875) and rank(P)=8. But what for?

F C 04/01/2013

Matt J on 2 Apr 2013

But what for?

Well, it shows you that there are bad ways to compute determinants, even for integer matrices. The formula det(P)=prod(eig(P)) clearly doesn't work here very well, again because of precision issues.

Feng Cheng Chang

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4 Answers

Answer by Ahmed A. Selman on 1 Apr 2013
Accepted answer

I don't think details are required since

A=magic(n)-ones(n)*((1+n*n)/2)

is changed into an antisymmetric matrix, any such A matrix must satisfy (basic math.. etc)

det(A) = -1^n * det(A)

since n is odd, det(A) must be zero (thus, A is singular). Changing A from magic(n) to (magic(n)-ones(n)*((1+n*n)/2) ) as mentioned in the question is enough to destroy the symmetry of A.

Yet, since this is too basic, and it works the same for magic(n) with n is odd or even, (also, produces antisymmetric), I'm afraid you already know this. I tried (quickly, to be honest) other means like the nice arguments above, but didn't got anything useful so I thought to share, it might help. Regards.

7 Comments

Ahmed A. Selman on 2 Apr 2013

If (A) was a square, i.e., (n by n), and antisymmetric matrix, with (n) is odd, the relation is valid. How is that basic math? Please be patient to read the following:

The definition of a determinant of any square matrix A (size is n by n, for any integer n > 0 ) is

   det(A)= S( (i=1 to n) a(i,j) F(i,j) ); 

Here, (S) refers to summation from (i=1 to n); a(i,j) is the matrix element of (A), and F(i,j) is the co-factor of (A), found from

   F(i,j)=(-1)^(i+j)*Mij

and (Mij) being the minor of (A).

Now, let (b) be any real constant. By definition,

   det(bA)=b^n *det(A). 

This comes true if we substituted (b) in the basic, first definition of determinant of (A).

Now, a matrix (A) is called (symmetric) iff (if and only if) the condition was satisfied:

   transpose (A) = A,

or, by definition of transpose (I don't believe proof is needed), then:

   Aij=Aji

But if

   Aij= - Aji 

the matrix (A) is said to be (antisymmetric, or skew-symmetric, the same are equal last time I checked). Now, for any odd (n) value, an (n by n) square, antisymmetric matrix, we can and will use

b=-1 , so, from

   det(bA)= b^n det(A) 
   det(-A)=(-1)^n det(A), 

the L.H.S. for such equality is the same as det(A) - again from the definition of determinant and transpose...Thus

    LHS= det(-A)=det(A) 

but the R.H.S. of the last equality was

   RHS= (-1)^n det(A); 

equaling RHS with LHS,

   det(A)=(-1)^n det(A) 

which can be true iff

   det(A)=0. 

ANd for any matrix (A), if the determinant was (ZERO) then (A) is said to be (singular) matrix.

Thank you, Feng and Matt, for all the earlier, useful comments.

Ahmed A. Selman on 2 Apr 2013

And this basic, primitive derivation, is found (must be found) in any textbook dealing with matrices and determinant properties. I did find it on Wolfram search that:

   det(-A)=(-1)^n det(A)

from: <http://mathworld.wolfram.com/Determinant.html >

and found that for antisymmetric matrix A then

   Aij= - Aji 

from

< http://mathworld.wolfram.com/AntisymmetricMatrix.html>

The rest is, however, a plane and direct substitution.

Matt J on 8 Apr 2013

Ahmed,

We established several Comments ago that Aij=-Aji is not satisfied for the modified magic square matrix.

There may be a way to extend the determinant equation to the weirder kind of asymmetry that this matrix exhibits, but it looks like it would take some work. Showing that ones(n,1) is a null-vector of the matrix seems to me like the quicker proof, not to mention that it also covers even-valued n.

Ahmed A. Selman
Answer by Jonathan Epperl on 1 Apr 2013

The row-sum, column-sum and diag-sum of a magic square are all the same, and the magic square uses all the integeres 1:n^2. Thus, the sum of all elements must be n^2*(n^2+1)/2, and each row, column, diag sum must be n*(n^2+1)/2.

Now look at what you wrote, multiply it from the right by ones(n,1), and you'll see that you will get zero. Voila, thus the matrix is singular.

2 Comments

Feng Cheng Chang on 1 Apr 2013

Dear Jonathan Epperl:

Thank you for your reply.

The first paragragh is the well-known definition or description of magic square matrix.

The second paragragh is unclear to me how to get the matrix singular. The test statement is listed here again for your reference

   det(magic(n)-ones(n)*(1+n*n)/2), for any odd n.

So I cannot accept your answer at this time.

F C Chang 3/30/2013

Jonathan Epperl on 5 Apr 2013

Just to fill that hole in your knowledge: A square Matrix A is singular if and only if

  • inv(A) does NOT exist
  • det(A)==0
  • The range of A is not all of R^n
  • THE KERNEL OF A IS NONTRIVIAL
  • ...

That last point there means that if you can find a nonzero vector v such that A*v==0, then you have proven that A is singular, and ones(n,1) is such a vector.

Jonathan Epperl
Answer by Matt J on 1 Apr 2013
Edited by Matt J on 1 Apr 2013

Let x=ones(n,1)/n and P be the perturbed matrix. You can verify that

P*x=0

proving that P is singular.

5 Comments

Feng Cheng Chang on 1 Apr 2013

Dear Matt J,

Thank you again for your reply.

I do not think the calculation of det(P) is numerically unstable since elements of P in the case are all integers.

By the way I got the result det(magic(9)-41)=0, instead of =-0.321.

F C Chang 4/1/2013

Matt J on 1 Apr 2013

I do not think the calculation of det(P) is numerically unstable since elements of P in the case are all integers.

You don't know that MATLAB computes determinants in steps that always result in integers. In fact, I've just learned that MATLAB uses LU factorization, which is consistent with the results Walter and I have been getting,

>> P=magic(9)-41; [L,U]=lu(P); det(L)*det(U)
ans =
     -0.3211
Feng Cheng Chang on 2 Apr 2013

Dear Matt J,

The result I got (by using OCTAVE within machine precession) is 0, but not -0.3211. I hope someone can verify it.

F C 04/01/2013

Matt J
Answer by Feng Cheng Chang on 6 Apr 2013

Dear Readers:

I have already accepted the expected simple answer from Ahmed A Selman, so I would like to conclude this "Question-Answer" contribution.

Also I would like to thank Walter Roberson, Matt J and Jonathan Epperl for the further extensive discussion related to the subject.

It gives me the great opportunity to find the interesting result unexpectedly. Hope anyone can see it.

For any n x n non-singular square matrix A, I can make it singular by perturbing the same amount to each of the matrix elements. That is

    det(A-ones(n)/sum(sum(inv(A))) == 0 

I would like to extend "the making a matrix singular" further to include the perturbation of any single element, any row or coulumn elements, or even the entire matrix elements in the prescribed distribution. Please watch my contribution in MATLAB Files Exchange in the near future.

For now, please try the followings:

    for n=1:9, A=rand(n); P=A-1/sum(sum(inv(A))); detP=det(P); n,A,P,detP, end;

Best wishes,

Feng Cheng Chang 04/06/2013

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Feng Cheng Chang

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