## Making a magic square matrix singular

### Feng Cheng Chang (view profile)

on 1 Apr 2013
Accepted Answer by Ahmed A. Selman

### Ahmed A. Selman (view profile)

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?

Matt J

### Matt J (view profile)

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

### Feng Cheng Chang (view profile)

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

### Matt J (view profile)

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.

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### Ahmed A. Selman (view profile)

on 1 Apr 2013

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.

Ahmed A. Selman

### Ahmed A. Selman (view profile)

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

### Ahmed A. Selman (view profile)

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

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

Matt J

### Matt J (view profile)

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.

### Jonathan Epperl (view profile)

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.

Feng Cheng Chang

### Feng Cheng Chang (view profile)

on 1 Apr 2013

Dear Jonathan Epperl:

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.`

F C Chang 3/30/2013

Jonathan Epperl

### Jonathan Epperl (view profile)

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.

on 1 Apr 2013
Edited by Matt J

### Matt J (view profile)

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.

Feng Cheng Chang

### Feng Cheng Chang (view profile)

on 1 Apr 2013

Dear Matt J,

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

### Matt J (view profile)

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

### Feng Cheng Chang (view profile)

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

### Feng Cheng Chang (view profile)

on 6 Apr 2013

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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