Making a magic square matrix singular
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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?
Ahmed A. Selman on 1 Apr 2013
I don't think details are required since
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.
More Answers (3)
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.