The determinant of even-magic square matrix is 0.

For example, this 4x4 magic square matrix:

{{16, 2, 3, 13}, 
 {5, 11, 10, 8},
 {9, 7, 6, 12},
 {4, 14, 15, 1}}

And this 6x6 matrix:

   {{35, 1, 6, 26, 19, 24}, 
    {30, 5, 7, 21, 23, 25}, 
    {31, 9, 2, 22, 27, 20},
    {8, 28, 33, 17, 10, 15}, 
    {3, 32, 34, 12, 14, 16}, 
    {4, 36, 29, 13, 18, 11}}

So is true for 8x8,10x10...

Is there anything special here? Is it possible to recognize this kind of matrix and conclude that its determinant is zero?

Reference for magic square matrix: http://mathworld.wolfram.com/MagicSquare.html

  • 1
    $\begingroup$ What is your question exactly? Not all $2n\times 2n$ magic squares are singular. Is there something special about the way you constructed the above examples? $\endgroup$ – Casteels Feb 6 '15 at 12:42
  • $\begingroup$ @Casteels Oh, really? I generate the magic square by a Mathematica function. And I tested several 2nx2n matrix, the determinant are all zeros. Of course, my test is by no means exhaustive. Can you give one example of none-zero magic square? Thank you. $\endgroup$ – Nick Feb 6 '15 at 14:02
  • $\begingroup$ most $4\times 4$ magic squares are singular, so it's not surprising that a bunch of random examples would only yield singular squares. $\endgroup$ – Casteels Feb 6 '15 at 14:09
  • $\begingroup$ There are some nonsingular $4\times 4$ examples here. $\endgroup$ – Casteels Feb 6 '15 at 14:12
  • $\begingroup$ @Casteels Thank you for the link. That post asked really good question. I should delete my question. $\endgroup$ – Nick Feb 6 '15 at 14:19

It is known that there are essentially $880$ different $4\times 4$ magic squares, of which $240$ are nonsingular. See

Dave Pountney, On Powers of Magic Square Matrices.

Some magic squares created with special methods are also known to be singular. For instance, your $4\times4$ example has the property that the sum of any pair of entries that are symmetric about the center of the matrix is equal to the magic constant (i.e. $a_{i,j}+a_{n+1-i,n+1-j}=n^2+1$ for each $(i,j)$). It's easy to explain why this is singular. See, for instance,

R. Bruce Mattingly (2000), Even Order Regular Magic Squares Are Singular, The American Mathematical Monthly, 107(9): 777-782.

  • 1
    $\begingroup$ Thank you for your explanation and references. I also find the words from your first link interesting: ‘The magical squares, however wonderful soever they may seem, are what I cannot value myself upon, but am rather ashamed to have it known I have spent any part of my time in employment that cannot possibly be of any use to myself or others.’ Benjamin Franklin $\endgroup$ – Nick Feb 6 '15 at 15:19

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