# Find all $3$ by $3$ magic square matrices $M$ such that $M^2$ is also a magic matrix

A magic square matrix $M$ is a square matrix with real entries such that the sum of the entries in each column, each row, and each main diagonal is the same.

The problem is to characterize all 3x3 magic square matrices $M$ such that $M^2$ is also a magic square matrix.

Supposedly, there is an elegant way to do this besides writing out variables and bashing out the multiplication, but I haven't yet found such a solution.

• I am not sure if it will work but maybe if you check if there is a relation between magic matrices and their eigenvalue decomposition and that way you could easily find your solution – Ziad Fakhoury Apr 8 '17 at 18:37

## 1 Answer

First we look at the properties of the matrix and its eigenvalues. Now since the matrix $M$ has rows that add up to the same number, lets say $s$, then we know that $$\begin{pmatrix} 1\\1\\1 \end{pmatrix}$$ is an eigenvector with eigenvalue $s$.

If we take the transpose of the matrix $M^T$ we can then apply the same principle and find that $s$ is a repeated eigenvalue.

Now for the third eigenvalue we notice that the sum of the trace is equal to the sum of the eigenvalues. However we know the sum of the trace to be $s$ so doing this would mean that we get the thrid eigenvalue to be $-s$.

Now by eigenvalue decomposition we know that $M^2 = S\Lambda^2S^{-1}$ and if it were magic its sums, $s_2$ its eigenvalues would be $s_2$ repeated twice and $-s_2$. However, notice how $\Lambda^2$ will have only positive entries. Therefore there does not exist a magic square $M$ such that $M^2$ is also magic.

• If all eigenvalues of $M$ are identical, then both $M$ and $M^2$ are magic. While I think that your proof can be modified to work, as it stands it is wrong. – user1551 Apr 8 '17 at 19:09
• Indeed it's wrong because I found the class of magic squares which work, and its nonempty. – Joshua Benabou Apr 8 '17 at 19:11