Basis for space of $3 \times 3$ magic squares

Question

Given the space

$$V = \left\{ A \in \mathbb R^{3 \times 3} : A \text{ is magic square matrix} \right\}$$

I am trying to find a basis for the space. I tried to represent matrix elements with $A(2,2)$ , since each row/col/diag is equal to $3*A(2,2)$ ($A(2,2)$ is the element in the middle)

Answer 1

Consider a matrix $$A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}.$$ If $A$ is a magic square, then the sum of the rows must be equal, hence \begin{align*} a + b + c &= d + e + f \\ a + b + c &= g + h + i. \end{align*} This sum must also be the sum of the columns: \begin{align*} a + b + c &= a + d + g \\ a + b + c &= b + e + h \\ a + b + c &= c + f + i. \end{align*} Finally, it must be the sum of the two diagonals: \begin{align*} a + b + c &= a + e + i \\ a + b + c &= c + e + g. \end{align*} This gives us $8$ homogeneous linear equations, which we can write as a matrix equation like so: $$\begin{pmatrix} 1 & 1 & 1 & -1 & -1 & -1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & -1 & -1 & -1 \\ 0 & 1 & 1 & -1 & 0 & 0 & -1 & 0 & 0 \\ 1 & 0 & 1 & 0 & -1 & 0 & 0 & -1 & 0 \\ 1 & 1 & 0 & 0 & 0 & -1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & -1 \\ 1 & 1 & 0 & 0 & -1 & 0 & -1 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix}a \\ b \\ c \\ d \\ e \\ f \\ g \\ h \\ i\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}.$$ Solve this, using Gaussian elimination, for $a, b, c, d, e, f, g, h, i$ in general, making sure to take into account free parameters. If you write out the solution $A$ in terms of these free parameters, you will automatically have $A$ as a linear combination of a set of magic squares, which form a basis for all $3 \times 3$ magic squares (the parameters will form the coefficients of these linear combinations).

Good luck!

Answer 2

There are only $8$ different magic squares with elements ranging from 1 to 9.

Just find one of them and look at the permutations of elements preserving the magic in your square.

Other magic squares are multiples of the above by positive integers.

If you are willing to consider real numbers instead of natural numbers then you may multiply by any scalar to generate new magic squares.