An $n \times n$ magic square $A = [ a_{ij}]$ with a magic sum $\mu$ is said to be reflective if
$$a_{ij}+a_{n+1-i, n+1-j}=\frac{2\mu}{n}, \forall i,j = 1, \ldots n.$$
If $\mu=0$, we say that it is a reflexive magic square.
Let $FMS(n)$ and $0FMS(n)$ denote the set of all $n\times n$ refletive magic squares and zero refletive magic squares respectively.
For $ \geq 4$, the dimension of $0FMS()$ is $\frac{^2 −3+2}{2}$ and the dimension of $FMS()$ is $\frac{^2 − 3 + 4}{2}$. I need the proof of this given theorem. Is it by using the rank nullity theorem?
Yes, one possible approach is by using rank nullity theorem.
The proof can be found in for example, here.
The idea of proof is as follows:
The dimension of $FMS(n)$ is equal to the dimension of $0FMS(n) +1$ as it has one more element in the basis by definition since we get to choose the diagonal entries and we can focus on the problem of finding dimension of $0FMS(n)$.
We associate each non-diagonal entries of magic square with a variable and write down the constraints that they have to satisfy and form a homogeneous linear system.
Constraints that are linearly independent are then identified and we can compute the rank.
Dimension of $0FMS(n)$ is then the nullity of this homogeneous linear system which is then computed by using number of columns subtract away the rank of the matrix which is the number of linearly independent constraints.