I consider a diagonal $L$-by-$L$ matrix $\tilde M$ with nonzero diagonal entries $\tilde M_{k}=\pm 1$, for instance \begin{align} \left(\begin{array}{cccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 &-1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 &0 &-1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 &0 &0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 &0 &0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 &0 &0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 &0 &0 & 0 & 0 & 0 & 1\\ \end{array}\right) \end{align} where $\tilde M_k=1$ for $1\leq k\leq L/4$ and for $k>L/4$ it repeats some regular pattern, such as $(-1,-1,1)$ in this example.
If I now perform a discrete Fourier transform to the new matrix $M$ with entries \begin{align} M_{xy}=\frac{1}{L}\sum^L_{k=1}\tilde M_k e^{i\frac{2\pi k(x-y)}{L}}\,, \end{align} I can study the eigenvalues of submatrices $M(L_s)$ of size $L_s$-by-$L_s$ where I restrict to the first $L_s$ columns and rows.
Numerically, I find that $L_s/4$ of the eigenvalues of $M(L_s)$ are given by $+1$, while the remaining eigenvalues are the same as if I had just transformed the part of the matrix containing the pattern. All of this is approximately true, but becomes exact in the limit $L\to\infty$.
How could I prove this?