Htykuut

Consider the following matrix structure:

$$
M = begin{pmatrix}
a & d & c & c \
d & a & b & b \
c & b & a & e \
c & b & e & a
end{pmatrix}
$$

It is a real, symmetric matrix.
I'm looking into physical properties (spin correlations) of a model that are described by a matrix of this structure.

Qn: Does anyone recognize this matrix as any special type of matrix?

It has the following (weird) property:

Consider its inverse, $M^{-1}$. Its elements are extremely complicated, for example the (3,3) element is:

$$
(M^{-1})_{3,3} = frac{a left(a^{2} - d^{2}right) left(left(a^{2} - c^{2}right) left(a^{2} - d^{2}right) - left(a b - c dright)^{2}right)}{left(left(a^{2} - c^{2}right) left(a^{2} - d^{2}right) - left(a b - c dright)^{2}right)^{2} - left(left(a^{2} - d^{2}right) left(a e - c^{2}right) - left(a b - c dright)^{2}right)^{2}}
$$
and the (3,4) element is:
$$
(M^{-1})_{3,4} = - frac{a left(a^{2} - d^{2}right) left(left(a^{2} - d^{2}right) left(a e - c^{2}right) - left(a b - c dright)^{2}right)}{left(left(a^{2} - c^{2}right) left(a^{2} - d^{2}right) - left(a b - c dright)^{2}right)^{2} - left(left(a^{2} - d^{2}right) left(a e - c^{2}right) - left(a b - c dright)^{2}right)^{2}}
$$

but their difference is extremely simple:
$$
(M^{-1})_{3,3} - (M^{-1})_{3,4} = boxed{frac{1}{a-e}}
$$

You can use Mathematica etc. to verify.

But I realized, this is just the same as considering the (3,4) subblock:
$$
M' = begin{pmatrix} a & e\e & a end{pmatrix}
$$
and its inverse is trivial:
$$
(M')^{-1} = frac{1}{(a+e)(a-e)} begin{pmatrix} a & -e\-e & a end{pmatrix}
$$

And the difference between the diagonal and off-diagonal elements:
$$
(M'^{-1})_{1,1} - (M'^{-1})_{1,2} = boxed{frac{1}{a-e}}
$$
is exactly the same as before!

Is this just a coincidence or is there some justification to just considering a subblock $M'$ instead of the full $M$ when taking the inverse?

I don't know the physics, but the phenomenon you described can be explained mathematically as follows. Suppose $M$ is an invertible matrix of the form $pmatrix{X&pv^T\ uq^T&Z}$, where $X,Z$ are invertible matrices of the same sizes and $p,q,u,v$ are some vectors. Then $M^{-1}$ is in the form of $pmatrix{ast&ast\ ast&S^{-1}}$, where
$$
S=Z-(uq^T)X^{-1}(pv^T)=Z-beta uv^T
$$
is the Schur complement of $Z$ in $X$ and $beta (=q^TX^{-1}p)$ is some scalar. Since $S$ is a rank-$1$ modifiction of $Z$, by Sherman-Morrison formula, $S^{-1}$ is also a rank-$1$ modification of $Z^{-1}$, with
$$
S^{-1}
=Z^{-1}+frac{beta Z^{-1}uv^TZ^{-1}}{1+beta v^TZ^{-1}u}
=Z^{-1}+gamma Z^{-1}uv^TZ^{-1}tag{1}
$$
for some scalar $gamma (=beta/(1+beta v^TZ^{-1}u))$.

In your case, $Z=pmatrix{a&e\ e&a}$ and $u=v=pmatrix{1\ 1}$. Therefore all elements of $Z^{-1}uv^TZ^{-1}$ are identical to each other. It follows from $(1)$ that $S^{-1}$ is obtained by shifting the entries of $Z^{-1}$ by the same constant. Therefore $
(M^{-1})_{33}-(M^{-1})_{34}=(S^{-1})_{11}-(S^{-1})_{12}=(Z^{-1})_{11}-(Z^{-1})_{12}$.