For reasons which the margin of this page is too small to hold, I have been reading parts of a recent paper by O. Selim

On submeasures on Boolean algebras, arXiv 1212.6822v3

and in Section 7 the following technical lemma is given (Lemma 7.5 in the paper)

Lemma (paraphrased) Let $S$ be the set of non-empty subsets of some fixed finite set $F$, and consider the matrix $A:S\times S\to {\mathbb Q}$ where $$ A_{I,J} = 1 \hbox{ if $I\cap J\neq \emptyset$, and } A_{I,J} = 0 \hbox{ if $I\cap J = \emptyset$.} $$ Then $A$ is invertible.

Selim gives a proof by induction that the columns of $A$ are linearly independent, but he says "we could not find a particularly enlightening proof". So my question is this: do we have a more conceptual argument to show this (real, symmetric) matrix is invertible? I suspect this will be routine for several MO regulars, but hope it is not too elementary or "too localized".

My vague thoughts are that one could view $A$ as the corner of a square matrix indicated by the power set of $F$, and then perhaps do some kind of Fourier transform on the group $\{0,1\}^{|F|}$. Or perhaps there is some kind of Möbius inversion at work here?

While I'm here, a question on terminology: the matrix $A$ is of course the adjacency matrix of a certain graph whose vertex set is $S$. Does this graph have an established name?

For reasons which the margin of this page is too small to hold, I have been reading parts of a recent paper by O. Selim

On submeasures on Boolean algebras, arXiv 1212.6822v3

and in Section 7 the following technical lemma is given (Lemma 7.5 in the paper)

Lemma (paraphrased) Let $S$ be the set of non-empty subsets of some fixed finite set $F$, and consider the matrix $A:S\times S\to {\mathbb Q}$ where $$ A_{I,J} = 1 \hbox{ if $I\cap J\neq \emptyset$, and } A_{I,J} = 0 \hbox{ if $I\cap J = \emptyset$.} $$ Then $A$ is invertible.

Selim gives a proof by induction that the columns of $A$ are linearly independent, but he says "we could not find a particularly enlightening proof". So my question is this: do we have a more conceptual argument to show this (real, symmetric) matrix is invertible?

[EDIT/UPDATE 2012-03-07: this was poorly phrased on my part; I was hoping to find some explanation that involved the lattice or group stucture on $\{0,1\}$, and which took advantage of the very particular structure of this matrix, although I am grateful for all answers received so far. In some sense I wanted to know: "what is the pattern?" or "what is the underlying algebraic mechanism?" -- the matrix is defined in terms of some incidence or order structure, so does that give some way to interpret invertibility of this matrix as part of a more general result? (I do not mean a result like "a matrix with non-zero determinant is invertible".)

Benjamin Steinberg's answer comes closest, at present, to what I was hoping for, but Benjamin Young's answer is also very suggestive and helpful.

I suspect this will be routine for several MO regulars, but hope it is not too elementary or "too localized".

[older comments/thoughts, left here for context]

My vague thoughts are that one could view $A$ as the corner of a square matrix indicated by the power set of $F$, and then perhaps do some kind of Fourier transform on the group $\{0,1\}^{|F|}$. Or perhaps there is some kind of Möbius inversion at work here?

While I'm here, a question on terminology: the matrix $A$ is of course the adjacency matrix of a certain graph whose vertex set is $S$. Does this graph have an established name?