Invertibility of a certain matrix indexed by the Hamming cube 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?
 A: The argument of Kim and Roush looks as follows after translating out the semigroup theory (and is essentially using a Mobius inversion idea).
Let $T\colon \mathbb Z^S\to \mathbb Z^S$ be the group homomorphism corresponding to left multiplication by $A$.  We show that in appropriate bases for the domain and codomain the matrix of $T$ is triangular with 1s on the diagonal and hence $A$ is invertible over $\mathbb Z$.  Let $e_X$ be the unit vector corresponding to a non-empty subset $X$ of $F$.  Put $e_{\emptyset}=0$ for convenience.  Let $b_X=e_F-e_{X^c}$ where $X^c$ is the complement of $X$.  Notice that $b_F=e_F$ and hence the $b_X$ form a basis for $\mathbb Z^S$.  
Now one computes $$Ab_X=A(e_F-e_{X^c})=\sum_{Y\subseteq X} e_Y.$$ If we use the $b_X$ with $X\in S$ as a basis for the domain of $T$, the $e_X$ with $X\in S$ as a basis for the codomain and total order $S$ by a topological sorting of $\subseteq$ then the matrix for $T$ with respect to these bases is triangular with 1s on the diagonal.  Thus $A$ is invertible over $\mathbb Z$.
A: For a vast generalization, see Exercise 3.96(a) of Enumerative Combinatorics, vol. 1, second ed. To get the posted problem, take $L$ to be the boolean algebra of all subsets of $F$ (ordered by inclusion), and set $F(u,s)=1$ if $u\neq\emptyset$ or $s=\emptyset$, and otherwise $F(u,s)=0$. (Note that I am using $F$ in two different ways: one is Yemen's use, and the other is the use in EC1.) Then in the row of the matrix $F(s\wedge t,s)$ indexed by $\emptyset$, every entry is 0 except in the column indexed by $\emptyset$. 
Hence the determinant remains the same if we remove the row and column and indexed by $\emptyset$, but this gives the matrix $A$. 
A: For any $i,j,k$, the automorphism group of $A$ is transitive on the set of pairs $(I,J)$ such that $|I|=i, |J|=j, |I\cap J|=k$.  Therefore the same is true of the inverse (if it exists).  That is, the $(I,J)$-th entry of the inverse is $f(i,j,k)$ for some function $f$. I'm too lazy, but I bet that by examining Benjamin's example the function $f(i,j,k)$ can be guessed rather easily.  Then we will have an explicit formula for the inverse.
Here's a WRONG guess:  The $(I,J)$-th entry of the inverse is 0 unless $|I\cup J|=n$ and otherwise is $(-1)^{n+k+1}$.
Here's a RIGHT guess: The $(I,J)$-th entry of the inverse is 0 if $|I\cup J|\lt n$ and otherwise equals $(-1)^{k+1}$.  I checked this up to n=8.
This is easy to prove by induction using Benjamin's recursive formula for the inverse.
A: After seeing very good proofs of this, I could not think other ways to prove than using induction.
I read O. Selim's proof, and I think it is possible to simplify their induction argument. 
We can associate each subset of $S$ to a binary expansion so that natural numbers from $0$ to $2^n-1$ will represent all subsets of $S$. The components of $2^n \times 2^n$ matrix $A_n$ where $|S|=n$ is then 
$$
A_{ij}= 1 \textrm{ if the binary expansions of $i$ and $j$ has 1 in common at some digit}
$$
$$A_{ij}=0 \textrm{ otherwise}
$$
where $i,j = 0, 1, \cdots , 2^n-1$. 
So, this matrix is basically one column and one row of zeros added to your original matrix, this does not change the rank. 
Let $E_n$ be the $2^n\times 2^n$ matrix with all 1's. 
Then we have the following block matrix form
$$
A_{n+1}=\begin{pmatrix}{A_n}&{A_n}\\\
{A_n}&{E_n}
\end{pmatrix}, \\ E_{n+1}-A_{n+1}=\begin{pmatrix}{E_n-A_n}&{E_n-A_n}\\\
{E_n-A_n}&{0}\end{pmatrix}
$$
We assume our induction hypothesis 
$$
\textrm{rank}A_n=2^n-1, \\ \textrm{rank}(E_n-A_n)=2^n
$$
After elementary row and column operations, we have
$$
\textrm{rank}A_{n+1}=\textrm{rank}\begin{pmatrix}{A_n}&{0}\\\
{0}&{E_n-A_n}
\end{pmatrix}, \\\ \textrm{rank}(E_{n+1}-A_{n+1})=\textrm{rank}\begin{pmatrix}{0}&{E_n-A_n}\\\
{E_n-A_n}&{0}
\end{pmatrix}
$$
Then we have 
$$
\textrm{rank}A_{n+1}=2^{n+1}-1, \\ \textrm{rank}(E_{n+1}-A_{n+1})=2^{n+1}
$$
Added) This method might also work for finding inverse matrix. Then we have to consider $n-1\times n-1$ minor of $A_n$ with row and column of all zeros deleted.  
A: This argument is motivated by some of the ideas in this paper of Dowling and Wilson --I think it may also be possible to extract the result directly from that paper somehow.  
Let $A'$ be formed by $A$ by adding an additional row and column of $0's$ to represent the empty set, and let $J$ be the $2^n \times 2^n$ matrix of all $1's$.  Then $J-A'$ can be thought of as the graph on all $2^n$ vertices where we connect two sets if they are disjoint.  We have 
$$det(J-A')=\sum_{\sigma} (-1)^{sgn(\sigma)},$$
where the sum is taken over all permutations such that $\sigma(I) \cap I=0$ for all subsets $I$.  But the only such permutation is the one where $\sigma(I)$ is the complement of $I$ (if you assign the sets from largest cardinality to smallest, at each step there's only one choice for $\sigma(I)$).  
Since $J-A'$ has full rank and $J$ has rank $1$, then $A'=J-(J-A')$ has rank at least $2^n-1$.  Dropping the row and column of $0's$, we have that $A$ has full rank.  
