This question is motivated by one of the problem set from this year's Putnam Examination. That is,

Problem. Let $S_1, S_2, \dots, S_{2^n-1}$ be the nonempty subsets of $\{1,2,\dots,n\}$ in some order, and let $M$ be the $(2^n-1) \times (2^n-1)$ matrix whose $(i,j)$ entry is $$M_{ij} = \begin{cases} 0 & \mbox{if }S_i \cap S_j = \emptyset; \\ 1 & \mbox{otherwise.} \end{cases}$$ Calculate the determinant of $M$. Answer: If $n=1$ then $\det M=1$; else $\det(M)=-1$.

I like to consider the following variant which got me puzzled.

Question. Preserve the notation from above, let $A$ be the matrix whose $(i,j)$ entry is $$A_{ij} = \begin{cases} 1 & \# (S_i \cap S_j) =1; \\ 0 & \mbox{otherwise.} \end{cases}$$ If $n>1$, is this true? $$\det(A)=-\prod_{k=1}^nk^{\binom{n}k}.$$

Remark. Amusingly, the same number counts "product of sizes of all the nonempty subsets of $[n]$" according to OEIS.


If $B$ is the matrix whose $(i,j)$ entry is $B_{ij} = q^{\#(S_i\cap S_j)}$ then does this hold? $$\det(B)=q^n(q-1)^{n(2^{n-1}-1)}.$$

  • 2
    Nice one! The proof I posted at still works, but you now have to replace $C$ by the matrix $\left( \left|S\right| \cdot \left[S \subseteq T\right] \right)_{\left(S, T\right) \in P \times P}$. – darij grinberg Dec 7 at 5:14
  • 1
    My answer below deals with the main question. The POSTSCRIPT is more interesting, though, as it does not directly follow from my approach for lack of zero entries. Wondering if the argument can be adapted to it. – darij grinberg Dec 7 at 6:13
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    Maybe, the eigenvalues of this matrix are equal to $\pm k$ with multiplicities $\binom{n-1}{k}$, $\binom{n-1}{k-1}$? And there is some clear pattern on how the eigenvectors look like?r – Fedor Petrov Dec 7 at 6:44
  • @FedorPetrov: That is a fine proposition worthy of pushing further that I like to see explored. – T. Amdeberhan Dec 9 at 17:50
up vote 17 down vote accepted

Here is a proof that follows my argument at as closely as possible. Sorry for its length, much of it due to exposition of matrix folklore. (If you know this, scroll all the way down to Theorem 1, or maybe to the two paragraphs preceding it.)

We are grown-ups and don't need to index the rows and the columns of a matrix by numbers. We can pick any set $P$, and define a $P \times P$-matrix (say, with rational entries) to be a family $\left(a_{p, q}\right)_{\left(p, q\right) \in P \times P} \in \mathbb{Q}^{P \times P}$ of rational numbers indexed by pairs $\left(p, q\right) \in P \times P$. When the set $P$ is finite, such a $P \times P$-matrix always has a well-defined determinant, which can be defined by \begin{equation} \det \left( \left(a_{p, q}\right)_{\left(p, q\right) \in P \times P} \right) = \sum_{\sigma \in S_P} \left(-1\right)^{\sigma} \prod_{p \in P} a_{p, \sigma\left(p\right)} . \end{equation} Here, $S_P$ denotes the set of permutations of $P$, and the notation $\left(-1\right)^{\sigma}$ stands for the sign of a permutation $\sigma$. If you think that this is problematic because the sign of a permutation is usually defined using a total order on $P$, then take a break and convince yourself that it does not actually depend on this total order. (This is proven in maximum detail in the solution of Exercise 5.12 of my Notes on the combinatorial fundamentals of algebra, 7th of November 2018.) Determinants of $P \times P$-matrices behave as nicely as determinants of "usual" $n \times n$-matrices, and arguably even better, since it is easier to define minors and cofactors when the rows and columns are indexed by an arbitrary finite set. Do keep in mind that the rows and the columns must be indexed by the same set; we cannot define the determinant of a $P \times Q$-matrix for different $P$ and $Q$, even when $\left|P\right| = \left|Q\right|$.

Of course, an $n \times n$-matrix (for a nonnegative integer $n$) is the same as an $N \times N$-matrix for $N = \left\{1,2,\ldots,n\right\}$. Thus, our notion of $P \times P$-matrices generalizes the classical notion of $n \times n$-matrices. This generalization "respects" standard concepts like determinants -- e.g., the determinant of an $n \times n$-matrix does not change if we instead consider it as an $N \times N$-matrix. Does this generalization exhibit genuinely new behavior or does it just add convenience? Arguably it's the latter, because conversely, you can turn any $P \times P$-matrix (for any finite set $P$) into an $n \times n$-matrix. Namely, if you have a finite set $P$ and a $P \times P$-matrix $A = \left(a_{p, q}\right)_{\left(p, q\right) \in P \times P} \in \mathbb{Q}^{P \times P}$, then you can set $n = \left|P\right|$ and fix a bijection $\phi : \left\{1,2,\ldots,n\right\} \to P$, and form the $n \times n$-matrix $A_\phi := \left(a_{\phi\left(i\right), \phi\left(j\right)}\right)_{\left(i, j\right) \in \left\{1,2,\ldots,n\right\} \times \left\{1,2,\ldots,n\right\}} \in \mathbb{Q}^{n \times n}$, and observe that $A_\phi$ is "basically the same as" $A$ except that the rows and the columns have been reindexed. We shall refer to this operation (going from $A$ to $A_\phi$) as numerical reindexing. So $P \times P$-matrices differ from $n \times n$-matrices only in how we index their rows and columns. Nevertheless they are useful, because often there is no canonical bijection $\phi : \left\{1,2,\ldots,n\right\} \to P$ (exhibit A: this Putnam problem), and we are algebraists and want to make as few non-canonical choices as possible.

Whenever $P$ is a finite set, we can multiply two $P \times P$-matrices: The product $AB$ of two $P \times P$-matrices $A = \left(a_{p, q}\right)_{\left(p, q\right) \in P \times P} \in \mathbb{Q}^{P \times P}$ and $B = \left(b_{p, q}\right)_{\left(p, q\right) \in P \times P} \in \mathbb{Q}^{P \times P}$ is defined to be the $P \times P$-matrix $\left(\sum_{r \in P} a_{p, r} b_{r, q} \right)_{\left(p, q\right) \in P \times P} \in \mathbb{Q}^{P \times P}$. This multiplication behaves just as nicely as usual multiplication of $n \times n$-matrices. In particular, $\det\left(AB\right) = \det A \cdot \det B$ whenever $A$ and $B$ are two $P \times P$-matrices. And of course, you can prove this by numerical reindexing, because numerical reindexing preserves both products and determinants (assuming, of course, that you pick one bijection $\phi : \left\{1,2,\ldots,n\right\} \to P$ and stick with it).

If you remember the definition of triangular $n \times n$-matrices, then its generalization to $P \times P$-matrices is straightforward: When $P$ is a totally ordered finite set, we say that a $P \times P$-matrix $A = \left(a_{p, q}\right)_{\left(p, q\right) \in P \times P}$ is lower-triangular if \begin{equation} a_{p, q} = 0 \qquad \text{for all } \left(p, q\right) \in P \times P \text{ satisfying } p < q ; \end{equation} and we say that a $P \times P$-matrix $A = \left(a_{p, q}\right)_{\left(p, q\right) \in P \times P}$ is upper-triangular if \begin{equation} a_{p, q} = 0 \qquad \text{for all } \left(p, q\right) \in P \times P \text{ satisfying } p > q . \end{equation} Again, lower-triangular $P \times P$-matrices become lower-triangular $n \times n$-matrices upon numerical reindexing, as long as we make sure to pick our bijection $\phi : \left\{1,2,\ldots,n\right\} \to P$ to be an order isomorphism (which, by the way, is unique). So we can derive properties of lower-triangular $P \times P$-matrices from analogous properties of lower-triangular $n \times n$-matrices by numerical reindexing. In particular, we can thus show that the determinant of a lower-triangular $P \times P$-matrix equals the product of its diagonal entries; i.e., if $A = \left(a_{p, q}\right)_{\left(p, q\right) \in P \times P}$ is a lower-triangular $P \times P$-matrix, then $\det A = \prod\limits_{p \in P} a_{p, p}$. The same holds for upper-triangular $P \times P$-matrices.

Now, let's not forget about the question.

Fix a positive integer $n$. Let $N$ be the set $\left\{1,2,\ldots,n\right\}$. Let $P$ be the set of all the $2^n - 1$ nonempty subsets of $N$. (We shall later totally order $P$, but for now $P$ is just a finite set.)

We shall also use the Iverson bracket notation, which will save us some ampersands and braces. This allows us to rewrite the definition of $A_{ij}$ in your question as $A_{ij} = \left[ \left|S_i \cap S_j\right| = 1 \right]$. But as we said, we drop the numerical indexing, and instead define a matrix whose rows and columns are indexed by $P$ (that is, by all the nonempty subsets of $N$). Thus, the claim of the exercise rewrites as follows:

Theorem 1. Let $A$ be the $P \times P$-matrix $\left( \left[ \left|S \cap T\right| = 1 \right] \right)_{\left(S, T\right) \in P}$. Then, \begin{equation} \det A = \left(-1\right)^{\left[n \neq 1\right]} \prod_{k=1}^n k^{\dbinom{n}{k}} . \end{equation}

The key to proving this is the following simple fact:

Lemma 2. Let $S \in P$ and $T \in P$. Then, \begin{equation} \left[ \left| S \cap T \right| = 1 \right] = \sum\limits_{U \in P} \left(-1\right)^{\left|U\right| - 1} \left|U\right| \cdot \left[ U \subseteq S \right] \left[ U \subseteq T \right] . \end{equation}

This, in turn, shall be derived from the following binomial identity:

Lemma 3. Let $k$ be a nonnegative integer. Then, \begin{equation} \sum_{i=1}^k \left(-1\right)^{i-1} i \cdot \dbinom{k}{i} = \left[k = 1\right]. \end{equation}

Proof of Lemma 3 (sketched). If $k = 0$, then this equality holds for obvious reasons (indeed, the left hand side is an empty sum and thus vanishes, while the right hand side vanishes since $k = 0 \neq 1$). Hence, for the rest of this proof, we WLOG assume that $k \neq 0$. Hence, $k$ is a positive integer. Thus, $k-1$ is a nonnegative integer.

It is well-known that $\sum\limits_{i=0}^n \left(-1\right)^i \dbinom{n}{i} = \left[n = 0\right]$ for every nonnegative integer $n$. Applying this to $n = k-1$, we obtain $\sum\limits_{i=0}^{k-1} \left(-1\right)^i \dbinom{k-1}{i} = \left[k-1 = 0\right] = \left[k = 1\right]$. Multiplying both sides of this equality by $k$, we obtain $k \sum\limits_{i=0}^{k-1} \left(-1\right)^i \dbinom{k-1}{i} = k \left[k = 1\right] = \left[k = 1\right]$ (where the last equality sign is easy to check directly). Hence, \begin{align} \left[k = 1\right] &= k \sum\limits_{i=0}^{k-1} \left(-1\right)^i \dbinom{k-1}{i} = \sum\limits_{i=0}^{k-1} \left(-1\right)^i \cdot \underbrace{k \dbinom{k-1}{i}}_{\substack{= \left(i+1\right) \dbinom{k}{i+1} \\ \text{(by a simple computation)}}} \\ &= \sum\limits_{i=0}^{k-1} \left(-1\right)^i \cdot \left(i+1\right) \dbinom{k}{i+1} = \sum\limits_{i=1}^{k} \left(-1\right)^{i-1} \cdot i \dbinom{k}{i} \end{align} (here, we have substituted $i-1$ for $i$ in the sum). This proves Lemma 3. $\blacksquare$

Proof of Lemma 2 (sketched). If a subset $U \in P$ does not satisfy $U \subseteq S$, then the Iverson bracket $\left[ U \subseteq S \right]$ is zero and thus renders the whole product $\left(-1\right)^{\left|U\right| - 1} \left|U\right| \cdot \left[ U \subseteq S \right] \left[ U \subseteq T \right]$ zero. Similarly, this product also is rendered zero when $U \in P$ does not satisfy $U \subseteq T$. Thus, the product $\left(-1\right)^{\left|U\right| - 1} \left|U\right| \cdot \left[ U \subseteq S \right] \left[ U \subseteq T \right]$ is zero unless $U \in P$ satisfies both $U \subseteq S$ and $U \subseteq T$. Hence, the sum $\sum\limits_{U \in P} \left(-1\right)^{\left|U\right| - 1} \left|U\right| \cdot \left[ U \subseteq S \right] \left[ U \subseteq T \right]$ does not change its value when we restrict it to those $U$ that satisfy both $U \subseteq S$ and $U \subseteq T$ (because all the addends that we lose are zero anyway). In other words, \begin{align*} \sum\limits_{U \in P} \left(-1\right)^{\left|U\right| - 1} \left|U\right| \cdot \left[ U \subseteq S \right] \left[ U \subseteq T \right] &= \sum\limits_{\substack{U \in P; \\ U \subseteq S \text{ and } U \subseteq T}} \left(-1\right)^{\left|U\right| - 1} \left|U\right| \cdot \underbrace{\left[ U \subseteq S \right]}_{=1} \underbrace{\left[ U \subseteq T \right]}_{=1} \\ &= \sum\limits_{\substack{U \in P; \\ U \subseteq S \text{ and } U \subseteq T}} \left(-1\right)^{\left|U\right| - 1} \left|U\right|. \end{align*} Now, let $K$ be the intersection $S \cap T$, and let $k = \left|K\right|$. Then, the sets $U \in P$ that satisfy both $U \subseteq S$ and $U \subseteq T$ are precisely the nonempty subsets of the $k$-element set $K$. Thus, we know how many such sets $U$ exist of each size: there are exactly $\dbinom{k}{1}$ ones of size $1$, exactly $\dbinom{k}{2}$ ones of size $2$, and so on, all the way up to $\dbinom{k}{k}$ ones of size $k$. Hence, \begin{align*} \sum\limits_{\substack{U \in P; \\ U \subseteq S \text{ and } U \subseteq T}} \left(-1\right)^{\left|U\right| - 1} \left|U\right| &= \dbinom{k}{1} \left(-1\right)^{1-1} \cdot 1 + \dbinom{k}{2} \left(-1\right)^{2-1} \cdot 2 + \cdots + \dbinom{k}{k} \left(-1\right)^{k-1} \cdot k \\ &= \sum_{i=1}^k \dbinom{k}{i} \left(-1\right)^{i-1} \cdot i = \sum_{i=1}^k \left(-1\right)^{i-1} i \cdot \dbinom{k}{i} \\ &= \left[k = 1\right] \qquad \left(\text{by Lemma 3}\right) \\ &= \left[\left|S\cap T\right| = 1\right] \end{align*} (since $k = \left|K\right| = \left|S\cap T\right|$). Combining what we have, we obtain \begin{equation} \sum\limits_{U \in P} \left(-1\right)^{\left|U\right| - 1} \left|U\right| \cdot \left[ U \subseteq S \right] \left[ U \subseteq T \right] = \sum\limits_{\substack{U \in P; \\ U \subseteq S \text{ and } U \subseteq T}} \left(-1\right)^{\left|U\right| - 1} \left|U\right| = \left[\left|S\cap T\right| = 1\right] . \end{equation} This proves Lemma 2. $\blacksquare$

Lemma 4. (a) We have $\sum\limits_{S \in P} \left|S\right| = n 2^{n-1}$.

(b) We have $\sum\limits_{S \in P} \left( \left|S\right| - 1 \right) = n 2^{n-1} - 2^n + 1$.

(c) We have $\prod\limits_{S \in P} \left(-1\right)^{\left|S\right| - 1} = \left(-1\right)^{\left[n \neq 1\right]}$.

(d) We have $\prod\limits_{S \in P} \left|S\right| = \prod\limits_{k=1}^n k^{\dbinom{n}{k}}$.

Proof of Lemma 4. (a) Fix $i \in \left\{1,2,\ldots,n\right\}$. Then, exactly $2^{n-1}$ subsets of $\left\{1,2,\ldots,n\right\}$ contain $i$ (since we can freely choose which of the other $n-1$ elements such a subset should contain). Of course, all of these $2^{n-1}$ subsets are nonempty, and thus belong to $P$. Hence, exactly $2^{n-1}$ subsets $S \in P$ contain $i$. Hence, the sum $\sum\limits_{S \in P} \left[i \in S\right]$ has exactly $2^{n-1}$ addends equal to $1$, while all its other addends are $0$. Therefore, this sum equals $2^{n-1}$. In other words, we have \begin{equation} \sum\limits_{S \in P} \left[i \in S\right] = 2^{n-1} . \label{} \tag{1} \end{equation}

Now, forget that we fixed $i$. We thus have proven \eqref{} for each $i \in \left\{1,2,\ldots,n\right\}$.

But we have $\left|S\right| = \sum\limits_{i=1}^n \left[i \in S\right]$ for each subset $S$ of $\left\{1,2,\ldots,n\right\}$ (because the sum $\sum\limits_{i=1}^n \left[i \in S\right]$ contains exactly $\left|S\right|$ many addends equal to $1$, whereas all its other addends are $0$). Summing up this equation over all $S \in P$, we obtain \begin{align} \sum\limits_{S \in P} \left|S\right| &= \sum\limits_{S \in P} \sum\limits_{i=1}^n \left[i \in S\right] = \sum\limits_{i=1}^n \underbrace{\sum\limits_{S \in P} \left[i \in S\right]}_{\substack{= 2^{n-1} \\ \text{(by \eqref{})}}} \\ &= \sum\limits_{i=1}^n 2^{n-1} = n 2^{n-1} . \end{align} This proves Lemma 4 (a).

(b) We have $\left|P\right| = 2^n - 1$ (since the set $P$ consists of all the $2^n$ subsets of $\left\{1,2,\ldots,n\right\}$ apart from the empty set). Now, \begin{align} \sum\limits_{S \in P} \left( \left|S\right| - 1 \right) = \underbrace{\sum\limits_{S \in P} \left|S\right|}_{\substack{= n 2^{n-1} \\ \text{(by Lemma 4 (a))}}} - \underbrace{\sum\limits_{S \in P} 1}_{= \left|P\right| = 2^n - 1} = n 2^{n-1} - \left(2^n - 1 \right) = n 2^{n-1} - 2^n + 1 . \end{align} This proves Lemma 4 (b).

(c) If $n \neq 1$, then $n \geq 2$ (since $n$ is a positive integer), and therefore both $2^{n-1}$ and $2^n$ are even integers. Hence, if $n \neq 1$, then $n 2^{n-1} - 2^n + 1 \equiv 1 \mod 2$. On the other hand, if $n = 1$, then $n 2^{n-1} - 2^n + 1 = 1 \cdot 2^{1-1} - 2^1 + 1 = 0 \equiv 0 \mod 2$. Combining the previous two sentences, we conclude that $n 2^{n-1} - 2^n + 1 \equiv \left[n \neq 1\right] \mod 2$. Hence, $\left(-1\right)^{n 2^{n-1} - 2^n + 1} = \left(-1\right)^{\left[n \neq 1\right]}$. Now, \begin{align} \prod\limits_{S \in P} \left(-1\right)^{\left|S\right| - 1} &= \left(-1\right)^{\sum\limits_{S \in P} \left( \left|S\right| - 1 \right)} = \left(-1\right)^{n 2^{n-1} - 2^n + 1} \qquad \left(\text{by Lemma 4 (b)}\right) \\ &= \left(-1\right)^{\left[n \neq 1\right]} . \end{align} This proves Lemma 4 (c).

(d) Recall that $P$ is the set of all nonempty subsets of $\left\{1,2,\ldots,n\right\}$. Hence, among the elements of $P$ are exactly $\dbinom{n}{1}$ subsets of size $1$, exactly $\dbinom{n}{2}$ subsets of size $2$, and so on, and finally exactly $\dbinom{n}{n}$ subsets of size $n$ (and no other subsets). Therefore, the product $\prod\limits_{S \in P} \left|S\right|$ has exactly $\dbinom{n}{1}$ factors equal to $1$, exactly $\dbinom{n}{2}$ factors equal to $2$, and so on, and finally exactly $\dbinom{n}{n}$ factors equal to $n$ (and no other factors). Therefore, this product equals $\prod\limits_{k=1}^n k^{\dbinom{n}{k}}$. This proves Lemma 4 (d). $\blacksquare$

Proof of Theorem 1 (sketched). Define two $P \times P$-matrices \begin{align*} B &= \left( \left(-1\right)^{\left|T\right| - 1} \left|T\right| \cdot \left[ T \subseteq S \right] \right)_{\left(S, T\right) \in P \times P} \qquad \text{and} \\ C &= \left( \left[ S \subseteq T \right] \right)_{\left(S, T\right) \in P \times P} . \end{align*}

Lemma 2 says that $A = BC$. (More precisely, Lemma 2 says that the $\left(S, T\right)$-th entry of $A$ equals the corresponding entry of $BC$ for all $S \in P$ and $T \in P$; but this yields $A = BC$.) Thus, $\det A = \det\left(BC\right) = \det B \cdot \det C$. Now, it remains to find $\det B$ and $\det C$.

Here we shall finally endow our set $P$ with a total ordering. Namely, we pick any total order on $P$ with the property that every pair of two subsets $S \in P$ and $T \in P$ satisfying $S \subseteq T$ must satisfy $S \leq T$. There are many total orders that do the trick here; for example, we can order them by increasing size (resolving ties arbitrarily). Now, it is easy to see that the $P \times P$-matrix $B$ is lower-triangular. Since the determinant of a lower-triangular $P \times P$-matrix equals the product of its diagonal entries, we thus conclude that \begin{align} \det B &= \prod_{S \in P} \left( \left(-1\right)^{\left|S\right| - 1} \left|S\right| \underbrace{\left[ S \subseteq S \right]}_{=1} \right) = \prod_{S \in P} \left( \left(-1\right)^{\left|S\right| - 1} \left|S\right| \right) \\ &= \underbrace{\left(\prod_{S \in P} \left(-1\right)^{\left|S\right| - 1}\right)}_{\substack{= \left(-1\right)^{\left[n \neq 1\right]} \\ \text{(by Lemma 4 (c))}}} \left( \prod_{S \in P} \left|S\right| \right) = \left(-1\right)^{\left[n \neq 1\right]} \left( \prod_{S \in P} \left|S\right| \right) . \end{align}

Likewise, the $P \times P$-matrix $C$ is upper-triangular. Since the determinant of an upper-triangular $P \times P$-matrix equals the product of its diagonal entries, we thus conclude that \begin{equation} \det C = \prod_{S \in P} \underbrace{\left[ S \subseteq S \right]}_{=1} = 1 . \end{equation} Now, \begin{align} \det A &= \det B \cdot \underbrace{\det C}_{=1} = \det B = \left(-1\right)^{\left[n \neq 1\right]} \underbrace{\left( \prod_{S \in P} \left|S\right| \right)}_{\substack{= \prod\limits_{k=1}^n k^{\dbinom{n}{k}} \\ \text{(by Lemma 4 (d))}}} \\ &= \left(-1\right)^{\left[n \neq 1\right]} \prod_{k=1}^n k^{\dbinom{n}{k}} . \end{align} This proves Theorem 1. $\blacksquare$

In hindsight, the above proof could have been restated without introducing $P \times P$-matrices, since picking a total order on a finite set $P$ is equivalent to picking a bijection $\phi : \left\{1,2,\ldots,n\right\} \to P$. But I prefer to give the exercise an "invariant" formulation, and $P \times P$-matrices are worth exposing anyway.

A few more telegraphic remarks:

  • The decomposition $A = BC$ constructed in the above proof of Theorem 1 is an LU-decomposition.

  • The proof I found first was the same recursive argument appearing a few times in the AoPS thread linked above. But the structure of the recursion foreshadowed the existence of a slicker proof -- the block-row-reduction operations felt like multiplying by a lower-triangular (in a properly chosen order) matrix. So I dug deeper and found the above.

  • This reminds me of Lemma 7.1 of Chris Godsil, An Introduction to the Moebius Function.

  • 1
    Lemma 3 is trivial since $i\binom ki=k\binom{k-1}{i-1}$, – Zhi-Wei Sun Dec 7 at 6:21
  • 5
    @Zhi-WeiSun: I know it is. This is a semi-expository writeup that is supposed to be readable by anyone who understands the problem. – darij grinberg Dec 7 at 6:38
  • @darijgrinberg: Thank you for the detailed work! – T. Amdeberhan Dec 9 at 4:36
  • @darij BRAVO for using Iverson notation in a matrix definition ? While avoiding to create a name names it adds clarity. From a computer point of view we could say "The implementation IS the name." – Jérôme JEAN-CHARLES yesterday

Darij already gave a great elementary detailed exposition. I wanted to remark that all such kinds of results (including the postscript) are special cases of a classical result that (at least in the generality of semilattices) goes back to Lindström in Determinants on semilattices. For this question we are in the special case of the Boolean lattice.

  • Oh, are you applying Lindström's result with $\cup$ as $\wedge$? – darij grinberg Dec 7 at 15:17
  • @darijgrinberg: isn't he using $\cap$ as $\wedge$ in the obvious way? – Sam Hopkins Dec 7 at 16:04
  • @SamHopkins: Because $\cap$ is not always a nonempty subset. This is what confused me, too. But I think $\cup$ can be used perfectly well. – darij grinberg Dec 7 at 16:15
  • @darijgrinberg: another way to resolve that might be to add a row of all $1$s to the bottom of the matrix (thinking of the last row/column as corresponding to the empty set) by choosing the corresponding $f_i$ to be the constant function $1$. When I expand the determinant about the last row, I will just get the determinant of the $(2^n-1)\times(2^n-1)$ submatrix in the upper left because every other $(2^n-1)\times(2^n-1)$ determinant will have a column of all zeros. – Sam Hopkins Dec 7 at 16:21
  • 1
    @darijgrinberg Yes, for the postscript, I was thinking the nonempty subsets with $\cup$. We can take the function $f(A,B)=q^{-|A\cup B|}$ and then multiply the row(or column) of A by $q^{|A|}$. – Gjergji Zaimi Dec 7 at 18:11

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