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I am interested in symmetric matrices where the $(i,j)$ entry is $\min(i,j)$, i.e.,

$$\begin{pmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 1 & 2 & 2 & \dots & 2 & 2 \\ 1 & 2 & 3 & \dots & 3 & 3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & 2 & 3 & \dots & n-1 & n-1 \\ 1 & 2 & 3 & \dots & n-1 & n \end{pmatrix}$$

Note that these symmetric matrices are related to Lehmer matrices. Have these symmetric matrices earned a name?


Motivation

These matrices are interesting because they are the Gramians of binary triangular matrices, e.g.,

$$ \begin{bmatrix}1 & 1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1 & 1\\ 0 & 0 & 1 & 1 & 1\\ 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}^\top \begin{bmatrix}1 & 1 & 1 & 1 & 1\\ 0 & 1 & 1 & 1 & 1\\ 0 & 0 & 1 & 1 & 1\\ 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2 & 2\\ 1 & 2 & 3 & 3 & 3\\ 1 & 2 & 3 & 4 & 4\\ 1 & 2 & 3 & 4 & 5\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0\\ 1 & 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0\\ 1 & 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 1 & 0\\ 1 & 1 & 1 & 1 & 1 \end{bmatrix}^\top$$

Thus, the determinant of these symmetric matrices is $1$.


Some findings

  • In 2002, Mario Catalani studied these matrices

  • In 2006, Rajendra Bhatia called these $\min$ matrices

  • In 2016, Mika Mattila & Pentti Haukkanen called the $n \times n$ matrix of this form the MIN matrix of the set $\{1,2,\dots,n\}$

  • In 2023, Darij Grinberg studied a generalization of these matrices.


Related

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    $\begingroup$ These matrices recently occurred here: math.stackexchange.com/a/4694871/42969 $\endgroup$
    – Martin R
    Commented May 8, 2023 at 11:12
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    $\begingroup$ @RodrigodeAzevedo The matrix was used in my answer here: math.stackexchange.com/questions/1614111/… $\endgroup$
    – River Li
    Commented May 14, 2023 at 14:32
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    $\begingroup$ @RodrigodeAzevedo A related matrix is the one whose $(i,j)$-entry is $n + 1 - \max(i,j)$. For $n=4$, $$\left[ \begin {array}{cccc} 4&3&2&1\\ 3&3&2&1 \\ 2&2&2&1\\ 1&1&1&1\end {array} \right]$$ $\endgroup$
    – River Li
    Commented May 14, 2023 at 14:46

1 Answer 1

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This is not to suggest a terminology, but to inform a relationship of the matrix to (algebraic) geometry after the comment. Thank you for your interest.

As one of the links shows, the inverse of our matrix is the negative of $$G=\begin{pmatrix} -2 & 1 \\ 1 & -2 & 1 \\ & 1 & -2 & 1 \\ &&&\ddots\\ &&& 1 & -2 &1\\ &&&& 1 & -1 \end{pmatrix}.$$ This matrix is classically known to arise in projective geometry as follows. Recall that for a 2-dimensional complex manifold $Z$ and its point $z\in Z$, we can perform blowing up at $z$: it creates a new manifold $Z_1$ with a map $\pi_1\colon Z_1\rightarrow Z$, which has the property that $Z_1-\pi_1^{-1}(z)\simeq Z-\{z\}$ via $\pi_1$ (bijective, isomorphic) and $E_1:=\pi_1^{-1}(z)\simeq \mathbb{P}^1$ is isomorphic to a projective line. In other words, $Z_1$ is the manifold which is identical to $Z$ except that the center $z\in Z$ is replaced by an exceptional curve $E_1\subset Z_1$.

Now, starting from $Z=\mathbb{C}^2$, we repeat blowings up $n$ times and get maps $$X=Z_n \stackrel{\pi_n}{\rightarrow} Z_{n-1}\stackrel{\pi_{n-1}}{\rightarrow} \cdots \stackrel{\pi_2}{\rightarrow} Z_1\stackrel{\pi_1}{\rightarrow} Z.$$ The center of $\pi_i$ is chosen to be a general (i.e., not special) point of the previous exceptional curve $E_{i-1}\subset Z_{i-1}$. (In fact, the choice is not very important.) The resulting $X$ has $n$ exceptional curves, which form a basis of the middle dimensional homology group $H_2(X,\mathbb{Z})$.

As a general theory via the Poincar'{e} duality, we have the intersection pairing $$H_2(X,\mathbb{Z})\times H_2(X,\mathbb{Z})\rightarrow \mathbb{Z}$$ which gives a symmetric bilinear form on $H_2(X,\mathbb{Z})$. Some computations show that the form values at the basis are $$(E_i,E_j)=\delta_{1,|i-j|},\ (E_n^2)=-1,\ (E_k^2)=-2\ (1\leq i,j\leq n, 1\leq k\leq n-1),$$ which can be seen also as a consequence of basic theorems in intersection theory. Therefore, the Gram matrix with respect to the basis $E_1,\dots,E_n$ coincides with $G$ above.

Finally, we note that some of the properties of the matrix are closely related to theorems in geometry: the intersection pairing is unimodular by Poincar'{e} duality, which gives that the determinant is $\pm 1$. Also, the (negative-) definiteness of the matrix can be geometrically proved from the Hodge index theorem.

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  • $\begingroup$ Thank you! The matrix $-G$ is very close to the Laplacian of a linear graph. Can you think of any graph-theoretic ways of looking at such matrices? $\endgroup$ Commented May 8, 2023 at 21:11
  • $\begingroup$ @RodrigodeAzevedo The linear graph you mentioned is what we call "the dual graph of exceptional curves on $X$." The difference between $-G$ and graph Laplacian seems to be the existence of ambient space, but I have no idea how to interpret it in graph theory. $\endgroup$
    – Ayaka
    Commented May 9, 2023 at 1:53
  • $\begingroup$ Consider solving the Laplace equation with a Neumann boundary condition on one side, and Dirichlet BC on the other, with a point source (delta) right hand side. To satisfy the PDE in the interior, the solution must be two straight lines that meet in a cusp at the delta function, and the change in slope at the cusp must be one. To satisfy the Neumann BC, the line in the Neumann side must have zero slope. Discretizing and identifying columns of the matrix with the greens function, and we arrive at this matrix $\endgroup$
    – Nick Alger
    Commented May 17, 2023 at 7:35
  • $\begingroup$ @NickAlger Thank you. That is worthy of an answer, IMHO $\endgroup$ Commented May 17, 2023 at 7:37

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