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12 votes
3 answers
784 views

Can the equation $1+z+z^2=z^n$ for natural $n$ have multiple complex roots $z$?

The question is stated in the title of this post. It is easy to see that, if $z$ is a multiple root of $p_n(z):=1+z+z^2-z^n$, then $(n-2)z^2+(n-1)z+n=0$, so that we can successively express $z^2,\dots,...
Iosif Pinelis's user avatar
16 votes
4 answers
3k views

How many minors I need to check to conclude all minors will vanish ?

Given a $m \times n$ matrix $n>m$, I was trying to check if all its $m \times m$ minor vanish. I remember hearing that one really does not need to check all possible minors in order to conclude ...
Vagabond's user avatar
  • 1,795
17 votes
2 answers
1k views

The GCD-matrix: generalizing a result of Smith?

Let $M$ be the $n\times n$ matrix, known as the GCD matrix, of entries $M_{ij}=\gcd(i,j)$. In the paper H J S Smith, On the value of a certain arithmetical determinant, Proc. London Math. Soc. 7:208-...
T. Amdeberhan's user avatar
12 votes
3 answers
2k views

Representability of matroids over $\mathbb R$

Let $M$ be a matroid, for example viewed as being given by a finite set $X$ and a rank function $d : P(X) \to {\mathbb N}$ such that 1) $d(\varnothing)=0$, $d(\lbrace x \rbrace)=1$, for all $x \in X$,...
Andreas Thom's user avatar
  • 25.5k
6 votes
1 answer
257 views

Families of subsets whose characteristic vectors are spanning sets

Let $X$ be a finite set and $\mathbb CX$ be a vector space with basis $X$. If $Y\subseteq X$ is a subset, then by the characteristic vector of $Y$ I mean $\sum_{y\in Y}y$. My question is: ...
Benjamin Steinberg's user avatar
6 votes
1 answer
588 views

A numerical matrix of power sum polynomials

Let $p_i=x_1^i+x_2^i+\cdots+x_m^i=\sum_{k=1}^mx_k^i$ be the power sum polynomials. Then, the determinant of the $m\times m$ Hankel matrix $M_m=(p_{i+j-2})$, for $1\leq i,j\leq m$, has a neat ...
T. Amdeberhan's user avatar
5 votes
0 answers
150 views

monomer-dimer tiling of a Young diagram

As a modest start, I propose the below problem for a special set of partitions. Perhaps it is known. Let $\lambda_n=(n,n-1,\dots,2,1)$ be the staircase partition and its corresponding Young diagram $...
T. Amdeberhan's user avatar
5 votes
1 answer
213 views

Matrix-valued periodic Fibonacci polynomials

Consider the Fibonacci polynomials $f_n(x)$, defined by the recursion $f_n(x)=xf_{n-1}(x)-f_{n-2}(x)$ with initial values $f_0(x)=0$ and $f_1(x)=1$. It is well known that the values of these ...
Johann Cigler's user avatar
5 votes
0 answers
190 views

Yet, another generalization of Catalan determinants

The discussion on this page is motivated by Johann Cigler's MO question. My intention arose from a possible generalization of Cigler's matrix $$A_{n,m}=\left( \binom{2m}{j-i+m}-\binom{2m}{m-i-j-1} \...
T. Amdeberhan's user avatar
3 votes
0 answers
207 views

On a variation of the Vandermonde matrix

The ubiquitous Vandermonde matrix, of entries $(x_i^{j-1})_{i,j}^{1,n}$, and its determinant $$\prod_{i<j}^{1,n}(x_j-x_i)$$ have found many utilities in Combinatorics and Physics, among other ...
T. Amdeberhan's user avatar
3 votes
1 answer
165 views

The inverse of a symbolic matrix (with reciprocal binomials) has Laurent entries

Recalling the $q$-binomials (Gaussian polynomials). Let $[n]_q!=\prod_{j=1}^n\frac{1-q^j}{1-q}$ and $\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}$. Now, consider the $n\times n$ matrix $\mathbf{M}...
T. Amdeberhan's user avatar
3 votes
0 answers
130 views

Where does this identity involving sums of Hankel-like determinants over partitions come from?

For a partition $\lambda=( \lambda_1,\dots,\lambda_n)\vdash n$ with $\lambda_1\ge\dots\ge\lambda_n\ge0$ and any function $f:\mathbb Z\to\mathbb C$, define a Hankel-like $n\times n$ matrix $$M_f(\...
Wolfgang's user avatar
  • 13.4k