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Start with a palindromic sequence of integers $(a_0, a_1, \ldots, a_{n+1})$, i.e. $a_j=a_{n+1-j}$, and put $a_j:=0$ for $j<0$ and $j>n+1$. You may readily guess that the choice of the binomial coefficients $a_j=\binom{n+1}j$ has been at the origin for this question.

So we consider the $n \times n$-matrix $B$ whose $\left(i, j\right)$-th entry is $a_{2j-i}$ for all $i, j \in \left\{ 1, 2, \ldots, n \right\}$.

E.g. for $n=4$ and the sequence $(1,2,3,3,2,1)$, this would be $$ B = \left(\begin{array}{rrrrr} 2& 3& 1 & 0 \\ 1 & 3& 2& 0 \\ 0 & 2& 3 & 1 \\ 0 & 1 & 3& 2 \end{array}\right).$$

Now what about the eigenvalues of such a matrix? In the case of binomial coefficients, they turned out to be integers in the cited question, and more precisely $2^1, 2^2, \ldots, 2^n$. In the general case, one might expect the characteristic polynomial of $B$ to be irreducible, apart from the trivial right eigenvector $(1,\dots,1)$ with eigenvalue $s:=\sum_{j=0}^{n/2} a_j$ for even $n$. But I have found that it always splits into two (generally irreducible) factors of degrees $\lfloor\frac {n+1}2\rfloor$ and $\lfloor\frac {n-1}2\rfloor$. And of course I wonder why.

Moreover, it appears that the absolute terms of those two factoring polynomials have an integer ratio $R$, which is in fact just a linear combination of the $a_j$'s: $$R=\sum_{j\in\mathbb Z} (-1)^{j+1}a_{\lfloor\frac {n}2\rfloor+2j}.$$ This is about the easiest way to write this finite sum without having to distinguish between even and odd $n$. In fact, for odd $n=2k+1$, half of the terms cancel out by palindromicity, making this $$R=-a_k +2 a_{k-2}-2 a_{k-4}+2 a_{k-6}-+\cdots,$$ while for even $n=2k$, all the different $a_j$ (thus, say, $a_0,...,a_k$) occur with signs $-++--++-\cdots$, viz. $$R=-a_k + a_{k-1}+ a_{k-2}- a_{k-3}-++-\cdots.$$

Some examples for small $n$, writing the initial sequence as $(a,b,c,b,a)$ etc.: for $n=3$, we have $$P(x)=\Bigl[x-\color{blue}{b}\Bigr]\Bigl[x^2-(b+c)x+\color{blue}{b}\underbrace{(2a-c)}_R\Bigr] $$

For $n=4$, $$P(x)=\overbrace{\Bigl[x-(a+b+c)\Bigr]}^{\text{trivial}}\Bigl[x+\color{blue}{(a-b)}\Bigr]\Bigl[x^2-cx+\color{blue}{(a-b)}\underbrace{(a+b-c)}_R\Bigr] $$

For $n=5$, $$P(x)=\Bigl[x^2+(-a+b+c)x+\color{blue}{(-ab+bc-ad)}\Bigr]\cdot\\ \Bigl[x^3-(a+b+c+d)x^2+(ab-bc+bd+cd)x+\color{blue}{(-ab+bc-ad)}\underbrace{(2b-d)}_R\Bigr] $$

For $n=6$, $$P(x)=\overbrace{\Bigl[x-(a+b+c+d)\Bigr]}^{\text{trivial}}\cdot\Bigl[x^2 + (a - c)x + \color{blue}{(a^2-b^2 + bc-ad)}\Bigr]\cdot\\ \Bigl[x^3-(b+d)x^2+(b^2-c^2+cd-ab)x+\color{blue}{(a^2-b^2 + bc-ad)} \underbrace{(-a+b+c-d)}_R\Bigr] $$

As a by-product it follows that if the determinant of $B$ is divided by $R$ (or, for even $n$, by $Rs$), we remain with a square.

Sadly, there do not seem to exist matrices $M_1$ and $M_2$ of respective sizes $\lfloor\frac {n+1}2\rfloor\times \lfloor\frac {n+1}2\rfloor$ and $\lfloor\frac {n-1}2\rfloor\times \lfloor\frac {n-1}2\rfloor$ with entries depending in an "easy" (i.e. linear) way of the $a_j$ such that for a suitable matrix $V$ $$B=V^{-1}\pmatrix{M_1& 0\\ 0 & M_2}V\quad \text{ resp. (for even $n$)}\quad B=V^{-1}\pmatrix{s&0&0\\ 0 &M_1&0\\0& 0 & M_2}V. $$

So how to prove this reducibility of the characteristic polynomial and the conjectured ratio of the absolute terms?

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  • $\begingroup$ You are correct that I had a bug in my calculations. Corrected Sage code $\endgroup$ Mar 11 at 0:00
  • $\begingroup$ There is a similar behaviour if the columns are shifted by $3$ instead of $2$. The entries $B_{i,j}=a_{3j-i}$ are now from a palindromic sequence of odd length $2n+3$. Sage code. For even $n$, both factors have the same coefficients in the $a_i$ up to some signs. For odd $n$, no integer quotient of CTs. $\endgroup$
    – Wolfgang
    Mar 13 at 20:04

1 Answer 1

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I have found an answer for the reducibility part of the question. In fact, I found that any "palindromic matrix" $M=((a_{i,j}))_{i,j\in\{1,..n\}}$ (i.e. with 180° rotational symmetry $a_{i,j}=a_{n+1-i,n+1-j}$) has a splitting characteristic polynomial, and moreover, the suggested matrix decompositions do exist.
After all, this looks quite elementary, though it does not seem to be well-known (or not known at all?) Thanks go to Peter Taylor for the idea with the Sage code. Sage is so much more efficient than Pari/GP when it comes to factoring polynomials with more than about 4 variables, so I'd never have been able to spot the patterns with Pari. (And the zero entries in the original matrices, a special case of this, didn't help either.)

For even order $n=2k$, let's write $$M=\pmatrix{A&B^{\circ}\\ B &A^\circ}=M^\circ.$$ For odd order $n=2k+1$, let's write $$M=\pmatrix{A&v&B^{\circ}\\ w&c&w^{\leftarrow}\\ B &v^{\uparrow}&A^\circ}=M^\circ.$$

Here, $A$ and $B$ are general $k\times k$ matrices with integer entries, $v$ is a column vector of length $k$, $w$ is a row vector of length $k$, $v^{\uparrow}$ and $w^{\leftarrow}$ denote the vectors with mirrored entries, thus reflected vertically/horizontally, $A^\circ={A^{\uparrow}}^{\leftarrow}={A^{\leftarrow}}^{\uparrow}$ a matrix rotated by 180°, and $c\in\mathbb Z$ is the central entry. (Note that for the vectors, we could as well write $v^\circ$ and $w^\circ$.)

For even $n=2k$, define matrices $M_1$ and $M_2$ of size $ k$ simply by $$M_1:= A-B^{\uparrow},\quad M_2:= A+B^{\uparrow}.$$ For odd $n=2k+1$, define matrices $M_1$ and $M_2$ of respective sizes $ k$ and $ k+1$ by $$M_1:= A-B^{\uparrow},\quad M_2:= \pmatrix{A+B^{\uparrow}&2v\\ w&c}.$$ Then in both cases, the characteristic polynomial splits as $\chi_M(x)=\chi_{M_1}(x)\chi_{M_2}(x)$.
Here is some Sage code, where each matrix is followed by its charpoly.

Note that if we replace $M$ by $M^T$, then $M_1$ is just transposed, but $M_2$ becomes for odd $n$ a different matrix $\pmatrix{A^T+(B^T)^{\leftarrow}&2w^T\\ v^T&c}=\pmatrix{A+B^{\uparrow}&v\\ 2w&c}^T$ with the same characteristic polynomial.

Coming back to the actual question, it remains to prove that $\frac{\det M_2}{\det M_1}$ has the conjectured form (removing the factor $\sum a_i$ for even $n$). For even $n$, I now think that the form $$ B=V^{-1}\pmatrix{s&0&0\\ 0 &M_1&0\\0& 0 & M_2}V$$ does also exist, but I have a hard time finding the general form of $M_1$.

The first of the characteristic polynomials are $$p=x+a_0-a_1 \Longrightarrow M_1=(-a_0 + a_1)\quad\ \ \ \quad\\ p=x^2 + (a_0 - a_2)x + a_0^2 - a_1^2 + a_1a_2- a_0a_3 \\ \Longrightarrow M_1=\pmatrix{ -a_0 + a_1 &a_0\\ -a_0 + a_1 - a_2 + a_3 & -a_1 + a_2 }$$ (up to transposition), so far almost trivial. But the next one is already unfeasible to do by hand: $$p=x^3 - a_3x^2 + \bigl[(a_0-a_2) (a_2-a_3 ) + ( a_0-a_1) ( a_4- a_3 )\bigr]x\\ - a_0^3 + a_0^2(a_1 + a_4)+a_0(a_1^2+ a_3^2 - a_2a_3 - 2a_1a_4)- a_1^3 + a_1^2a_4+ a_1a_2^2 - a_1a_2a_3. $$ I have asked another question of independent interest about how to recover such a matrix from its characteristic polynomial.

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