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?

  • $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.