Show that these matrices are invertible for all $p>3$

I am working on a paper which will extend a result in my thesis and have boiled one problem down to the following: show that the symmetric matrix $$M_p$$, whose definition follows, is invertible for all odd primes $$p$$. Letting $$p>3$$ be prime and $$\ell = \frac{p-1}{2}$$, we define $$M_p = \begin{pmatrix} 2ij - p - 2p\left\lfloor\frac{ij}{p}\right\rfloor\end{pmatrix}_{1\leq i,j\leq \ell}$$

Examples:

1. For $$p=5$$ we have $$M_5 = \begin{pmatrix} -3 & -1 \\ -1 & 3 \end{pmatrix}$$ and $$\det(M_5) = -1\cdot 2\cdot 5$$.

2. For $$p=7$$ we have $$M_7 = \begin{pmatrix} -5 & -3 & -1 \\ -3 & 1 & 5 \\ -1 & 5 & -3 \end{pmatrix}$$ and $$\det(M_7) = 2^2 \cdot 7^2$$.

3. For $$p=11$$ we have $$M_{11} = \begin{pmatrix} -9 & - 7 & -5 & -3 & -1 \\ -7 & -3 & 1 & 5 & 9 \\ -5 & 1 & 7 & -9 & -3 \\ -3 & 5 & -9 & -1 & 7 \\ -1 & 9 & -3 & 7 & -5 \end{pmatrix}$$ and $$\det(M_{11}) = -1\cdot 2^4\cdot 11^4$$.

Though this (seemingly) nice formula that we see above fails for primes greater than 19, though the determinant has been checked to be non-zero for primes less than 1100. (My apologies if this question is not as motivated or as well discussed as is desired. If there are any questions or if further clarification is needed just let me know!)

• Is it important that $p$ be a prime number ? Wouldn't it be sufficient that it is odd ? Mar 9, 2022 at 20:03
• Great question... this formula is coming from some Jacobian (change of variables) in my research and if you carry it through for primes the formula condenses down to this. If you do it for odd numbers the formula ends up being a bit less nice but seems true for those also. Like you said though if we use this formula for odd numbers the determinant seems to be non-zero as well. (maybe not 9? Though there is something to correct this). Mar 9, 2022 at 20:14
• No need for the apology: This is a very well formulated question for this forum. Mar 10, 2022 at 8:57
• Most matrices are invertible. Mar 10, 2022 at 10:48
• @GerryMyerson OTOH, most matrices that come up in practice aren't most matrices. Mar 10, 2022 at 19:11

Experimentally, we have the following formula for $$p$$ prime: $$\det(M_p)=(-1)^{(p^2-1)/8}(2p)^{(p-3)/2}h_p^-\;,$$ where $$h_p^-$$ is the minus part of the class number of the $$p$$-th cyclotomic field, itself essentially equal to a product of $$\chi$$-Bernoulli numbers. I have not tried to prove this, but since there are many determinant formulas for $$h_p^-$$ in the literature, it should be possible.

• As you correctly guessed this exact determinant in the problem is considered here ams.org/journals/proc/1955-006-02/S0002-9939-1955-0069207-2 Mar 10, 2022 at 13:13
• @VladMatei Thank you for providing the reference as well! There were some useful papers which cited this one also. Much appreciated! Apr 1, 2022 at 15:54
• This is a wonderful answer and one that would have taken me (as someone who hasn't thought too much about specific class numbers) much more time to figure out. So thank you for looking into this and writing your response! It was extremely helpful! Apr 1, 2022 at 16:04

Here are some elements to solve the question.

1st step. Extend $$M_p$$ as a $$(2\ell)\times(2\ell)$$-matrix $$N_p$$, with the same definition of entries. By the way the entries may be writen as $$a_{j,k}=p(2\left\{\frac{jk}p\right\}-1)$$ where $$\{x\}=x-\lfloor x\rfloor$$ is the fractional part of $$x$$. Remark that $$a_{j,p-k}=-a_{j,k}$$ and likewise $$a_{p-j,k}=-a_{j,k}$$. Remark that $$a_{j,k}$$ depends only upon the product $$jk\mod p$$, thus can be written $$a_{jk}$$ instead.

Therefore $$N_p=\begin{pmatrix} M_p & -M_pF \\ -FM_p & FM_pF \end{pmatrix}=\binom{I}{-F}M_p\begin{pmatrix} I & -F \end{pmatrix}$$ where $$F$$ is the anti-unit matrix, $$f_{j,k}=\delta_j^{\ell+1-k}$$.

2nd step. From the formula above, $$N_p$$ and $$M_p$$ have the same rank. Thus $$M_p$$ is invertible if and only if $$N_p$$ has rank $$\ell$$. This rank is unchanged by a permutation of rows. Thus we examine instead the matrix $$N_p'$$ obtained by the involution of rows given by the inversion $$j\mapsto j^{-1}$$ $$\mod p$$. That way, the entries of $$N_p'$$ are the numbers $$a_{j^{-1}k}$$.

3rd step. The matrix $$N_p'$$ is diagonalizable, its eigenpairs being explicit, because this is a group matrix (entries of the form $$a_{g^{-1}h}$$) for the multiplicative group $${\mathbb F}_p^\times$$ (the non-zero elements of $${\mathbb Z}/p$$). Recall that because $${\mathbb Z}/p$$ is a field, this group is isomorphic to the additive group $${\mathbb Z}/2\ell$$ : there exists an element $$\theta\in{\mathbb Z}/p$$ of order exactly $$2\ell$$, and an isomorphism is $$\psi(k)=\theta^k$$ ($$k\mod2\ell$$, $$\theta^k\mod p$$).

The eigenvectors $$v^\omega$$ have coordinates $$v^\omega_k=\omega^{\psi^{-1}(k)},\qquad 1\le k\le2\ell,$$ where $$\omega$$ is any complex solution of $$\omega^{2\ell}=1$$. The corresponding eigenvalue is $$\lambda_\omega=\sum_{r=0}^{2\ell-1}a_{\psi(r)}\omega^r=:Q_p(\omega).$$

4th step. There remains to see that among these $$2\ell$$ eigenvalues, $$\ell$$ of them only vanish, so that $$N_p$$ has rank $$2\ell-\ell=\ell$$. I leave this as an open question (though I am rather optimistic). At least, the fact that $$\ell$$ of them vanish is clear, because we have $$\theta^\ell=-1$$, hence $$\psi(k+\ell)=-\psi(k)$$. There follows that $$a_{\psi(k+\ell)}=-a_{\psi(k)}$$, so that $$Q_p$$ factorizes: $$Q_p(X)=(X^\ell-1)R_p(X),\qquad R_p(X)=\sum_{r=0}^{\ell-1}a_{\psi(r)}X^r.$$ Thus $$\lambda_\omega=0$$ for every root of $$\omega^\ell=1$$. Thus there remains only to prove that if instead $$\omega^\ell=-1$$ (the other roots of unity), one has $$R_p(\omega)\ne0$$. I have done a few calculations for small prime numbers $$p$$, which make me optimistic.

Edit. One can conclude at least when $$\ell$$ is a power of $$2$$ (that is when $$p$$ is a Fermat prime). Because then $$X^\ell+1$$ is irreducible over $$\mathbb Q$$, while $$R_p\in{\mathbb Z}[X]$$ has degree $$\ell-1$$, hence they cannot share a root.

In the general case, the question reduces to whether some cyclotomic polynomial $$\Phi_m$$ with $$m|\ell$$ can be such that $$\Phi_m(-X)|R_p(X)$$. Notice that $$\Phi_m$$ is reciprocal, while $$R_p$$ is not, thus this divisibility will imply some other ones, ...

• Thank you for your response! I was able to use Cohen's observation but your answer was helpful and enlightening. I appreciate you taking the time to answer in such detail! Apr 1, 2022 at 15:56