On determinants of a certain class of matrices

Question

$\newcommand{\M}{\mathcal{M}}$Recently I encountered a certain class of matrices whose determinants behave in an interesting manner. Define $\M(n,k)$ for positive integers $n,k$ with $k\leq n$ to be the real $n\times n$ matrix with all $1$s on the diagonal, all $1$s for $k-1$ entries to the right of the diagonal on each row, and $0$s everywhere else. Note that if there are less than $k-1$ entries to the right of the diagonal, then the $1$s carry over to the leftmost columns. For example: $$\M(4,2)=\begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\1&0&0&1\end{bmatrix}\quad\text{and}\quad\M(3,1)=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}.$$

At first, I believed that $\M(n,k)$ would always be nonsingular when $k<n$, but this turned out to be false. The smallest example where this conjecuture fails is $\M(4,2)$, whose determinant is $0$. After even more numerical testing I have been led to believe the following new conjecture:

The determinant $\det\M(n,k)$ is $0$ if and only if $\gcd(n,k)>1$. If $n,k$ are coprime, then $\det\M(n,k)=k$.

I have tested for all $k\leq n$ up to $n=9$ with the help of a computer without any counterexamples. Does anyone have an idea how this conjecture might be proven?

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Here are some partial results. It is trivial that $\det\M(n,k)$ is zero when $k=0$ or $n$, and that it is $1$ when $k=1$. I can prove the result when $k=2$ as well: Let $U(n)$ be the $n\times n$ matrix with all entries $0$ except a $1$ at the bottom leftmost corner. Then, we have the recurrence relation $$\det\M(n,2)=\det\big(\M(n-1,2)-U(n-1)\big)+(-1)^{n+1}\det\big(\M(n-1,2)^t-U(n-1)^t\big)$$ which of course simplifies to $$\det\M(n,2)=(1+(-1)^{n+1})\det\big(\M(n-1,2)-U(n-1)\big).$$ Upon verifying $\det\M(2,2)=0$, it's clear that this proves the conclusion for $k=2$ if we assume that $\det\big(\M(n-1,2)-U(n-1)\big)=1$. Call the LHS $x(n-1)$. It is easy to compute that $x(2)=1$, from which it follows by induction $x(n)=1$ for all $n$, since $x(n+1)=x(n)$.

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Unfortunately, I am at a complete loss in any of the cases other than $k=0,1,2,n$. It seems like (for sufficiently large $n$) as $k$ increases from $3$ until $\lfloor n/2\rfloor$, the proof going by the same thinking as my proof for $k=2$ would get increasingly complicated until it becomes hopeless to even attempt. (I could be wrong, of course.) Not to mention the general case for any positive integer $n$.

Does anyone have an idea how to attack the general problem? Thanks in advance!

Answer 1

$\newcommand{\M}{\mathcal{M}}$Thanks to @achillehui's suggestion, I am able to prove the conjecture, and I present the proofs below. For any $n\times n$ circulant matrix $C$ with column vectors being the permutations of a certain vector $c=(c_0,c_1,\dots,c_{n-1})^T$, and with the associated polynomial $p(x)=c_0+c_1x+\dots+c_{n-1}x^{n-1}$, the following is known.

The determinant of $C$ is given by $\displaystyle\prod_{j=0}^{n-1}p(\zeta_n^j)$, where $\zeta_n:=\exp(2\pi i/n)$ is the primitive $n$th root of unity.

The matrices $\M(n,k)$ are circulant, with associated polynomial $p(x)=1+x+\dots+x^{k-1}$. Note that whenever $x\neq 1$, we have $p(x)=(x^k-1)/(x-1)$, so we may write $$\det\M(n,k)=\prod_{j=0}^{n-1}p(\zeta_n^j)=p(1)\prod_{j=1}^{n-1}\frac{\zeta_n^{jk}-1}{\zeta_n^j-1}.$$ In the case where $\gcd(n,k)=:d>1$, we know that $\zeta_n^k$ is no longer a primitive $n$th root of unity, but it is a primitve $(n/d)$th root of unity. Since $d>1$, the integer $n/d$ lies in the set $\{1,2,\dots,n-1\}$ from which $j$ takes values in the product, so there is a term in the product where $j=n/d$. For this term, the numerator is $\zeta_n^{jk}-1=\zeta_{n/d}^{n/d}-1=0$, so that the whole product is $0$ and the determiant is $0$. Conversely, when $d=1$, $\zeta_n^k$ will always be a primitive $n$th root of unity, and their $j$th powers will never be $1$ (for $1\leq j\leq n-1$). Hence the product is nonzero.

The stronger conjecture that $\det\M(n,k)=k$ when $\gcd(n,k)=1$ can be proven in this way as well. It is trivial that $p(1)=k$, so it suffices to show the product $\prod(\zeta_n^{jk}-1)/(\zeta_n^j-1)$ is $1$. The reason this is true is because of cancellation. The element $k$ in the additive group $\mathbb Z/n\mathbb Z$ (which is isomorphic to the multiplicative group of roots of unity) always generates the group whenever $n,k$ are coprime. Furthermore, each element is generated in a unique way. Therefore, as $j$ takes all nonzero values in $\mathbb Z/n\mathbb Z$, every element is represented exactly once, so the cancellation is perfect, proving the result.

Answer 2

If you use row operations, you can see where the periodicity comes in. As an example take $M(5,3)$; subtract the first row from the two bottom ones, then add the second row, etc.:

$$\begin{pmatrix}1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\\{\color{red}1}&0&0&1&1\\{\color{red}1}&{\color{red}1}&0&0&1\end{pmatrix}\mapsto\begin{pmatrix}1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\\0&{\color{red}-\color{red}1}&{\color{red}-\color{red}1}&1&1\\0&0&\color{red}-\color{red}1&0&1\end{pmatrix}\mapsto\begin{pmatrix}1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\\0&0&0&2&1\\0&0&-1&0&1\end{pmatrix}$$

Let us just track the $1$s in the lower left corner as you perform row subtraction or addition. You'll see that the triangle of $1$s moves right with a period of $k$ as seen below (omitting the $0$s):

$$k=2,\qquad 1\mapsto\quad-1\quad\mapsto 1$$ $$k=3,\qquad \begin{matrix}1\\1&1\end{matrix}\quad\mapsto\quad \begin{matrix}-1&-1\\&-1\end{matrix}\quad\mapsto\quad\begin{matrix}&1\\-1\end{matrix}\quad\mapsto\quad\begin{matrix}1\\1&1\end{matrix}$$

$$k=4,\qquad\begin{matrix}1&&\\1&1&\\1&1&1\end{matrix}\quad\mapsto\quad\begin{matrix}-1&-1&-1\\&-1&-1\\&&-1\end{matrix}\quad\mapsto\quad\begin{matrix}&&1\\-1&-1\\&-1\end{matrix}\quad\mapsto\quad\begin{matrix}&&1\\&1&1\\-1\end{matrix}\quad\mapsto\quad\begin{matrix}1\\1&1\\1&1&1\end{matrix}$$

When the $1$s have moved by $k$ steps, they leave a matrix of $0$s behind. Since $\det\begin{pmatrix}A&B\\O&C\end{pmatrix}=\det A\det C$, and since the upper left matrix $A$ has determinant 1, it follows that $\det M(n,k)=\det M(n-k,k)$. So what's left is to check what happens when this triangle of $1$s meets the rest of the matrix as it moves right. The end result is one of $M(k,k)$, $M(k+1,k)$,..., $M(2k-1,k)$. Clearly if $n$ is a multiple of $k$ we reach the matrix $M(k,k)$ of all $1$s which has determinant $0$. I believe you can take it from here.