Determinant of a symmetric Toeplitz matrix

Question

Let $A_n = (a_{ij})$ be an $n\times n$ matrix such that $a_{ii} = 0, a_{ij} = 2$ when $|j − i| = 1$, and $a_{ij} = 1$ otherwise. The question is to find the determinant in terms of $n$. I computed the first six terms, depending on $n$, but, unfortunately, no clear relationship was found. Here they are: \begin{align} \det A_1&=0,\\ \det A_2&=-4,\\ \det A_3&=8,\\ \det A_4&=-7,\\ \det A_5&=0,\\ \det A_6&=7. \end{align} Laplace expansion turned out to be useful only if $n$ is known. How can one derive a formula for $\det A_n$ in terms of $n$?

Answer 1

Its probably the best one can do, is to write a recursion relation

$$ n^2 \det (n)+\left(2 n^2+3 n+2\right) \det (n-2)+\left(2 n^2+n+1\right) \det (n-1)+(n+1)^2 \det (n-3)=0 $$

Answer 2

Here is a partial answer. Call your matrix $A_n$. We claim that $$ \det A_n= \begin{cases} 1-n &\text{ if }\ n\equiv1\mod 6,\\ -4 &\text{ if }\ n\equiv2\mod 6,\\ n+5 &\text{ if }\ n\equiv3\mod 6,\\ -(n+3) &\text{ if }\ n\equiv4\mod 6,\\ 0 &\text{ if }\ n\equiv5\mod 6,\\ n+1 &\text{ if }\ n\equiv0\mod 6. \end{cases} $$ Proof. We shall prove this claim below when the remainder of $n$ divided by $6$ is not $2$.

When the remainder is $5$, $\det A_n=0$ because $$ (1,1,0,-1,-1,\underbrace{0,1,1,0,-1,-1}_{\text{repeat}},\ldots)^T\in\ker A_n. $$

When the remainder is $1,3,4,6$, write $A_n=B_n+ee^T$, where $e$ is the all-one vector and $B_n$ is the symmetric tridiagonal Toeplitz matrix whose main diagonal entries are $-1$s and whose super/sub-diagonal entries are ones. By Laplace expansion, it is easy to see that $\det B_n=-\det B_{n-1}-\det B_{n-2}$. Solving the recurrence relation with the initial conditions $\det B_1=-1$ and $\det B_2=0$, we get $$ \det B_n=\begin{cases}-1\\ 0\\ 1\end{cases} \text{ when } n\equiv\begin{cases}1\\ 2\\ 3\end{cases}\mod3.\tag{1} $$ It is straightforward to verify that $$ B_n^{-1}e= \begin{cases} (-1,\underbrace{0,2,3,2,0,-1}_{\text{repeat}},\ldots)^T &\text{ if }\ n\equiv1\mod 6,\\ (2,3,2,\underbrace{0,-1,0,2,3,2}_{\text{repeat}},\ldots)^T &\text{ if }\ n\equiv3\mod 6,\\ (1,2,2,1,\underbrace{0,0,1,2,2,1}_{\text{repeat}},\ldots)^T &\text{ if }\ n\equiv4\mod 6,\\ (\underbrace{0,1,2,2,1,0}_{\text{repeat}},\ldots)^T &\text{ if }\ n\equiv0\mod 6. \end{cases} $$ It follows that $$ e^TB_n^{-1}e= \begin{cases} n-2 &\text{ if }\ n\equiv1\mod 6,\\ n+4 &\text{ if }\ n\equiv3\mod 6,\\ n+2 &\text{ if }\ n\equiv4\mod 6,\\ n &\text{ if }\ n\equiv0\mod 6. \end{cases}\tag{2} $$ Hence the assertion follows from $(1), (2)$ and Sherman-Morrison formula $$ \det A_n = \det(B_n + ee^T) = (1+e^TB_n^{-1}e) \det B_n. $$ This completes our proof when the remainder of $n$ divided by $6$ is not $2$. When the remainder is $2$, by using the fact that $\ker A_{6k+5}\cap \ker B_{6k+5}$ is nonempty, we can actually prove that $\det A_n=-4$ by showing that the Schur complement of $A_2$ in $A_n$ has determinant $1$. However, this will make the proof too long and so we shall omit this part here.