Determinant of an $n\times n$ Toeplitz matrix

Question

Let $A = (a_{ij}) \in R^{n\times n}$. Find the determinant if:

$$a_{ij}= |i-j|$$

So we have the symmetric matrix \begin{bmatrix} 0 & 1 & 2 & 3 & 4 & \dots & n-1 \\ 1 & 0 & 1 & 2 & 3 & \dots & n-2 \\ 2 & 1 & 0 & 1 & 2 & \dots & n-3 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ n-1 & n-2 & n-3 & \dots & \dots & \dots & 0 \end{bmatrix}

But i can't find a way to diagonalize the matrix nor find the determinant by Laplace expansion.

Any ideas ??

looks like $(-1)^n n 2^{n-1}$ could be the answer... See here: http://oeis.org/A085750 — Fabian, Nov 18 '15 at 21:26
Such matrices are called Toeplitz matrices, and there exist "fast" algorithms for computing their determinants. — hardmath, Nov 18 '15 at 21:33
But I've to use "regular methods" (row or columns operations, Laplace expansion..) to argument an answer, in fact i had to calculate the determinants of $a_{ij}$ = $max(i,j)$ and $a_{ij}$ = $min(i,j)$ and I used those methods, do you think it's possible to calculate that matrix in a similar way ?? — C. Alexander, Nov 18 '15 at 21:47
@Fabian The sign’s wrong in your proposed solution. It’ll be $(-1)^{n-1}$. — amd, Nov 18 '15 at 22:16
@amd: I think, it should be $n\mapsto n-1$ in my suggestion. — Fabian, Nov 19 '15 at 05:51

score 7 · Accepted Answer · edited Feb 04 '19 at 21:55

If we do row reduction, i.e. $R_n-R_{n-1}$, $R_{n-1}-R_{n-2}$, up to $R_2-R_1$, and then again do $R_n-R_{n-1}$, $R_{n-1}-R_{n-2}$, up to $R_3-R_2$ and finally $R_2+R_1$, then we get \begin{align} &\begin{vmatrix} 0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\ 1 & 0 & 1 & 2 & 3 & \cdots & n-2 \\ 2 & 1 & 0 & 1 & 2 & \cdots & n-3 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ n-1 & n-2 & n-3 & \cdots & \cdots & \dots & 0 \end{vmatrix}\\ \ \\ =&\begin{vmatrix} 0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\ 1 & -1 & -1 & -1 & -1 & \cdots & -1 \\ 1 & 1 & -1 & -1 & -1 & \cdots & -1 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1&1&1&\cdots&\cdots&-1&-1 \\ 1 & 1 & 1 & \cdots & \cdots & 1 & -1 \end{vmatrix}\\ \ \\ =&\begin{vmatrix} 0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\ 1 & 0 & 1 & 2 & 3 & \cdots & n-1 \\ 0 & 2 & 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0&0&0&\cdots&2&0&0 \\ 0 & 0 & 0 & \cdots & \cdots & 2 & 0 \end{vmatrix}\\ \end{align} Now we calculate the determinant and its minors all by the last row: \begin{align} =&\begin{vmatrix} 0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\ 1 & 0 & 1 & 2 & 3 & \cdots & n-1 \\ 0 & 2 & 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0&0&0&\cdots&2&0&0 \\ 0 & 0 & 0 & \cdots & \cdots & 2 & 0 \end{vmatrix}\\ \ \\ =&-2\begin{vmatrix} 0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\ 1 & 0 & 1 & 2 & 3 & \cdots & n-1 \\ 0 & 2 & 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0&0&0&\cdots&\cdots&2&0 \\ \end{vmatrix}\\ \ \\ &=(-1)^{n-2}2^{n-2}\begin{vmatrix}0&n-1\\1&n-1 \end{vmatrix}\\ \ \\ &=(-1)^{n-1}2^{n-2}(n-1). \end{align}

Appreciate it, in fact I did the first row reduction but I didn't see a clear path from there. Thanks to everyone for the help — C. Alexander, Nov 19 '15 at 06:42
Nice answer. +1. I was wondering whether this can be done using definition of determinants (as summation over permutations, etc - in this case it will be over all derangements) — Shailesh, Nov 26 '15 at 14:24
That sounds complicated: all diagonals but one are nonzero, so it's all about the cancelations. — Martin Argerami, Nov 27 '15 at 02:39

score 2 · Answer 2 · edited Feb 04 '19 at 21:52

Call your matrix $A$ and let $J$ be the upper triangular nilpotent Jordan block of size $n$ (i.e. the superdiagonal of $J$ contains ones and all other entries are zero). Then $A=U+U^T$, where $U=J+2J^2+\ldots+(n-1)J^{n-1}$. Now, note that $(I-J)U=J+J^2+\ldots+J^{n-1}$ is the strictly upper triangular matrix of ones. Therefore $$ (I-J)U(I-J^T)= \pmatrix{ -1&&&&1\\ &-1&&&1\\ &&\ddots&&1\\ &&&-1&1\\ &&&&0 } $$ and hence \begin{align*} B &:= (I-J)A(I-J^T) = (I-J)(U+U^T)(I-J^T) \\ &= (I-J)U(I-J^T) + ((I-J)U(I-J^T))^T \\ &= \pmatrix{ -2&&&&1\\ &-2&&&1\\ &&\ddots&&1\\ &&&-2&1\\ 1&1&\cdots&1&0 }. \end{align*} Since $\det(I-J)=1$, we get $\det B=\det A$. Now, adding one half of each of the first $n-1$ rows of $B$ to the last row, we get $$ \det B= \det\pmatrix{ -2&&&&1\\ &-2&&&1\\ &&\ddots&&1\\ &&&-2&1\\ &&&&\frac{n-1}2 }=\frac{(-2)^{n-1}(n-1)}2. $$

Determinant of an $n\times n$ Toeplitz matrix

2 Answers2

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