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determinant of matrix
jokerjoestar   7
N Mar 31, 2025 by loup blanc
Calculate the determinant of the matrix below: \[ A = \begin{bmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 3 & 4 & \cdots & n & 1 \\ 3 & 4 & 5 & \cdots & 1 & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{bmatrix} \]
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jokerjoestar
Mar 30, 2025
loup blanc
Mar 31, 2025
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determinant of matrix
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jokerjoestar
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jokerjoestar
#1 h  Mar 30, 2025, 3:32 PM • 1 Y
Y by MihaiT
Calculate the determinant of the matrix below: \[ A = \begin{bmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 3 & 4 & \cdots & n & 1 \\ 3 & 4 & 5 & \cdots & 1 & 2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & n & 1 & \cdots & n-3 & n-2 \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{bmatrix} \]
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paxtonw
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paxtonw
#2 h  Mar 30, 2025, 4:31 PM
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Let A be the given n x n matrix.

The last two rows are cyclic shifts of each other, meaning one is just a shifted version of the other.

So, the last two rows are linearly dependent.

A matrix with two linearly dependent rows has determinant 0.

By the way.. can anyone tell me why people ask these questions? Is it for school, work, etc? Or for fun.
This post has been edited 1 time. Last edited by paxtonw, Mar 30, 2025, 4:33 PM
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loup blanc
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loup blanc
#3 h  Mar 30, 2025, 7:14 PM
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When $n=10$, $\det(A)=-55\times 10^8$.
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Soupboy0
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Soupboy0
#4 h  Mar 30, 2025, 7:20 PM
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When $n = 2, \det{A} = 3$
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vanstraelen
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vanstraelen
#5 h  Mar 30, 2025, 7:46 PM
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$\det A_{n}=-\frac{n(n+1)}{2} \cdot n^{n-2}$.
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ysharifi
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ysharifi
#8 h  Mar 30, 2025, 10:37 PM
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If you switch the rows (2, n), (3,n-1), ..., you'll get a circulant matrix and there's a formula for the determinant of circulant matrices. In your case, the answer is $(-1)^{\lfloor n/2 \rfloor}\frac{n^{n-1}(n+1)}{2}.$
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ddot1
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ddot1
#9 h  Mar 31, 2025, 2:04 AM • 1 Y
Y by MihaiT
paxtonw wrote:
The last two rows are cyclic shifts of each other, meaning one is just a shifted version of the other.

So, the last two rows are linearly dependent.
The last two rows can't be linearly dependent, since that would mean that one is a multiple of the other.

As others mentioned, there's a nice formula for the determinant of a circulant matrix, where each row is a shift of the previous:
$$A=\begin{pmatrix}a_1& a_2&a_3&\cdots& a_{n-1}&a_n\\ a_n& a_1&a_2&\cdots& a_{n-2}&a_{n-1}\\ a_{n-1}& a_n&a_1&\cdots& a_{n-3}&a_{n-2}\\ \cdots\\ a_2& a_3&a_4&\cdots &a_n &a_1 \end{pmatrix}.$$It turns out that it has eigenvalues $$\begin{pmatrix}1\\ \zeta\\ \zeta^2\\ \vdots \\ \zeta^{n-1} \end{pmatrix},$$where $\zeta$ is any $n$th root of unity. The corresponding eigenvectors are $a_1+a_2\zeta+a_3\zeta^2+\cdots+a_n\zeta^{n-1},$ so that gives a nice factorization for the determinant: $$\det A=\prod_{\zeta^n=1}\left(a_1+a_2\zeta+a_3\zeta^2+\cdots+a_n\zeta^{n-1}\right).$$
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loup blanc
3563 posts
loup blanc
#10 h  Mar 31, 2025, 1:46 PM • 2 Y
Y by MS_asdfgzxcvb, MihaiT
Here $A$ is an anti-circulant matrix; to find a circulant matrix it is enough to consider $JA$, where $J$ is the permutation
$J_{1,1}=1$, $J_{i,n-i+2}=1$ for $i=2,\cdots, n$ and the others $J_{i,j}$ are $0$ (note that $J^2=I_n$).
Indeed, $JA=I+2P+3P^2+\cdots+nP^{n-1}$, where $P$ is the cycle $(n,n-1,\cdots,2,1)$. Let $f(x)=1+2x+\cdots+nx^{n-1}$.
Let $u=\exp(2i\pi/n)$; then $spectrum(P)=\{u^k;k=0,\cdots,n-1\}$ and $spectrum(JA)=(\lambda_k)_k=(f(u^k))_k$.
$\lambda_0=n(n+1)/2$. $f(x)=(\dfrac{x(1-x^n)}{1-x})'=\dfrac{(1-(n+1)x^n)(1-x)+x-x^{n+1}}{(1-x)^2}$ and if $x^n=1$ and $x\not= 1$, then
$f(x)=\dfrac{-n}{1-x}$. Up to the signum, $\det(JA)=\Pi_k \lambda_k=\pm \dfrac{n(n+1)n^{n-1}}{2\Pi_{k>0}(1-u^k)}$.
Exercise: $\Pi_{k>0}(1-u^k)= n$. Finally $\det(JA)=\pm \dfrac{n(n+1)n^{n-2}}{2}$.
This post has been edited 4 times. Last edited by loup blanc, Mar 31, 2025, 1:52 PM
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