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In my linear algebra class, my professor gave us this determinant for practice:

$$det\pmatrix {1 & 2 & 3 & \dots & n \\ 2 & 3 & 4 & \dots & 1 \\ \vdots & \vdots & \vdots & \dots & \vdots \\ n & 1 & 2 & \dots & n-1}$$

Where the $i$-th row or column is the set $$\{1,2,\dots,n\}$$ shifted $i-1$ times to the left.

He recommended we add rows $2$ through $n$ to row $1$. And then factor out the constant. This yields:

$$\frac{n(n+1)}{2}det\pmatrix {1 & 1 & 1 & \dots & 1 \\ 2 & 3 & 4 & \dots & 1 \\ \vdots & \vdots & \vdots & \dots & \vdots \\ n & 1 & 2 & \dots & n-1}$$

However, I cannot for the life of me find a useful way to simplify this further. I have tried adding the first column to each subsequent column, subtracting the first row from each subsequent row, but nothing seems to simplify the matrix. Perhaps, there is an inductive solution, but I haven't been able to find one simply from the $2\times2$ and $3\times3$ cases.

Any help would be appreciated.

Peter Kagey
  • 5,052

1 Answers1

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We use the fact that multiplying row $i$ by $\lambda \neq 0$ then adding it to row $j$ does not change the determinant of the matrix.

Also, multiplying row $i$ by $\lambda \neq 0$ also multiples the determinant by $\lambda$. Thus,

\begin{align} \ & det\pmatrix{1 & 2 & 3 & \dots & n \\ 2 & 3 & 4 & \dots & 1 \\ 3 & 4 & 5 & \dots & 2 \\ \vdots & \vdots & \vdots & & \vdots \\ n & 1 & 2 & \dots & n-1} \\ \ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 \\ 2 & 3 & 4 & \dots & 1 \\ 3 & 4 & 5 & \dots & 2 \\ \vdots & \vdots & \vdots & & \vdots \\ n & 1 & 2 & \dots & n-1} && \text{as you have} \\ \ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 1 & 2 & \dots & n-3 & -2 & -1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 1 & 2-n & \dots & -3 & -2 & -1\\0 & 1-n & 2-n & \dots & -3 & -2 & -1} \\ \ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 0 & 0 & \dots & 0 & -n & 0 \\ 0 & 0 & 0 & \dots & -n & -n & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & -n & \dots & -n & -n & 0\\0 & -n & -n & \dots & -n & -n & 0} \\ \ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 0 & 0 & \dots & 0 & -n & 0 \\ 0 & 0 & 0 & \dots & -n & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & -n & \dots & 0 & 0 & 0\\0 & -n & 0 & \dots & 0 & 0 & 0} \\ \ & = \frac{(-1)^nn^{n-1}(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 0 & 0 & \dots & 0 & 1 & 0 \\ 0 & 0 & 0 & \dots & 1 & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & 1 & \dots & 0 & 0 & 0\\0 & 1 & 0 & \dots & 0 & 0 & 0} \\ \ & = \frac{(-1)^nn^{n-1}(n+1)}{2}\cdot (1 \cdot 1 \cdot 1 \cdot \; \cdots \cdot 1 \cdot -1)(-1)^{\lceil n/2 \rceil +1} \\ \ & = \frac{(-1)^{n+\lceil n/2 \rceil}n^{n-1}(n+1)}{2} \end{align}

where the second to last line uses the permutation formula for the determinant.

The permutation formula gives the stated result because, as I have explained in the comment section, you have to choose one entry from each row and column and compute their product.

But in the last row, we have to choose the $1$ in order to not get a product of $0$. Similarly with the second to last row, up to the third row.

So far, we have already chosen one entry from each of columns $2$ to $n-1$, thus we need to choose an entry from columns $1$ and $n$ in the remaining two rows. In the second row, we can only choose the $-1$ at the far right to avoid a $0$ product, meaning we can only choose the $1$ in the first column for the first row.

This product is $1\cdot 1\cdot 1\dots 1\cdot -1 \cdot 1=-1$

The parity of this permutation is even if $n=1,2,5,6,9,10,\dots$ and is odd if $n=3,4,7,8,11,12,\dots$, so $sgn (\sigma) = (-1)^{\lceil n/2 \rceil +1}$, giving the determinant of the reduced matrix to be $-1 \cdot (-1)^{\lceil n/2 \rceil +1} = (-1)^{\lceil n/2 \rceil}$

  • I was unfamiliar with the permutation formula for the determinant. However, it makes sense to simplify it into this form if that is possible – Orren Ravid Apr 18 '18 at 03:38
  • Formally stated, if a square matrix $A$ has entries $a_{ij}$ where $1≤i,j≤n$ then $$det(A)=\sum_{\sigma \in S_n} \text{sgn} (\sigma) \prod_{i=1}^na_{i,\sigma (i)}$$ where $S_n$ is the permutation group on ${ 1,2,3,\dots,n }$ – glowstonetrees Apr 18 '18 at 05:18
  • The above formula might be quite complicated, but intuitively it really just means you choose one entry from each row and column, multiply them together, multiply by the parity of permutation, and sum over all such possibilities. – glowstonetrees Apr 18 '18 at 05:21
  • For example, consider the familiar $n=3$ case, so $$A=\begin{pmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \ \end{pmatrix}$$ – glowstonetrees Apr 18 '18 at 05:23
  • The way we would usually compute the determinant is to first multiply the entries on the main diagonal $a_{11}a_{22}a_{33}$. This corresponds to the identity permutation $\sigma =()$ because $1$ is matched with $1$, $2$ with $2$ and $3$ with $3$, so $sgn(\sigma )=1$ – glowstonetrees Apr 18 '18 at 05:26
  • And then we have $a_{11}a_{23}a_{32}$. Here, the $1$ is matched with $1$, but the $2$ and $3$ are switched around, so it corresponds to the permutation $\sigma = (23)$. Since it is represented by $1$ (an odd number) swap, so $sgn(\sigma )=-1$ and it becomes $-a_{11}a_{23}a_{32}$. And so on. – glowstonetrees Apr 18 '18 at 05:28