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I want to solve the following determinant:

$D_n= \begin{vmatrix} a_n & a_{n-1} & \cdots & a_2 & x\\ a_n & a_{n-1} & \cdots & x & a_1\\ a_n & a_{n-1} & \cdots & a_2 & a_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ a_n & x & \cdots & a_2 & a_1\\ a_n & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$

The idea I had was to 1) get the $a_n$ in front of the determinant, which gets me to:

$D_n= a_n \begin{vmatrix} 1 & a_{n-1} & \cdots & a_2 & x\\ 1 & a_{n-1} & \cdots & x & a_1\\ 1 & a_{n-1} & \cdots & a_2 & a_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & x & \cdots & a_2 & a_1\\ 1 & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$

Then I multiplied the nth row with $-1$ and added it to all the other ones, which gives me:

$D_n= a_n \begin{vmatrix} 0 & 0 & \cdots & 0 & x-a_1\\ 0 & 0 & \cdots & x-a_2 & 0\\ 0 & 0 & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & x-a_{n-1} & \cdots & 0 & 0\\ 1 & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$

So, I got a triangular determinant, but with the opposite diagonal, not the main one. How do I transform it into a real triangular determinant? Obviously, my idea to replace every adjacent row (1st and nth, (n-1)th and 2nd...) is wrong? Any ideas?

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  • $\begingroup$ Exchanging two adjacent rows multiplies the determinant by $-1$. Now exchange $n$th row with $(n-1)$th, then $(n-1)$th with $(n-2)$th until the last row becomes the first. This needs $n-1$ permutations. Then iterating the procedure we see that the appropriate factor is $(-1)^{ (n-1) +(n -2)+\ldots +1}= (-1)^{n(n-1)/2}$. $\endgroup$ Jan 12, 2016 at 23:40
  • $\begingroup$ But is it really $\begin{vmatrix} a_n & a_{n-1} & \cdots & a_2 & x\\ a_n & a_{n-1} & \cdots & x & a_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ a_n & x & \cdots & a_2 & a_1\\ a_{n} & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$, or rather $\begin{vmatrix} a_n & a_{n-1} & \cdots & a_2 & x\\ a_n & a_{n-1} & \cdots & x & a_1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ a_n & x & \cdots & a_2 & a_1\\ x & a_{n-1} & \cdots & a_2 & a_1\\ \end{vmatrix}$? $\endgroup$ Jan 12, 2016 at 23:43
  • $\begingroup$ @Startwearingpurple Would the solution then be $(-1)^{n(n-1)/2}(x-a_n!)$? $\endgroup$
    – Quant
    Jan 12, 2016 at 23:44
  • $\begingroup$ @AndreasCaranti If you mean that there is a mistake in the original text of the problem, there isn't. I also double-checked just in case. $\endgroup$
    – Quant
    Jan 12, 2016 at 23:45
  • $\begingroup$ I am not sure about my interpretation of your notation. The answer will be $(-1)^{n(n-1)/2}a_n \prod_{k=1}^{n -1}(x-a_k)$. $\endgroup$ Jan 12, 2016 at 23:47

2 Answers 2

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We can simply calculate the determinant of an opposite (lower) triangular matrix:

Let $J_n$ be the $n \times n$ matrix with $1$ on the anti-diagonal and $0$ otherwise (i.e. $J_ne_i = e_{n+1-i}$ for all $1 \leq i \leq n$, where $e_1, \dotsc, e_n$ denotes the standard basis). Given any $m \times n$-matrix $A$ the matrix $AJ_n$ originates from $A$ by vertically mirroring its colums from the middle, i.e. swapping the first column with the last, the second with the second last, etc.

If $A$ is an $n \times n$ square matrix then we get from $J_n^2 = I_n$ that $$ \det(A) = \det(J_n) \det(AJ_n). $$ In the case of $D_n$ we get that $D_n J_n$ is the $a_n$-scalar multiple a lower triangular matrix with diagonal entries $x-a_1, x-a_2, \dotsc, x-a_{n-1}, 1$, so $$ \det(D_n) = \det(J_n) \det(D_n J_n) = \det(J_n) a_n (x-a_1) \dotsm (x-a_{n-1}). $$

So the only difference is that we need to know $\det(J_n)$. Because $J_n$ is a permutation matrix, corresponding to $\sigma_n \in S_n$ with $\sigma(i) = n+1-i$, we have $\det(J_n) = \mathrm{sgn}(\sigma_n)$. Notice that \begin{align*} \sigma_{2n} &= (1 \;\; 2n) (2 \;\; n-1) \dotsm (n \;\; n+1) \\ \sigma_{2n+1} &= (1 \;\; 2n+1) (2 \;\; n-1) \dotsm (n \;\; n+2). \end{align*} So we can just count the number of transpositions used and get that $$ \mathrm{sgn}(\sigma_n) = \begin{cases} \phantom{-}1 & \text{if $n \equiv 0,1 \bmod 4$}, \\ -1 & \text{if $n \equiv 2,3 \bmod 4$}, \end{cases} = (-1)^{n(n-1)/2}. $$

So alltogether we have $$ \det(D_n) = (-1)^{n(n-1)/2} a_n (x-a_1) \dotsm (x-a_{n-1}). $$

(The nice thing about this is that now that we have calculated $\det(J_n) = (-1)^{n(n-1)/2}$ we can use this to calculate the determinant of opposite triangular and opposite diagonal matrices more ore less in the usual way.)

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You have gone almost through most of the way! The determinant then simply is:$$D=(-1)^{n+1}(x-a_1)(-1)^{n+1}(x-a_2)(-1)^{n+1}(x-a_3)\cdots (-1)^{n+1}(x-a_{n-1})(-1)^{n+1}=(x-a_1)(x-a_2)\cdots (x-a_{n-1})$$

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