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?
Best Answer
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.)