Can't get a lower triangular nxn determinant

Question

I'm trying to solve the following nxn determinant:

$ \displaystyle \begin{vmatrix} 0 & 1 & 1 &\ldots &1\\ 1 & a_1& 0& \ldots & 0\\ 1 & 0 &a_2 &\ldots &0 \\ \vdots & \vdots & \vdots &\ddots & \vdots \\ 1 & 0& 0 & \ldots & a_n \\ \end{vmatrix} $

I've tried subtracting the last row from all the others except the first in order to get a lower triangle determinant but I still don't know how to get rid of the 1 on the bottom left corner. Any hints?

What about subtracting $1/a_i$ times row $i$ from the first row? — Jean-Claude Arbaut, Jan 03 '22 at 14:01

score 2 · Answer 1 · answered Jan 03 '22 at 17:59

2

The determinant $d$ is unchanged if you replace the first row by $$ \mathbf{r}_1 \leftarrow \mathbf{r}_1 - \frac{1}{a_1} \mathbf{r}_2 - \ldots - \frac{1}{a_N} \mathbf{r}_{N+1} $$ (if none of the $a$'s are null)

It follows that $$ d= \begin{vmatrix} -\sum a_n^{-1} & 0 & 0 & & 0\\ 1 & a_1 & & & \\ 1 & & a_2 & & \\ \vdots & & & \ddots & \\ 1 & & & & a_N \end{vmatrix}= -(\sum a_n^{-1})(\prod a_n) $$

answered Jan 03 '22 at 17:59

Steph

3,665

1

+1, but it's not the end of the story. If any $a_i=0$, then make it nonzero and $a_i\to0$. Since the determinant is a polynomial in the matrix elements, it's also continuous. Now, find out the polynomial: the formula you found can be simplified to $-\sigma_{n-1}=-\sum_i\prod_{j\ne i} a_j$. And this one is valid for any $a_i$, zero or nonzero. – Jean-Claude Arbaut Jan 04 '22 at 09:23
very nice simplification ! – Steph Jan 04 '22 at 14:56
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Can't get a lower triangular nxn determinant

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