Calculating the determinant of this matrix

Question

Given this (very) tricky determinant, how can we calculate it easily?

$$\begin{pmatrix} \alpha + \beta & \alpha \beta & 0 & ... & ... & 0 \\ 1 & \alpha + \beta & \alpha \beta & 0 & ... & 0 \\ 0 & 1 & \alpha + \beta & \alpha \beta & ... & ... \\ ... & ... & ... & ... & ... & 0 \\ ... & ... & .... & ... & ... & \alpha \beta \\ 0 & 0 & 0 & ... & 1 & \alpha + \beta \\ \end{pmatrix} \in M_{n\times n}$$

EDIT:

I have to prove it is equal to $\frac{{\alpha}^{n+1} - {\beta}^{n+1}}{\alpha - \beta}$

Any help is appreciated, I just could not find a trick to ease it up!

Show that the result works for $n=1$ and $n=2$ and fulfills the recursion worked out by vadim123. You can also systematically solve the recursion to arrive at the result. By the way, the solution is also known as Dickson Polynomial of the second kind $E_n(\alpha+\beta,\alpha\beta)=\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}$. — ccorn, May 06 '13 at 20:24

score 6 · Answer 1 · answered May 06 '13 at 20:02

Let $D_n$ represent the determinant of this $n\times n$ matrix. Expanding on the first column, we see that $D_n=(\alpha+\beta)D_{n-1}-\alpha\beta D_{n-2}$, where the second is found by expanding on the first row of the resulting minor. The recurrence begins with $D_1=\alpha+\beta, D_2=(\alpha+\beta)^2-\alpha\beta=\alpha^2+\alpha\beta+\beta^2$.

To prove that this recurrence equals $\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}$ one can use induction.

I edited it to what I need to prove! please see it! – TheNotMe May 06 '13 at 20:04 — TheNotMe, May 06 '13 at 20:04

Riccardo · Answer 2 · 2013-05-07T07:43:57.290

You can consider factorizing the matrix using L-U factorization. Let A be your matrix, $L$ the matrix $$\begin{pmatrix} p_1 & 0 & 0 & ... & ... & 0 \\ 1 & p_2 & 0 & 0 & ... & 0 \\ 0 & 1 & p_3 & 0 & ... & ... \\ ... & ... & ... & ... & ... & 0 \\ ... & ... & .... & 1 & p_{n-1} & 0 \\ 0 & 0 & 0 & ... & 1 & p_n \\ \end{pmatrix}$$

and U the matrix

$$\begin{pmatrix} 1 & q_1 & 0 & ... & ... & 0 \\ 0 & 1 & q_2 & 0 & ... & 0 \\ 0 & 0 & 1 & q_3 & ... & ... \\ ... & ... & ... & ... & ... & 0 \\ ... & ... & .... & 0 & 1 & q_{n-1} \\ 0 & 0 & 0 & ... & 0 & 1\\ \end{pmatrix}$$

where

$p_{11} = a_{11}; \\ q_1= a_{12}/p_1; \\ p_2= a_{22} - a_{21}q_1; \\ q_2 = a_{23} / p_2; \\ p_3= a_{33}-a_{32}q_2; \\ \dots \\ \dots \\ q_{n-1}= a_{n-1,n}/p_{n-1}; \\ p_n = a_{nn} - a_{n,n-1}q_{n-1};$

then by Binet formula, you can note that $\det A = \det L \det U= \det L = \prod_{i=1}^n p_i$

(used the fact that $L,U$ are upper/lower triangular)

then you can use induction. hope it helps

Calculating the determinant of this matrix

2 Answers2

Linked

Related