# Induction proof: $\det(M) = \prod_{1 \le j \le n} (x_j - x_i)$

Following problem:

Let $\mathbb{K}$ be a Field and

$M = \begin{pmatrix} 1 & x_1 & \ldots & x_1^{n-1} \\ \vdots & \vdots & & \vdots \\ 1 & x_n & \ldots & x_n^{n-1} \end{pmatrix} \in Mat(n, \mathbb{K})$

Proove with induction that: $$\det(M) = \prod_{1 \le i < j \le n} (x_j - x_i)$$

Unfortunately I am totally lost here. I have confirmed $n = 1$ and for the rest I got the following:

$$\prod_{1 \le i < j \le n + 1} (x_j - x_i) = \prod_{1 \le i < j \le n} (x_j - x_i) \prod_{1 \le i < n + 1} (x_{n+1} - x_i) = \det(M_n) \prod_{1 \le i < n + 1} (x_{n+1} - x_i)$$

From there on I don't know how to continue.

Any ideas? :S Thanks!

• @Winther You are right. My bad I will fix this. Thanks. – FunkyPeanut Apr 7 '15 at 2:22
• Also the formula you write is not correct. $(x_{n+1}-x_n)$ should be $(x_{n+1} - x_1)(x_{n+1} - x_2)\ldots (x_{n+1} - x_n)$. Anyway you cannot prove it (just) this way. You need to use what the determinant of the matrix $M_{n+1}$ is and relate this to the determinant of $M_{n}$. btw the matrix you have is known as the Vandermonde matrix – Winther Apr 7 '15 at 2:23
• @Winther Ah - I see. Sure totally overlooked the fact that there are too product-signs involved... thanks. I'll fix it, too. – FunkyPeanut Apr 7 '15 at 2:32
• Did you try subtracting row $R_{n+1}$ from each row $R_i$, $1 \le i \le n$, then expanding through the first column? – goldenratio Apr 7 '15 at 2:32
• This question has a step by step way to prove it. If you manage to get to point d) then you can apply induction to arrive at the result. – Winther Apr 7 '15 at 2:36