Proof for a formula of a special real determinant

Question

For some $x_1,\cdots,x_n$ in $\mathbb{R}$ we define: $$A_n:= \left(\begin{matrix} 1 &x_1 &x_1^2 &\cdots &x^{n-1}_1\\ 1 &x_2 &x_2^2 &\cdots &x_2^{n-1}\\ \vdots &\vdots &\vdots & &\vdots\\ 1 &x_n &x_n^2 &\cdots &x^{n-1}_n \end{matrix}\right) \in \mathbb{R}^{n\times n} $$ I now need to show that: $$\det(A_n)=\prod_{1 \le i < j \le n}(x_j-x_i)$$ I tried subtracting the $n$-th row from all others to get: $$\det(A_n)=\det\left(\begin{matrix} 0 &x_1-x_n &x_1^2-x_n^2 &\cdots &x^{n-1}_1-x_n^{n-1}\\ 0 &x_2-x_n &x_2^2-x_n^2 &\cdots &x_2^{n-1}-x_n^{n-1}\\ \vdots &\vdots &\vdots & &\vdots\\ 0 &x_{n-1}-x_n &x_{n-1}^2-x_n^2 &\cdots &x_{n-1}^{n-1}-x_n^{n-1}\\ 1 &x_n &x_n^2 &\cdots &x^{n-1}_n \end{matrix}\right)$$ With the usage of Laplaces Expansion utilising the numerous $0$s I was able to write: $$ \begin{align*} \det(A_n) &=(-1)^{n+1}\cdot1\cdot\det \left(\begin{matrix} x_1-x_n &x_1^2-x_n^2 &\cdots &x^{n-1}_1-x_n^{n-1}\\ x_2-x_n &x_2^2-x_n^2 &\cdots &x_2^{n-1}-x_n^{n-1}\\ \vdots &\vdots & &\vdots\\ x_{n-1}-x_n &x_{n-1}^2-x_n^2 &\cdots &x_{n-1}^{n-1}-x_n^{n-1}\\ \end{matrix}\right)\\ &=(-1)^{n+1}\det\left(x_i^j-x_n^j\right)_{\substack{1\le i \le n-1\\1\le j \le n-1}} \end{align*} $$ Because I'm allowed to use the fact that $$x^n-y^n=(x-y)\sum_{k=0}^{n-1}x^ky^{n-1-k}$$ I then used the multilinearity of the determinant ($(x_i-x_n)$ is independant of $j$): $$ \begin{align*} \det(A_n) &=(-1)^{n+1}\det\left((x_i-x_n)\sum_{k=0}^{j-1} x_i^k\,x_n^{j-1-k}\right)_{i,j}\\ &=(-1)^2(-1)^{n-1}\prod_{1\le i < n} (x_i-x_n) \det\left(\sum_{k=0}^{j-1} x_i^k\,x_n^{j-1-k}\right)_{i,j}\\ &=\prod_{1\le i < n} (x_n-x_i)\det\left(\sum_{k=0}^{j-1} x_i^k\,x_n^{j-1-k}\right)_{i,j} \end{align*} $$ This reminds me an awful lot of an induction; for that I'd need to show that: $$\det(A_{n-1})=\det\left(\sum_{k=0}^{j-1} x_i^k\,x_n^{j-1-k}\right)_{i,j}$$ But unfortunately at this point I do not know how to proceed. I probably got stuck, trying to solve it this way, with the consequence of overlooking an easier proof :P Does anyone see an/the solution and is willing to guide me in the right direction?

Thanks for checking in and reading all that above :)

~Cedric

Answer 1

Take the case $n = 4$ as an example. Let $(x_1,x_2,x_3,x_4) = (x,y,z,t)$, we have

$$\begin{align} \left|\begin{matrix} 1 & x & x^2 & x^3 \\ 1 & y & y^2 & y^3 \\ 1 & z & z^2 & z^3 \\ 1 & t & t^2 & t^3 \\ \end{matrix}\right| &\stackrel{\color{blue}{[1]}}{=} \left|\begin{matrix} 0 & x-t & x^2-t^2 & x^3-t^3 \\ 0 & y-t & y^2-t^2 & y^3-t^3 \\ 0 & z-t & z^2-t^2 & z^3-t^3 \\ 1 & t & t^2 & t^3 \\ \end{matrix}\right|\\ &\stackrel{\color{blue}{[2]}}{=} (-)^{n-1}\left|\begin{matrix} x-t & x^2-t^2 & x^3-t^3 \\ y-t & y^2-t^2 & y^3-t^3 \\ z-t & z^2-t^2 & z^3-t^3 \end{matrix}\right|\\ &\stackrel{\color{blue}{[3]}}{=} (t-x)(t-y)(t-z)\left|\begin{matrix} 1 & x+t & x^2 + tx + t^2\\ 1 & y+t & y^2 + ty + t^2\\ 1 & z+t & z^2 + tz + t^2 \end{matrix}\right|\\ &\stackrel{\color{blue}{[4]}}{=} (t-x)(t-y)(t-z) \left|\begin{matrix}1 & x & x^2\\ 1 & y & y^2 \\ 1 & z & z^2\end{matrix}\right| \end{align}$$

• $\color{blue}{[1]}$ - subtract $n^{th}$ row from $k^{th}$ row for $k = 1,\ldots,n-1$.
• $\color{blue}{[2]}$ - Laplace expand against column $1$.
• $\color{blue}{[3]}$ - extract common factors from each row.
• $\color{blue}{[4]}$ - substract $x_n$ times $(k-1)^{th}$ column form $k^{th}$ column for $k = 2, \ldots, n - 1$.

Can you see the pattern?