# Proof for a formula of a special real determinant

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

• That's a Vandermonde determinant. – Lord Shark the Unknown Oct 26 '18 at 17:17
• The matrix on the right hand side of your last equation is actually the product $A_{n-1} B$ for some triangular matrix $B$. – darij grinberg Oct 27 '18 at 2:59
• Thanks! Going to look into that! – C. Brendel Oct 27 '18 at 4:14

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?

• I think that's what I've looked for! Going to try that later in school :) – C. Brendel Oct 27 '18 at 4:14
• Yes exactly what I needed! Thanks! – C. Brendel Oct 27 '18 at 9:22