I f $$A_{(n+1)\times(n+1)}= \begin{pmatrix}x_{1}^{n}& x_{1}^{n-1} &\ldots& x_{1}& 1 \\ x_{2}^{n}& x_{2}^{n-1} &\ldots& x_{2}& 1\\ \vdots & \vdots & &\vdots &\vdots \\ x_{n}^{n}& x_{n}^{n-1} &\ldots& x_{n}& 1 \\ 1&1&\ldots&1&1\end{pmatrix}$$ be a matrix, i like to calculate the $det (A)$. I think that this matrix is rather than like vandrmonde matrix, what is your idea?
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i think $det(A)$ is a product of $(x-x_{i})$ 's. – Ali Qurbani Feb 18 '16 at 10:30
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Classic Vandermonde matrix has rows of elements in geometric progression, starting from 1.
You can transform this matrix in a Vandermonde one by multiplying each row $i$ by $x_i^{-n}$, so that the row becomes $$ 1\quad \left(\frac{1}{x_i}\right)\quad \left(\frac{1}{x_i}\right)^2\quad \left(\frac{1}{x_i}\right)^3\cdots\left(\frac{1}{x_i}\right)^n $$
The formula should follow easily. When you compute the determinant with the Vandermonde formula, remember that you have scaled the matrix.
EDIT, yeah, do like Donkey_2009 says.
Giovanni Resta
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There's no need to multiply by $x_i^{-n}$; just swapping the rows and columns so the $1$s appear at the left hand side will put it into that form. – John Gowers Feb 18 '16 at 10:54
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$\det(A)$ will be a degree ${n+1}\choose 2$ polynomial in $x_1,\dots,x_n$. (Why?)
Now observe that for $i\ne j$, if $x_i=x_j$ then $\det(A)=0$ by properties of the determinant. Therefore, $(x_i-x_j)$ is a factor of that polynomial for each $i\ne j$.
Furthermore, if $x_i=1$ for any $i$ then $\det(A)=0$. So $(x_i-1)$ is a factor of $\det(A)$ for each $i$.
Putting all this together, we see that $\prod_{i=1}^n(x_i-1)\prod_{j<i}(x_i-x_j)$ is a factor of $\det(A)$. But what is the degree of that polynomial? And what does that tell us?
John Gowers
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