Editing Vandermonde matrix (section)

===Third proof: row and column operations===

The third proof is based on the fact that if one adds to a column of a matrix the product by a scalar of another column then the determinant remains unchanged.

So, by subtracting to each column – except the first one – the preceding column multiplied by <math>x_0</math>, the determinant is not changed. (These subtractions must be done by starting from last columns, for subtracting a column that has not yet been changed). This gives the matrix 
:<math>V = \begin{bmatrix}
1&0&0&0&\cdots&0\\
1&x_1-x_0&x_1(x_1-x_0)&x_1^2(x_1-x_0)&\cdots&x_1^{n-1}(x_1-x_0)\\
1&x_2-x_0&x_2(x_2-x_0)&x_2^2(x_2-x_0)&\cdots&x_2^{n-1}(x_2-x_0)\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&x_n-x_0&x_n(x_n-x_0)&x_n^2(x_n-x_0)&\cdots&x_n^{n-1}(x_n-x_0)\\
\end{bmatrix}</math>
Applying the [[Laplace_expansion|Laplace expansion formula]] along the first row, we obtain <math>\det(V)=\det(B)</math>, with
:<math>B = \begin{bmatrix}
x_1-x_0&x_1(x_1-x_0)&x_1^2(x_1-x_0)&\cdots&x_1^{n-1}(x_1-x_0)\\
x_2-x_0&x_2(x_2-x_0)&x_2^2(x_2-x_0)&\cdots&x_2^{n-1}(x_2-x_0)\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
x_n-x_0&x_n(x_n-x_0)&x_n^2(x_n-x_0)&\cdots&x_n^{n-1}(x_n-x_0)\\
\end{bmatrix}</math>
As all the entries in the <math>i</math>-th row of <math>B</math> have a factor of <math>x_{i+1}-x_0</math>, one can take these factors out and obtain
:<math>\det(V)=(x_1-x_0)(x_2-x_0)\cdots(x_n-x_0)\begin{vmatrix}
1&x_1&x_1^2&\cdots&x_1^{n-1}\\
1&x_2&x_2^2&\cdots&x_2^{n-1}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&x_n&x_n^2&\cdots&x_n^{n-1}\\
\end{vmatrix}=\prod_{1<i\leq n}(x_i-x_0)\det(V')</math>,
where <math>V'</math> is a Vandermonde matrix in <math>x_1,\ldots, x_n</math>. Iterating this process on this smaller Vandermonde matrix, one eventually gets the desired expression of <math>\det(V)</math> as the product of all <math>x_j-x_i</math> such that <math>i<j</math>.