Value of Vandermonde Determinant/Formulation 1/Proof 1

Theorem
The Vandermonde determinant of order $n$ is the determinant defined as follows:


 * $V_n = \begin{vmatrix}

1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1} \end{vmatrix}$

Its value is given by:
 * $\displaystyle V_n = \prod_{1 \le i < j \le n} \left({x_j - x_i}\right)$

Proof
Let $V_n = \begin{vmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\ 1 & x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1} \end{vmatrix}$.

By Multiple of Row Added to Row of Determinant, we can subtract row 1 from each of the other rows and leave $V_n$ unchanged:


 * $V_n = \begin{vmatrix}

1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\ 0 & x_2 - x_1 & x_2^2 - x_1^2 & \cdots & x_2^{n-2} - x_1^{n-2} & x_2^{n-1} - x_1^{n-1} \\ 0 & x_3 - x_1 & x_3^2 - x_1^2 & \cdots & x_3^{n-2} - x_1^{n-2} & x_3^{n-1} - x_1^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & x_{n-1} - x_1 & x_{n-1}^2 - x_1^2 & \cdots & x_{n-1}^{n-2} - x_1^{n-2} & x_{n-1}^{n-1} - x_1^{n-1} \\ 0 & x_n - x_1 & x_n^2 - x_1^2 & \cdots & x_n^{n-2} - x_1^{n-2} & x_n^{n-1} - x_1^{n-1} \end{vmatrix}$

Similarly without changing the value of $V_n$, we can subtract, in order, $x_1$ times column $n-1$ from column $n$, $x_1$ times column $n-2$ from column $n-1$, and so on, till we subtract $x_1$ times column $1$ from column $2$.

The first row will vanish all apart from the first element $a_{11} = 1$.

On all the other rows, we get, with new $i$ and $j$:
 * $\displaystyle a_{ij} = \left({x_i^{j-1} - x_1^{j-1}}\right) - \left({x_1 x_i^{j-2} - x_1^{j-1}}\right) = \left({x_i - x_1}\right) x_i^{j-2}$:


 * $V_n = \begin{vmatrix}

1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & x_2 - x_1 & \left({x_2 - x_1}\right) x_2 & \cdots & \left({x_2 - x_1}\right) x_2^{n-3} & \left({x_2 - x_1}\right) x_2^{n-2} \\ 0 & x_3 - x_1 & \left({x_3 - x_1}\right) x_3 & \cdots & \left({x_3 - x_1}\right) x_3^{n-3} & \left({x_3 - x_1}\right) x_3^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & x_{n-1} - x_1 & \left({x_{n-1} - x_1}\right) x_{n-1} & \cdots & \left({x_{n-1} - x_1}\right) x_{n-1}^{n-3} & \left({x_{n-1} - x_1}\right) x_{n-1}^{n-2}\\ 0 & x_n - x_1 & \left({x_n - x_1}\right) x_n & \cdots & \left({x_n - x_1}\right) x_n^{n-3} & \left({x_n - x_1}\right) x_n^{n-2} \end{vmatrix}$

For all rows apart from the first, the $k$th row has the constant factor $\left({x_k - x_1}\right)$.

So we can extract all these as factors, and from Determinant with Row Multiplied by Constant, we get:


 * $V_n = \prod_{k=2}^n \left({x_k - x_1}\right) \begin{vmatrix}

1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & x_2 & \cdots & x_2^{n-3} & x_2^{n-2} \\ 0 & 1 & x_3 & \cdots & x_3^{n-3} & x_3^{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 1 & x_{n-1} & \cdots & x_{n-1}^{n-3} & x_{n-1}^{n-2}\\ 0 & 1 & x_n & \cdots & x_n^{n-3} & x_n^{n-2} \end{vmatrix}$

From Determinant with Unit Element in Otherwise Zero Row, we can see that this directly gives us:


 * $V_n = \prod_{k=2}^n \left({x_k - x_1}\right) \begin{vmatrix}

1 & x_2 & \cdots & x_2^{n-3} & x_2^{n-2} \\ 1 & x_3 & \cdots & x_3^{n-3} & x_3^{n-2} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_{n-1} & \cdots & x_{n-1}^{n-3} & x_{n-1}^{n-2}\\ 1 & x_n & \cdots & x_n^{n-3} & x_n^{n-2} \end{vmatrix}$

and it can be seen that:
 * $\displaystyle V_n = \prod_{k=2}^n \left({x_k - x_1}\right) V_{n-1}$

$V_2$, by the time we get to it (it will concern elements $x_{n-1}$ and $x_n$), can be calculated directly using the formula for calculating a Determinant of Order 2:


 * $V_2 = \begin{vmatrix}

1 & x_{n-1} \\ 1 & x_n \end{vmatrix} = x_n - x_{n-1}$

The result follows.