Value of Vandermonde Determinant/Formulation 1/Proof 2

== Theorem ==
== Proof ==
Proof by induction:
Let the Vandermonde determinant be presented in the following form:
:$V_n = \begin {vmatrix}
{x_1}^{n - 1} & {x_1}^{n - 2} & \cdots & x_1 & 1 \\
{x_2}^{n - 1} & {x_2}^{n - 2} & \cdots & x_2 & 1 \\
 \vdots & \vdots & \ddots & \vdots & \vdots \\
{x_n}^{n - 1} & {x_n}^{n - 2} & \cdots & x_n & 1 \\
\end {vmatrix}$
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j}$
$\map P 1$ is true, as this just says $\begin {vmatrix} 1 \end {vmatrix} = 1$.
=== Basis for the Induction ===
$\map P 2$ holds, as it is the case:
:$V_2 = \begin {vmatrix}
a_1 & 1 \\
a_2 & 1
\end {vmatrix}$
which evaluates to $V_2 = a_1 - a_2$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds V_k = \prod_{1 \mathop \le i \mathop < j \mathop \le k} \paren {x_i - x_j}$
Then we need to show:
:$\ds V_{k + 1} = \prod_{1 \mathop \le i \mathop < j \mathop \le k + 1} \paren {x_i - x_j}$
=== Induction Step ===
This is our induction step:
Take the determinant:
:$V'_{k + 1} = \begin{vmatrix}
 x^k & x^{k - 1} & \cdots & x^2 & x & 1 \\
 {x_2}^k & {x_2}^{k - 1} & \cdots & {x_2}^2 & x_2 & 1 \\
 \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
 {x_{k + 1} }^k & {x_{k + 1} }^{k - 1} & \cdots & {x_{k + 1} }^2 & x_{k + 1} & 1
\end{vmatrix}$
where it is seen that $x_1$ has been replaced in $V_{k + 1}$ with $x$.
Let the Expansion Theorem for Determinants‎ be used to expand $V'_{k + 1}$ in terms of the first row.
It can be seen that it is a polynomial in $x$ whose degree is no greater than $k$.
Let that polynomial be denoted $\map f x$.
Let any $x_r$ be substituted for $x$ in the determinant.
Then two of its rows will be the same.
From Square Matrix with Duplicate Rows has Zero Determinant, the value of such a determinant will be $0$.
Such a substitution in the determinant is equivalent to substituting $x_r$ for $x$ in $\map f x$.
Thus it follows that:
:$\map f {x_2} = \map f {x_3} = \cdots = \map f {x_{k + 1} } = 0$
So $\map f x$ is divisible by each of the factors $x - x_2, x - x_3, \ldots, x - x_{k + 1}$.
All these factors are distinct, otherwise the original determinant is zero.
So:
:$\map f c = \map C {x - x_2} \paren {x - x_3} \cdots \paren {x - x_k} \paren {x - x_{k + 1} }$
As the degree of $\map f x$ is no greater than $k$, it follows that $C$ is independent of $x$.
From the Expansion Theorem for Determinants‎, the coefficient of $x^k$ is:
:$\begin{vmatrix}
 {x_2}^{k - 1} & \cdots & {x_2}^2 & x_2 & 1 \\
 \vdots & \ddots & \vdots & \vdots & \vdots \\
{x_{k + 1} }^{k - 1} & \cdots & {x_{k + 1} }^2 & x_{k + 1} & 1
\end{vmatrix}$
By the induction hypothesis, this is equal to:
:$\ds \prod_{2 \mathop \le i \mathop < j \mathop \le k + 1} \paren {x_i - x_j}$
This must be our value of $C$.
So we have:
:$\ds \map f x = \paren {x - x_2} \paren {x - x_3} \dotsm \paren {x - x_k} \paren {x - x_{k + 1} } \prod_{2 \mathop \le i \mathop < j \mathop \le k + 1} \paren {x_i - x_j}$
Substituting $x_1$ for $x$, we retrieve the proposition $\map P {k + 1}$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j}$
== Sources ==
* : $\text{I}, \ \S 1.4: \ 1.4.5$
Category:Value of Vandermonde Determinant