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| \documentclass[a4paper]{article}
\usepackage{amsmath}
\usepackage[margin=1in]{geometry}
\title{Formula of Verdermonde Determinant}
\author{Common Knowledge}
\date{\today}
\begin{document}
\maketitle
\begin{description}
\item[Approach] Induction
\item[Target] ``Kill'' the last row and column.
\end{description}
\section{First Attempt}
\begin{equation*}
\begin{vmatrix}
1&\lambda_1 & \cdots & \lambda_1^{n - 1} \\
1&\lambda_2 & \cdots & \lambda_2^{n - 1} \\
\vdots & \vdots & \ddots & \vdots \\
1&\lambda_n & \cdots & \lambda_n^{n - 1}
\end{vmatrix}
=
\begin{vmatrix}
0&\lambda_1 - \lambda_n & \cdots & \lambda_1^{n - 1} - \lambda_n^{n - 1} \\
0&\lambda_2 - \lambda_n & \cdots & \lambda_2^{n - 1} - \lambda_n^{n - 1} \\
\vdots & \vdots & \ddots & \vdots \\
1&\lambda_n & \cdots & \lambda_n^{n-1}
\end{vmatrix} \quad (R_k \rightarrow R_k - R_n, k \neq n)
\end{equation*}
Question: Use $a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2} b + \cdots + a b^{n - 2} + b^{n - 1})$?
\begin{align*}
&\begin{vmatrix}
1&\lambda_1 & \cdots & \lambda_1^{n - 1} \\
1&\lambda_2 & \cdots & \lambda_2^{n - 1} \\
\vdots & \vdots & \ddots & \vdots \\
1&\lambda_n & \cdots & \lambda_n^{n - 1}
\end{vmatrix}\\
&= (-1)^{n + 1}
\begin{vmatrix}
\lambda_1 - \lambda_n & \lambda_1^2 - \lambda_n^2 & \cdots & \lambda_1^{n-1} - \lambda_n^{n-1} \\
\lambda_2 - \lambda_n & \lambda_2^2 - \lambda_n^2 & \cdots & \lambda_2^{n-1} - \lambda_n^{n-1} \\
\vdots & \vdots & \ddots & \vdots \\
\lambda_{n - 1} - \lambda_n & \lambda_{n - 1}^2 - \lambda_n^2 & \cdots & \lambda_{n - 1}^{n - 1} - \lambda_n^{n - 1}
\end{vmatrix}\\
&= (-1)^{n + 1} (\lambda_1 - \lambda_n) (\lambda_2 - \lambda_n) \cdots (\lambda_{n - 1} - \lambda_n)
\begin{vmatrix}
1&\lambda_1 + \lambda_n & \cdots & \lambda_1^{n - 2} + \lambda_1^{n - 3} \lambda_n + \cdots + \lambda_1 \lambda_n^{n - 3} + \lambda_n^{n - 2} \\
1&\lambda_2 + \lambda_n & \cdots & \lambda_2^{n - 2} + \lambda_2^{n - 3} \lambda_n + \cdots + \lambda_2 \lambda_n^{n - 3} + \lambda_n^{n - 2} \\
\vdots & \vdots & \ddots & \vdots \\
1&\lambda_{n - 1} + \lambda_n & \cdots & \lambda_{n - 1}^{n - 2} + \lambda_{n - 1}^{n - 3} \lambda_n + \cdots + \lambda_{n - 1} \lambda_n^{n - 3} + \lambda_n^{n - 2} \\
\end{vmatrix}
\end{align*}
It doesn't seem good. Let's try another way.
\section{Second Attempt}
We ``kill'' the last row by making the entries zero, except the first one.
\begin{align}
&\begin{vmatrix}
1&\lambda_1 & \cdots & \lambda_1^{n - 1} \\
1&\lambda_2 & \cdots & \lambda_2^{n - 1} \\
\vdots & \vdots & \ddots & \vdots \\
1&\lambda_n & \cdots & \lambda_n^{n - 1}
\end{vmatrix}\\
&=
\begin{vmatrix}
1&\lambda_1 - \lambda_n & \cdots & \lambda_1^{n - 1} - \lambda_n \lambda_1^{n - 2} \\
1&\lambda_2 - \lambda_n & \cdots & \lambda_2^{n - 1} - \lambda_n \lambda_2^{n - 2} \\
\vdots & \vdots & \ddots & \vdots \\
1&\lambda_{n - 1} - \lambda_n & \cdots & \lambda_{n - 1}^{n - 1} - \lambda_n \lambda_2{n - 1}{n - 2} \\
1& 0 & \cdots & 0
\end{vmatrix} \quad (C_k \rightarrow C_k - \lambda_n C_{k - 1}) \label{eq:col_op}\\
&= (-1)^{n + 1}
\begin{vmatrix}
\lambda_1 - \lambda_n & \cdots & (\lambda_1 - \lambda_n) \lambda_1^{n - 2} \\
\lambda_2 - \lambda_n & \cdots & (\lambda_2 - \lambda_n) \lambda_2^{n - 2} \\
\vdots & \ddots & \vdots \\
\lambda_{n - 1} - \lambda_n & \cdots & (\lambda_{n - 1} - \lambda_n) \lambda_{n - 1}^{n - 2}
\end{vmatrix}\\
&= (-1)^{n + 1} (\lambda_1 - \lambda_n) (\lambda_2 - \lambda_n) \cdots (\lambda_{n - 1} - \lambda_n)
\begin{vmatrix}
1 & \cdots & \lambda_1^{n - 2} \\
1 & \cdots & \lambda_2^{n - 2} \\
\vdots & \ddots & \vdots \\
1 & \cdots & \lambda_{n - 1}^{n - 2}
\end{vmatrix}\\
&= (\lambda_n - \lambda_1) (\lambda_n - \lambda_2) \cdots (\lambda_n - \lambda_{n - 1})
\begin{vmatrix}
1 & \cdots & \lambda_1^{n - 2} \\
1 & \cdots & \lambda_2^{n - 2} \\
\vdots & \ddots & \vdots \\
1 & \cdots & \lambda_{n - 1}^{n - 2} \\
\end{vmatrix}
\end{align}
\begin{itemize}
\item In \eqref{eq:col_op}, $k = n - 2,n - 3,\dots,3,2$. Start with $C_{n - 2}$ \emph{first}.
\item In the last step, I changed the factors from $\lambda_k - \lambda_n$ to $\lambda_n - \lambda_k$, where $k = 1,2,\dots,n - 1$, so that it is in the form of what you see in Wikipedia.
\end{itemize}
OK! I believe that you know what to do next to get the formula for Verdermonde determinant.
\begin{equation*}
\boxed{
\begin{vmatrix}
1&\lambda_1 & \cdots & \lambda_1^{n - 1} \\
1&\lambda_2 & \cdots & \lambda_2^{n - 1} \\
\vdots & \vdots & \ddots & \vdots \\
1&\lambda_n & \cdots & \lambda_n^{n - 1}
\end{vmatrix}
=
\prod_{1 \le i < j \le n} (\lambda_j - \lambda_i)
}
\end{equation*}
\end{document}
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