# Calculating Vandermonde determinant

I understand that the Vandermonde determinant

$$W(x_1, \ldots, x_n) = \left| \begin{array}{cccc} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2\\ \cdot & \cdot & \cdots & \cdot \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1}\\ \end{array} \right|$$

may be calculated by regarding the determinant as a polynomial $$W(x_1,\ldots, x_n) = P(x_n) = k_n \prod_{i=1}^{n-1} (x_n - x_i)$$ and then performing induction on $k_n = W(x_1, \ldots, x_{n-1})$.

However, I am not sure how we obtain this equality for $k_n$. Is it sufficient to say that $W(x_1, \ldots, x_{n-1})$ is the coefficient for the $x_n^{n-1}$ term in $P(x_n)$ when we expand the determinant with respect to the right-most column?

This question is also in part (c) of this post.

• The induction is clearly going to be on $n$; it cannot be on $k_n$ which is not a natural number. – Marc van Leeuwen Dec 6 '18 at 5:22

## 1 Answer

You are right. When expanding the determinant with respect to the last column, only the term $$\left| \begin{array}{cccc} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots & x_{n-1} \\ x_1^2 & x_2^2 & \cdots & x_{n-1}^2\\ \cdot & \cdot & \cdots & \cdot \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_{n-1}^{n-2}\\ \end{array} \right|x_n^{n-1}$$

is of degree $n-1$for $x_n$, thus $$\left| \begin{array}{cccc} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots & x_{n-1} \\ x_1^2 & x_2^2 & \cdots & x_{n-1}^2\\ \cdot & \cdot & \cdots & \cdot \\ x_1^{n-2} & x_2^{n-2} & \cdots & x_{n-1}^{n-2}\\ \end{array} \right| = k_n$$

• So just to be clear, we can assume from this that everything will distribute properly? Is there any particular theorem or principle that allows us to say this? – yepikhodov Aug 22 '14 at 15:34
• @yepikhodov What do you mean by "distribute properly"? Since we know the determinant is a polynomial of degree $n-1$ for $x_n$ and we know what all the roots are, we only need to determinant the coefficient for $x_n^{n-1}$ – Petite Etincelle Aug 22 '14 at 15:36
• I see. So we are only concerned about $k _n$ as a leading coefficient even though $k_n$ is factored out of the entire polynomial. – yepikhodov Aug 22 '14 at 15:41
• @yepikhodov yes, we only need to identify $k_n$ with $W(x_1, \cdots, x_{n-1})$ – Petite Etincelle Aug 22 '14 at 15:43