1
$\begingroup$

Here is my problem:

Given a vector x $\in \mathbb{R}^{n+1}$, the $(n + 1) \times (n + 1)$ matrix $V$ defined by

$$v_{ij} = \begin{cases} 1 & \text{if } j = 1 \\ x_i^{j - 1} & \text{for } j = 2, \dots, n + 1 \end{cases}$$

is called the Vandermonde matrix.

Suppose that $x_1, x_2, \dots, x_{n+1}$ are all distinct. Show that if c is a solution to $V$ x = 0, then the coefficients $c_1, c_2, \dots, c_{n+1}$ must all be zero, and hence $V$ must be nonsingular.

Each entry of $V$ c is an $n$th degree polynomial, so if $c_1, c_2, \dots, c_{n+1}$ were not all zero, then $x_1, x_2, \dots, x_{n+1}$ would have to be the roots of the polynomial. But it's perfectly possible for an $n$th degree polynomial to have $n$ roots, so I'm not sure how I can get this to work.

  • 0
    There are $n+1$ $x_i$s and BTW $(n+1)$ $c_j$s (typo?).2012-03-22
  • 0
    If you will write out the matrix explicitly on paper rather than in LaTeX, you might see that the entries of $Vc$ are the values of a polynomial $c_0+c_1x + c_2x^2 + \cdots + c_nx^n$ at $n+1$ different points $x_1, x_2, \ldots, x_{n+1}$.2012-03-22
  • 0
    Yeah, you're right. There are $n+1$ elements in **c**2012-03-22
  • 0
    @Dilip: What if $x_1, x_2, \dots, x_{n+1}$ are the roots of this polynomial? Then that doesn't prove anything.2012-03-22
  • 0
    @MattGregory: It does prove everything :). Since the polynomial is of degree $n$, it can't have $n+1$ zeros $x_1, \ldots, x_{n+1}$ besides it is the zero polynomial.2012-03-22
  • 0
    @martini: Oh you're right hehe. I was confused by the typo.2012-03-22
  • 0
    @MattGregory You should write your own answer or edit the question then, I think :)2012-03-22

1 Answers 1