UTM Ideals Varieties and Algorithm - Chapter 1 Section 5 Exercise 2

Problem:

Solution:

The hint basically spell out the solution. If the determinant is 0, then the columns are linearly dependent.

We can write, with $c_i$ not all zero.

$c_0\left(\begin{array}{c}1\\ \vdots \\ 1\end{array}\right) + c_1 \left(\begin{array}{c}a_1\\ \vdots \\ a_n\end{array}\right) + \cdots + c_n\left(\begin{array}{c}a_1^{n-1}\\ \vdots \\ a_n^{n-1}\end{array}\right) = 0$

These equations can be interpreted as $n$ roots of the polynomial $c_0 + c_1 x + \cdots + c_n x^{n-1}$.

A non-zero polynomial of degree $n -1$ cannot have $n$ roots, so we have a contradiction, the determinant is not 0.