Powers of Complex Numbers Are Linearly Independent

Question

Suppose I have a set of distinct, nonzero complex numbers $\{z_1\,,...\,z_k\}\,,$ is there a quick way of arguing that if $$a_1z_1^n+\cdots a_kz_k^n=0$$ for all $n\in\mathbb{Z}\,,$ then $$a_1=\cdots=a_k=0$$ ?

If none of them have the same modulus then I can divide the whole equation by the highest modulus and send $n\to\infty\,,$ showing term by term that each coefficient is zero, so I have reduced the problem to $|z_i|=1$ for all $i\,.$ If $k=2$ this is simple but for arbitrary $k$ I'm not sure what is an efficient way to do it.

Ah, I'm terribly sorry, I've misread that as "for some $n$". — Wojowu, Dec 19 '16 at 18:12

score 5 · Accepted Answer · answered Dec 19 '16 at 18:09

5

If $$a_1z_1^n+\cdots + a_kz_k^n=0$$ for $n = 0, 1, \dots, k-1$, then $$ \begin{bmatrix} 1 & 1 & \dots & 1\\ z_1 & z_2 & \dots & z_k\\ z_1^{2} & z_{2}^{2} & \dots & z_{k}^{2}\\ & & \ddots \\ z_1^{k-1} & z_{2}^{k-1} & \dots & z_{k}^{k-1} \end{bmatrix} \cdot \begin{bmatrix} a_1\\a_2\\\vdots\\a_k \end{bmatrix} = 0, $$ where the Vandermonde matrix has nonzero determinant, as the $z_i$ are distinct. I use the fact that the $z_i$ are non-zero to determine the first row.

answered Dec 19 '16 at 18:09

Andreas Caranti

70,862

1

If one doesn't know the Vandermonde matrix: The OP asks for linear independence of the rows; this is equivalent to linear independence of the columns; and that simply expresses that a degree $k-1$ polynomial with $k$ distinct roots is the zero polynomial – Hagen von Eitzen Dec 19 '16 at 18:12
Nice @Andreas! +1! – Arnaldo Dec 19 '16 at 18:15
@OP see also here for an example of the view by Hagen. – Bill Dubuque Dec 19 '16 at 18:41

Powers of Complex Numbers Are Linearly Independent

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