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Show that $\{x_1,...,x_n\}$, where $x_j=t^j$, is a linearly independent set in the space $C[a,b]$.

I think I can use properties of polynomials in $R[x]$ here, but I'm not sure.

Using $\sum_{i=1}^n c_i t^i$, $\forall t\in C[a,b]$?

Any help would be nice.

Desperate Fluffy
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Suppose $\sum\limits_{i=1}^{n}c_ix^i=0$. Then this polynomial has infinitely many roots. What do we know about polynomials with infinitely many roots?

voldemort
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  • That there's only one it could be, the polynomial $0$. – Desperate Fluffy Sep 08 '14 at 02:20
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    @DesperateFluffy: and that provides you with your answer. – voldemort Sep 08 '14 at 04:08
  • @voldemort why this polynomial has infinitely many roots? That does not have at most $n$ roots for the fundamental theorem of algebra? – user424241 Dec 19 '17 at 20:25
  • @user424241 because it holds $\forall t \in [a,b]$, so there are infinitely many roots. But we know that there are at most $n$ roots because this is a polynomial of degree $n$ (There are in fact exactly $n$ roots, neglecting multiplicity, over $\mathbb C$). This is a contradiction so we know this can only be the zero polynomial, which is the only polynomial with infinitely many roots. –  Dec 03 '19 at 21:12