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If I have the following vector space $ V, \text{{$e_0, e_1, e_2$}} \text{ where } e_0(x) = 1, e_1(x) = x \text{ and } e_2(x) = x^2$.I want to know the linear dependency of it how can I proceed? I thought of following the definition of linearly independent $$c_0e_0 + c_1e_1 + c_2e_2 = c_0+ c_1x + c_2x^2=0\iff c_0 = c_1 = c_2 = 0$$ but I can not mount a system because of the $x^2$

I know that $c_0 = c_1 = c_2 = 0$ is solution, but i want to know if there is another solution for the equation $c_0+ c_1x + c_2x^2=0$ with $c_0 \neq 0, c_1 \neq 0 \text{ and }c_2 \neq 0 $

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  • $\begingroup$ Keep in mind that the equation on the left has to hold for all $x$. $\endgroup$ – amd Apr 17 '18 at 17:28
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One way to show this is repeated differentiation. If $$c_0 + c_1 x + c_2 x^2 \equiv 0,$$ then \begin{align*} c_1 + 2c_2 x &\equiv 0, \\ 2c_2 &\equiv 0. \end{align*} From evaluating all these polynomials at $x = 0$, we obtain $c_0 = c_1 = c_2 = 0$.

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Note that for the zero polynomial property

$$c_0e_0 + c_1e_1 + c_2e_2 = c_0+ c_1x + c_2x^2=0 \quad \forall x\iff c_0 = c_1 = c_2 = 0$$

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