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Question

I understand that based on the axioms of vector spaces there needs to be a unique member the zero vector in V such that for all v element of V, v+0=v, but how do I find the appropriate function in the bove case?

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    $\begingroup$ Isn’t there a very obvious choice: take $a=b=c=0$. $\endgroup$ Commented Feb 10, 2019 at 21:17

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For all $f \in V$ we have $f+z=f$, so for all $x \in \mathbb{R}$: $(f+z)(x) = f(x)+z(x) = f(x)$. For example, $1 = f_1(x) = (f_1+z)(x) = f_1(x)+z(x) = 1 + z(x)$ and so $z(x) = 0$ for all $x \in \mathbb{R}$. The only function $z \in V$ with $z(x)=0$ for all $x \in \mathbb{R}$ is the one with $z(x) = 0 + 0x + 0x^2$. The coordinate vector of the zero-element for either ordered basis must be $(0, 0, 0)^T$.

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