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?
1 Answer
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 $(0, 0, 0)^T$.