My lecture notes define a real inner product on a real vector space $V$ as a map $\langle -\rangle :V^2 \to \mathbb{R}$ that satisfies the following axioms for all $v, w, z \in V$ and $\lambda \in \mathbb{R}$: $$\langle \lambda v + w, z\rangle = \lambda \langle v, z \rangle + \langle w, z\rangle \qquad \text{linearity in the first argument}$$ $$\langle v, w\rangle = \langle w, v\rangle \qquad \text{symmetry}$$ $$\langle v, v\rangle \geq 0 \quad \text{and} \quad \langle v, v \rangle = 0 \Leftrightarrow v = \textbf{0}.$$
There is a remark which states that it would be redundant to list the property of linearity in the second argument as an axiom as linearity in the first argument and symmetry already imply linearity in the second argument.
I was wondering whether the given set of properties is "minimal". Why is it necessary to demand $v = \textbf{0} \Rightarrow \langle v, v \rangle = 0$? Isn't that a consequence of linearity too, since $$\langle \textbf{0}, \textbf{0}\rangle = \langle 0 \cdot \textbf{0}, \textbf{0}\rangle = 0\cdot \langle \textbf{0}, \textbf{0}\rangle = 0?$$
It is true that $v=0\implies \langle v, v\rangle=0$ follows from the other axioms. However, the converse is not redundant: the map $(v, w)\mapsto 0$ satisfies all axioms except $\langle v, v\rangle=0\implies v=0$.