Q:

Vector spaces - Multiplying by zero vector yields zero vector.

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The following proof is solely based on vector space axioms.
Axiom names are italicised.
They are defined in Wikipedia (vector space).
Multiplying by zero vector yields zero vector.
Let $V$ be a vector space over a field $F$.
Let $a \in F$.
\begin{array}{lrll}
\text{By} & \dots & \text{we denote} & \dots \\
\hline
& (-a)       && \text{an additive inverse of $a$ in $F$.} \\
& 1          && \text{a multiplicative identity element in $F$.} \\
& 0          && \text{an additive identity element in $F$.} \\
& \mathbf{0} && \text{an additive identity element in $V$.} \\
\end{array}
We want to prove that
$$a \mathbf{0} = \mathbf{0}.$$
Proof.
\begin{align*}
 \mathbf{0}
 &= 1 \mathbf{0}                                    && \text{by }\textit{Identity element of scalar multiplication} \\
 &= (a + (-a)) \mathbf{0}                           && \text{by }\textit{Inverse elements of field addition} \\
 &= a \mathbf{0} + (-a) \mathbf{0}                  && \text{by }\textit{Distrib. of scalar mult. (field addition)} \\
 &= a (\mathbf{0} + \mathbf{0}) + (-a) \mathbf{0}   && \text{by }\textit{Identity element of vector addition} \\
 &= (a \mathbf{0} + a \mathbf{0}) + (-a) \mathbf{0} && \text{by }\textit{Distrib. of scalar mult. (vector addition)} \\
 &= a \mathbf{0} + (a \mathbf{0} + (-a) \mathbf{0}) && \text{by }\textit{Associativity of vector addition} \\
 &= a \mathbf{0} + (a + (-a)) \mathbf{0}            && \text{by }\textit{Distrib. of scalar mult. (field addition)} \\
 &= a \mathbf{0} + 0 \mathbf{0}                     && \text{by }\textit{Inverse elements of field addition} \\
 &= (a + 0) \mathbf{0}                              && \text{by }\textit{Distrib. of scalar mult. (field addition)} \\
 &= a \mathbf{0}                                    && \text{by }\textit{Identity element of field addition}
\end{align*}
QED

A:

Why not $a(\mathbf{0})=a(\mathbf{0}+\mathbf{0})=a(\mathbf{0})+a(\mathbf{0})$; now add the inverse of $a(\mathbf{0})$ to both sides? 

