[Math] Vector spaces – Multiplying by zero scalar yields zero vector

Question

Please rate and comment. I want to improve; constructive criticism is highly appreciated.
Please take style into account as well.

The following proof is solely based on vector space related axioms.
Axiom names are italicised.
They are defined in Wikipedia (see vector space article).

Vector spaces – Multiplying by zero scalar yields zero vector

\begin{array}{lrll}
\text{Let} & \dots & \text{be} & \dots \\
\hline
& F && \text{a field.} \\
& V && \text{a vector space over $F$.} \\
& 0 && \text{an identity element of addition of $F$.} \\
& \mathbf{0} && \text{an identity element of addition of $V$.} \\
& \mathbf{v} && \text{an arbitrary vector in $V$.} \\
\end{array}

$$\text{Then, }0\mathbf{v} = \mathbf{0}.$$

Proof. We will denote by $1$ an identity element of scalar multiplication;
we will denote by $(-\mathbf{v})$ an additive inverse of $\mathbf{v}$.
\begin{align*}
0\mathbf{v}
&= 0\mathbf{v} + \mathbf{0} && \text{by }\textit{Identity element of vector addition} \\
&= 0\mathbf{v} + (\mathbf{v} + (-\mathbf{v})) && \text{by }\textit{Inverse elements of vector addition} \\
&= (0\mathbf{v} + \mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Associativity of vector addition} \\
&= (0\mathbf{v} + 1\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Identity element of scalar multiplication} \\
&= ((0 + 1)\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Distributivity of scalar multiplication (field addition)} \\
&= ((1 + 0)\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Commutativity of field addition} \\
&= (1\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Identity element of field addition} \\
&= \mathbf{v} + (-\mathbf{v}) && \text{by }\textit{Identity element of scalar multiplication} \\
&= \mathbf{0} && \text{by }\textit{Inverse elements of vector addition} \\
\end{align*}
QED

Answer 1

To shorten the proof, we may write as suggested by André Nicolas,

Proof. Let $(V,+,\cdot)_F$ be a vector space over the field $F$. We wish to show that $\forall v\in V$ one has $0\cdot v=\mathbf{0}$, where $0$ is the zero scalar and $\mathbf{0}$ is the zero vector. Let $v$ be an element of the vector space $V$;

By one of the axioms of field addition, $$0\cdot v=(0+0)\cdot v.$$ Since scalar multiplication is distributive over addition, $$(0+0)\cdot v=0\cdot v+0\cdot v.$$ From the previous two equalities we conclude that $$0\cdot v=0\cdot v+0\cdot v.$$ Adding to both sides the inverse element for addition of $0\cdot v$, which we'll denote by $-0\cdot v$: $$0\cdot v+(-0\cdot v)=0\cdot v+0\cdot v+(-0\cdot v).$$ By the inverse axiom, $$\mathbf{0}=0\cdot v+\mathbf{0},$$ hence by the identity axiom, $$0\cdot v=\mathbf{0}.\tag*{$\square$}$$