linear algebra - Vector spaces - Multiplying by zero scalar yields zero vector

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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