# 4.3 Using the vector space axioms

There are some familiar properties of vector addition and scalar multiplication — like the fact that if you multiply a vector by the scalar zero, you get the zero vector — which aren’t listed in the axioms. Are they special to column vectors, or do they hold in every vector space?

To answer questions like this we can give a proof that uses only the vector space axioms, not the specific form of a particular vector space’s elements.

###### Lemma 4.3.1.

Let $V$ be a vector space and $\mathbf{v}\in V$. Then $0\mathbf{v}=\mathbf{0}_{V}$.

Be careful that you understand the notation here. $\textbf{0}_{V}$ means the special zero vector given in the definition of the vector space $V$, and $0\mathbf{v}$ means the vector v scalar multiplied by the scalar 0. They’re not obviously the same thing.

###### Proof.

$0\mathbf{v}=(0+0)\mathbf{v}=0\mathbf{v}+0\mathbf{v}$ (by axiom 8 of Definition 4.2.1). Axiom 4 says there’s an element $\mathbf{u}$ of $V$ such that $\mathbf{u}+0\mathbf{v}=\textbf{0}_{V}$, so add it to both sides:

 $\displaystyle\mathbf{u}+0\mathbf{v}$ $\displaystyle=\mathbf{u}+(0\mathbf{v}+0\mathbf{v})$ $\displaystyle\textbf{0}_{V}$ $\displaystyle=(\mathbf{u}+0\mathbf{v})+0\mathbf{v}$ axiom 2, definition of u $\displaystyle\mathbf{0}_{V}$ $\displaystyle=\mathbf{0}_{V}+0\mathbf{v}$ $\displaystyle\text{definition of }\mathbf{u}$ $\displaystyle\mathbf{0}_{V}$ $\displaystyle=0\mathbf{v}$ axiom 3.

###### Lemma 4.3.2.

Let $V$ be a vector space and let $\mathbf{x}\in V$. Then $\mathbf{x}+(-1)\mathbf{x}=\mathbf{0}_{V}$.

###### Proof.
 $\displaystyle\mathbf{x}+(-1)\mathbf{x}$ $\displaystyle=1\mathbf{x}+(-1)\mathbf{x}$ axiom 6 $\displaystyle=(1+-1)\mathbf{x}$ axiom 8 $\displaystyle=0\mathbf{x}$ $\displaystyle=\mathbf{0}_{V}$ Lemma 4.3.1.

We write $-\mathbf{x}$ for the additive inverse of $\mathbf{x}$ which axiom 2 provides, and $\mathbf{y}-\mathbf{x}$ as shorthand for $\mathbf{y}+-\mathbf{x}$. Here are two more proofs using the axioms.

###### Lemma 4.3.3.
1. 1.

Let $\lambda$ be a scalar. Then $\lambda\mathbf{0}_{V}=\mathbf{0}_{V}$.

2. 2.

Suppose $\lambda\neq 0$ is a scalar and $\lambda\mathbf{x}=\mathbf{0}_{V}$. Then $\mathbf{x}=\mathbf{0}_{V}$.

###### Proof.
1. 1.
 $\displaystyle\lambda\mathbf{0}_{V}$ $\displaystyle=\lambda(\mathbf{0}_{V}+\mathbf{0}_{V})$ axiom 3 $\displaystyle=\lambda\mathbf{0}_{V}+\lambda\mathbf{0}_{V}$ axiom 7

Axiom 2 tells there’s an additive inverse to $\lambda\mathbf{0}_{V}$. Adding it to both sides and using axiom 2, we get $\mathbf{0}_{V}=\lambda\mathbf{0}_{V}$.

2. 2.
 $\displaystyle\lambda\mathbf{x}$ $\displaystyle=\mathbf{0}_{V}$ $\displaystyle\lambda^{-1}(\lambda\mathbf{x})$ $\displaystyle=\lambda^{-1}\mathbf{0}_{V}$ $\displaystyle(\lambda^{-1}\lambda)\mathbf{x}$ $\displaystyle=\mathbf{0}_{V}$ axiom 5 and part 1 $\displaystyle 1\mathbf{x}$ $\displaystyle=\mathbf{0}_{V}$ $\displaystyle\mathbf{x}$ $\displaystyle=\mathbf{0}_{V}$ axiom 6.