4 Linear algebra 4.2 Vector spaces 4.4 Subspaces

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.

Let $\lambda$ be a scalar. Then $\lambda\mathbf{0}_{V}=\mathbf{0}_{V}$ .
2.

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

Proof.

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.

$\displaystyle\lambda\mathbf{x}$ $\displaystyle=\mathbf{0}_{V}$

$\displaystyle\lambda^{-1}(\lambda\mathbf{x})$ $\displaystyle=\lambda^{-1}\mathbf{0}_{V}$

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

$\displaystyle 1\mathbf{x}$ $\displaystyle=\mathbf{0}_{V}$

$\displaystyle\mathbf{x}$ $\displaystyle=\mathbf{0}_{V}$ axiom 6.

∎

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

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