4 Linear algebra

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 𝐯V. Then 0𝐯=𝟎V.

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

Proof.

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

𝐮+0𝐯 =𝐮+(0𝐯+0𝐯)
𝟎V =(𝐮+0𝐯)+0𝐯 axiom 2, definition of u
𝟎V =𝟎V+0𝐯 definition of 𝐮
𝟎V =0𝐯 axiom 3.

Lemma 4.3.2.

Let V be a vector space and let 𝐱V. Then 𝐱+(1)𝐱=𝟎V.

Proof.
𝐱+(1)𝐱 =1𝐱+(1)𝐱 axiom 6
=(1+1)𝐱 axiom 8
=0𝐱
=𝟎V Lemma 4.3.1.

We write 𝐱 for the additive inverse of 𝐱 which axiom 4 provides, and 𝐲𝐱 as shorthand for 𝐲+𝐱. Here are two more proofs using the axioms.

Lemma 4.3.3.
  1. 1.

    Let λ be a scalar. Then λ𝟎V=𝟎V.

  2. 2.

    Suppose λ0 is a scalar and λ𝐱=𝟎V. Then 𝐱=𝟎V.

Proof.
  1. 1.
    λ𝟎V =λ(𝟎V+𝟎V) axiom 3
    =λ𝟎V+λ𝟎V axiom 7

    Axiom 2 tells there’s an additive inverse to λ𝟎V. Adding it to both sides and using axiom 2, we get 𝟎V=λ𝟎V.

  2. 2.
    λ𝐱 =𝟎V
    λ1(λ𝐱) =λ1𝟎V
    (λ1λ)𝐱 =𝟎V axiom 5 and part 1
    1𝐱 =𝟎V
    𝐱 =𝟎V axiom 6.