Examples of 'almost' vector spaces where unitary law fails

I was looking at the definition on wikipedia of a vector space (similar/equivalent definitions are everywhere, but I thought I'd list it here for completion):

A vector space over a field $F$ is a set $V$ together with two binary operations that satisfy the eight axioms listed below. Elements of $V$ are called vectors. Elements of $F$ are called scalars. In the two examples above, our set consists of the planar arrows with fixed starting point and of pairs of real numbers, respectively, while our field is the real numbers. The first operation, vector addition, takes any two vectors $v$ and $w$ and assigns to them a third vector which is commonly written as $v + w$, and called the sum of these two vectors. The second operation takes any scalar a and any vector $v$ and gives another vector $av$. In view of the first example, where the multiplication is done by rescaling the vector $v$ by a scalar $a$, the multiplication is called scalar multiplication of $v$ by $a$.

To qualify as a vector space, the set $V$ and the operations of addition and multiplication must adhere to a number of requirements called axioms. In the list below, let $u$, $v$ and $w$ be arbitrary vectors in $V$, and $a$ and $b$ scalars in $F$.

1. Associativity of addition: $u + (v + w) = (u + v) + w$

2. Commutativity of addition: $u + v = v + u$

3. Identity element of addition: There exists an element $0 ∈ V$, called the zero vector, such that $v + 0 = v$ for all $v ∈ V$.

4. Inverse elements of addition: For every $v ∈ V$, there exists an element $−v ∈ V$, called the additive inverse of $v$, such that $v + (−v) = 0$

5. Compatibility of scalar multiplication with field multiplication: $a(bv) = (ab)v$

6. Identity element of scalar multiplication: $1v = v$, where $1$ denotes the multiplicative identity in $F$.

7. Distributivity of scalar multiplication with respect to vector addition: $a(u + v) = au + av$

8. Distributivity of scalar multiplication with respect to field addition: $(a + b)v = av + bv$

I was wondering if there were any examples of vector spaces where only the law for the identity element of scalar multiplication fails? The only one I could think of would be to redefine a true vector space's scalar multiplication to multiply the result by a constant factor. Are there others?

Solution 1:

Any “almost” vector space can be described as the direct sum $V = V_0 ⊕ V_1$ of abelian groups, where $V_1$ is an actual vector space and scalar multiplication by $1$ is the projection onto $\{0\}×V_1$: $$V_0⊕V_1 ∋ (w, v) ↦ 1⋅(w, v) = (0, v) ∈ V_0 ⊕ V_1$$ The decomposition can be found as follows:

Because of Axiom 4 and 1 for every $v ∈ V$ there is a unique $w ∈ V$ such that $v = 1⋅v + w$. Then Axiom 5 and 7 give us $1⋅v = 1⋅(1⋅v) + 1⋅w = 1⋅v + 1⋅w$. So again with Axiom 1 and 4 we have $1⋅w = 0$. It is easily shown, that then $$V_0 := \{v ∈ V | 1⋅v = 0\}$$ is an abelian group and $$V_1 := \{v ∈ V | 1⋅v = v\}$$ is a vector space.

Solution 2:

I think your suggestion "Multiply the result by a constant factor" does not satisfies the 5th condition. But I think the following works:

Put $F=\mathbb{Q}$, $V=\{a+b\sqrt{2} \mid a,b\in \mathbb{Q}\}$:

The scalar multiplication: $\lambda.(a+b\sqrt{2})=\lambda a$

The later is the usual product and the first is the new scalar product. Now $1.\sqrt{2}=0$.

So $V$ is not a unital $\mathbb{Q}$-module.