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?

  • $\begingroup$ Equivalently, you're asking if the unitary law follows from the other axioms. Try proving it from the rest of them, and see if you can construct a counterexample from there? $\endgroup$
    – djechlin
    Commented Aug 30, 2014 at 3:14

3 Answers 3


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.

  • $\begingroup$ I think your first sentence is doing yourself injustice: with $V_0$ being any Abelian group, this is much more general that the answer by Ali. You might want to state that (and that $V_1$ is a an actual vector space) right away, as well as the fact that $V_0\times V_1$ denotes a direct product of Abelian groups (without that the answer is a bit hard to follow). By the way my answer now contains this general example (but without claiming generality). +1 $\endgroup$ Commented Feb 16, 2015 at 11:06
  • $\begingroup$ @MarcvanLeeuwen: Thanks for the suggestion. I started writing when there was just Alis answer. I'll try to improve my answer. $\endgroup$
    – bodo
    Commented Feb 16, 2015 at 11:10

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.

  • $\begingroup$ This shows that multiplying the result by a constant factor does work, even using different factors on different basis vectors, if the factors in question are either 1 or 0. $\endgroup$ Commented May 1, 2014 at 23:25
  • $\begingroup$ I like this example. $\endgroup$
    – TonyK
    Commented Feb 16, 2015 at 10:37

One simple kind of example that can be made is to take any Abelian group, defining the additive structure, and to define any scalar multiplication to simply return the zero vector. This shows that without axiom 6. our space does not even have to remotely resemble an actual vector space.

Another option (in fact a generalization) is to start with an Abelian group $A$ that contains an actual vector space $V$ as a subgroup (can easily be done using direct sums), choose a fixed vector $v_0\in V$ and to define scalar multiplication $ax=a*v_0$ where $*$ denotes actual scalar multiplication in the subspace. In other words, scalar multiplication ignores its vector argument, and replaces it by a fixed vector $v_0$. By inspection the axioms 5., 7., and 8. involving scalar multiplication are all satisfied.

One may also ask where in the theory one uses the axiom 6. One very basic use is to combine addition and multiplication into the operation of linear combination; For instance a subspace can be defined to be any non empty set closed under linear combination, rather than under addition and multiplication separately. Bit without axiom 6., the sum $x+y$ is not a linear combination $ax+by$ of $x$ and $y$. Since linear combinations are fundamental to the theory, I think one does not get very far without axiom 6.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .