# Trivial Property of vector spaces

In several Linear Alegebra textbooks there is always a property of vector spaces listed that seems pointless to mention given it can be simplified. Can someone please tell me the significance of it that warrants it's mention.

Property in question: If a, b are two numbers, then (a+b) v = a v +b v .

This is the same as c being a number and v a vector, c v =(a+b) v with c=(a+b).

It seems completely pointless to mention this property. Maybe I am missing something? Why this property must exist.. Counter examples..

An explanation of why the number (a+b) is used vs. an arbitrary number, c, would be awesome, because it seems to be saying that the multiplication of a vector by a constant is just that, and that we can decompose the scalar into parts, or we can factor out a scalar from a vector say, 3 twice which gives 6 v , but this is equivalent to (3+3) v.

perhaps my problem lies in the memorization of the definition of a subspace in Lang's Linear Algebra.
'Let V be a vector space, and let W be a subset of V. We define W to be a subspace if W satisfies the following conditions:

(i) If v,w, are elements of W, their sum v+w is also an element of W.
(ii) If v is an element of W and c a number, then cv is an element of W.
(iii) The element O of Vis also an element of W.

Then W itself is a vector space. Indeed, properties VS1-VS8, being satisfied for all elements of V, are satisfied a fortiori for the elements of W.'

using this above quotation out of the same book, this is what leads me partly to the conclusion that the original edit of the property listed is accurate. I'm not saying we should throw out the property, I'm trying to see why it is so important that a+b was used instead of just c.

• Why do you think it is pointless? It's a fact that you use all the time when dealing with $\mathbb{R}^n$ so you want to make sure you can do it in a general vector space. Commented Mar 17, 2015 at 21:39
• @unintuit If you think it is unnecessary then try to prove it from the other axioms.
– user137731
Commented Mar 17, 2015 at 21:53
• Do you think the other axioms are pointless? Commented Mar 17, 2015 at 22:03
• Are you specifically referring to third paragraph in the question? Commented Mar 17, 2015 at 22:05
• I don't understand your "This is the same...". $a{\bf v} + b{\bf v}$ is the sum of two vectors. $c{\bf v} = (a+b) \bf v$ is not the sum of two vectors, it's the product of a number $c = a+b$ and a vector. Commented Mar 17, 2015 at 22:18

Raking my brain ... Can you even show that $0\mathbf{v}=\mathbf{0}$ without using this axiom? The one proof I recall makes use of the fact that $0+0=0$ and consequently we have, for all vectors $\mathbf{v}$ that $$0\mathbf{v}=(0+0)\mathbf{v}=0\mathbf{v}+0\mathbf{v}.$$ The group axioms (=the set of rules governing vector addition, the existence of the zero vector, and the existence of opposite vectors) then allow us to conclude that $0\mathbf{v}$ must be the zero vector $\mathbf{0}$ (=the neutral element of vector addition).

But we used this very axiom here in the second equality!

It seems likely to me that your difficulty comes from unfamiliarity with the process of defining mathematical structures by using axioms. A vector space, as defined by the set of axioms given in your textbook, may even be your first encounter with them (in which case my argument from above is incomplete, because you would still need to use the group axioms to reach the destination).

The point in Bill Dubuque's (+1) answer is that if we leave out this axiom from the list of requirements a very weird structure could claim that it should be treated as a vector space.

My favorite example of a weird structure, proving that another "obvious" axiom: $1\cdot\mathbf{v}=\mathbf{v}$ for all vectors $\mathbf{v}$ cannot be left out either, is the following. Let $V=\Bbb{R}^2$ as a set. Define the addition as usual, but define scalar multiplication by the rule $$a\cdot(x_1,x_2)=(ax_1,0).$$ All the other axioms save for $1\cdot\mathbf{v}=\mathbf{v}$ will be satisfied (you are invited to verify them).

The reason why we use axioms in the definitions is one of economy of thought. We prove results about all the systems satisfying the same set of axioms. Then when we meet another structure, and observe that it satisfies the given set of axioms, we can just take all the results we already have without needing to rederive and reprove them for the new structure as well.

Trying to address the question in the fourth paragraph. The point of this axiom is that it ties together the following three operations:

1. The addition of scalars, the plus sign in $(a+b)$ is just the usual addition of numbers
2. The addition of vectors, the plus sign in $a\mathbf{v}+b\mathbf{v}$ is the addition of vectors.
3. The multiplication of a vector by a scalar. In the equation $$a\mathbf{v}+b\mathbf{v}=(a+b)\mathbf{v}$$ the vector $\mathbf{v}$ is multiplied by three different scalars, $a$, $b$ and $a+b$.

Basically this axiom is telling that we can change the order of these operations. We can either first multiply the vector $\mathbf{v}$ by scalars $a$ and $b$, and then add the resulting vectors together. Or, we can first add the scalars together $a+b$, and then multiply the vector $\mathbf{v}$ with the resulting scalar. We require that the outcomes of the two calculations should always be the same. Irrespective of choice of $a,b,\mathbf{v}$.

• I'm stating that for any number c, one can divide this number into two parts, even if one part is zero. Then you could say with c=(a+b) and c=1=(1+0) equally, multiplying a vector v by c=(a+b) then c v = (a+b) v = a v + b v. In the case with c=1 we have c=1=(1+0), now multiplying c by a vector, v we have c v = 1 v = 1 v + 0 v . Commented Mar 18, 2015 at 6:44
• Could we factor out numbers from the components, say v = (2,2) such that we have 2 v with v = (1,1)? Commented Mar 18, 2015 at 6:56
• @unintuit Can't you see you are using the mentioned property? You may set $c=a+b$, but without the aforementioned axiom $(a+b)v \neq av+bv$. For the two sides to be equal, you must accept the axiom. Commented Mar 18, 2015 at 8:23

It is not hard to show that this axiom is independent of the rest, for example $\,V = \Bbb R\,$ equipped with scalar product $\ a\cdot r = a^2 r\$ satisfies all vector space axioms except $\,(a+b)r = ar + br.$