# Do we need to check all 10 axioms to verify that some set, call it $V$, is a vector space?

The definition of the vector space from the book:

Let $$\bf u, v$$ and $$\bf w$$ be vectors in the set $$V$$. The set $$V$$ is called a vector space if it satisfies the following 10 axioms.

1. The vector addition $$\bf u + v$$ is also in the vector space $$V$$.
2. Commutative law: $$\bf u + v = v + u$$
3. Associative law: $$\bf (u+v) + w = u + (v + w)$$
4. Neutral element. There is a vector called the zero vector in $$V$$ denoted by $$\bf O$$ which satisfies

$$\bf u + O = u$$ for every vector $$\bf u$$ in $$V$$

1. Additive inverse: $$\bf u + (-u) = O$$

2. Let $$k$$ be a real scalar. Then $$k\bf u$$ is also in $$V$$.

3. Let $$k$$ and $$c$$ be scalars, then $$k(c\mathbf{u}) = (kc)\mathbf{v}$$

4. Let $$k$$ be a real scalar. Then $$k(\mathbf{u + v}) = k\mathbf u + k\mathbf v$$

5. Let $$k$$ and $$c$$ be real scalars, then $$(k+c)\mathbf u = k\mathbf{u} + c\mathbf{u}$$

6. For every vector $$\bf u$$: $$1\mathbf{u} = \mathbf{u}$$

We say that if the elements of the set $$V$$ satisfy the above 10 axioms then $$V$$ is called a vector space and the elements are known as vectors.

Suppose we have some set, call it $$S$$, and we verified that axiom $$1$$ and $$6$$ hold.

Does it make sense to check other axioms? Isn't it self-evident that they will be satisfied? If it is not, could you provide me an example when axiom $$1$$ and $$6$$ hold but at least one other fail?

Take for example the subset $$V = \mathbb{R}_{\geq 0} \subset \mathbb{R}$$ with the usual addition and define the scalar multiplication via $$k \cdot x = \vert k \vert x$$. Then 1) and 6) hold, but 5) does for example not hold as we do not have negative real numbers in $$V$$.

Part I. I'm going to give a meta-answer here: most definitions in elementary mathematics (let's say "up to junior year in college") are pretty well established and tested. There are some bad books/papers out there, but if four different linear algebra books define "vector space" the same way, you can bet that there's a reason.

Now what about redundancy? Do you really need all those properties? You bet. If there were a way to prove, say, the 7th property from the others, every book would have only 9 rules, and then have a little theorem saying that a system that satisfies those 9 rules also satisfies this property (the 7th one from the original set).

Why is this? Because mathematicians, just like you, don't like having to confirm 15 things when confirming 10 is enough. So they do their very best to make the definitions minimal. As you've seen from the examples here, if you remove one of the properties, someone can usually come up a a system that has the other 9, but not the one you removed, i.e., they can show that the removed property is not a consequence of the other 9, so it's really necessary.

Part II. Do mathematicians really go through all those steps every time they encounter something and want to prove it's a vector space? Well, yes and no. After you've done a few of these proofs, it gets so easy that you can just say to yourself, "yep...that's gonna work out..." and not write down the details. But if someone claims that something unusual is a vector space, they'll usually sketch out the reasons for at least a bunch of the properties. And perhaps most important, folks tend to prove things like "If we have a set $$S$$ and consider the collection $$Q$$ of all formal sums of elements of $$S$$ with coefficients in $$F_2$$, the field of two elements, then $$Q$$ has a natural vector space structure." Then one can apply this to any set $$S$$, while only doing the "is a vector space" proof once.

Part III. Your particular text's definition of a vector space is kinda...weak, at least if you transcribed it faithfully. So my claim in part 1 is a little shaky. In particular, condition 5 should read $$u + (-u) = 0$$, rather than $$u + (-v) = 0$$, but I assume that this was just a miscopying on your part. The real version of condition $$5$$ is that there should be, for each vector $$u$$, exactly one other vector $$v$$ with the property that $$u + v = 0$$; that vector $$v$$ is denoted $$-u$$. As written, there's an assumption that your set comes with a negation operation, which wasn't mentioned anywhere. In fact, the whole definition's a bit of a mess, because the author didn't want to say that a vector space consisted of a set together with an operation (called vector addition), a field, and another operation (call scalar multiplication) from pairs $$(c, v)$$ in the field and the set to items in the set, and that these two operations had to have certain properties. Perhaps the author didn't want to define "operation" carefully, or maybe the author was just lazy. I can't say, but a text that defined vector space this way wouldn't be one of my treasured friends...

• You are right, condition 5 must read $𝑢+(−𝑢)=0$, I mistyped. Sep 21, 2019 at 12:09
• Great answer! Not to mention the fact that in the definition OP has given, it is basically assumed that scalar field is $\mathbb{R}$
– blue
Sep 21, 2019 at 12:12
• That's right. In fact a vector space, more formally, is a quadruple $(S, F, +, \cdot)$ where $S$ is a set ('the vectors'), $F$ is a field ('the scalars') (and it's reasonable at first to restrict to $F = \Bbb R$), $+$ is an operation that takes pairs of elements in $S$ to a new element of $S$, and $\cdot$ takes a scalar and a vector and produces a new vector, and where these operations satisfy 10 properties, etc. In actual practice, the addition and scalar multiplication operations are usually very simple and obvious, so this isn't as messy as it sounds. Sep 21, 2019 at 12:38
• Well...having taught math and math-related topics for 50 years does help a bit. Glad I was of some use to you. Sep 21, 2019 at 13:21
• First there were "vectors" as defined by Hamilton -- what we now call quaternions. They could not only be added and multiplied by a scalar, but could be "multiplied" as well, although the multiplication was non-commutative. I suppose that folks began noticing they were useful even without the multiplication operation, and then realized that other sets had similar properties.. and wanted to call them "vector-like spaces"; that probably degenerated to VS. (This is all wild conjecture, of course.) The things we now call vectors in physics probably came later... Sep 21, 2019 at 18:25

Consider the set $$S$$ of $$2 \times 2$$ real matrices and endow it with a $$+$$ law which is in fact the product of matrices. With that axiom 1 is satisfied.

And say the multiplication by a real is the usual one. Therefore axiom 6 is verified.

We know that product of matrices in not commutative. So axiom 2 is not fulfilled.

You can look at this post to dig further into the topic.

• But I'd thought that if we had, for example matrices $\bf A,B$, then $\bf A + B$ would mean matrix addition but not multiplication. Sep 21, 2019 at 11:45
• The trick is indeed to change the meaning ! This is the beauty of abstract algebra. Sep 21, 2019 at 11:46
• But if you let $+$ stand for usual addition, then the set that you've considered will be a vector space. Correct? Sep 21, 2019 at 11:59
• So in other words, whether a particular set is a vector space, depends on how you define $+$ and $\cdot$? Am I getting this correctly? Sep 21, 2019 at 12:00
• The answer is yes to both of your questions. Sep 21, 2019 at 12:05

The simple answer is yes. If you want to rely on the properties, you need to know they exist. The definition given is slightly informal.

What is defined here is a real vector space - ie a vector space in which the scalars are real numbers. In fact for a vector space in general, we can choose scalars from any field.

The definition starts with $$V$$ being an abelian group, and uses $$+$$ for the group operation. Sometimes you know $$V$$ is an abelian group because of where it has come from. This is items $$1$$ to $$5$$

Then there is the action of a field (here the real numbers) on $$V$$ which we call scalar multiplication. And this action respects both the group structure of $$V$$ and also the field structure of the scalars - which means that everything behaves as expected. Note that the field has to come from somewhere, and its properties are assumed already to be known - they are not listed in the definition.

Now $$6$$ just says that the multiplication exists and is not a property of the scalar multiplication as such. $$7$$ deals with multiplication in the field, $$8$$ with addition in the group, $$9$$ with addition in the field and $$10$$ with the multiplicative identity being an identity for the vector space as well as the field.

Once you know all the properties work together, you can use them without particularly thinking about them.