Redundancy in definition of vector space?

[Aziza]

According to my book, a vector space V is a set endowed with two properties:
-closure under addition
-closure under scalar multiplication

and these two properties satisfy eight axioms, one of which is:
"for all f in V there exists -f in V such that f+(-f)=0"

But then isn't this axiom redundant in describing a vector space, since we already specified that V is closed under scalar multiplication? I mean, just by closure under multiplication, we know that if f is in V, -f must be in V since -f = (-1)*f, and (-1) is a scalar..

 

[Useful nucleus]

The statement:
-f =(-1)*f
is not trivial and requires a proof. The axiom is not redundant, it provides the basis for proving the above statement.

 

[homeomorphic]

I guess that's true. In more technical terms, if you assume that the set is only an abelian monoid, rather than an abelian group, it follows automatically that it is a group if you have the scalar field acting on it in this way (using multiplicative group action that is distributive over vector addition). I use these technical terms because the technical terms summarize all the axioms in my brain where they are stored in a more organized way, so that I can wrap my head around them more easily. If this doesn't help you now, it will if you study abstract algebra. In my mind, I think of a vector space as an abelian group with a field acting on it in a distributive way. That's a lot easier to remember than 8 axioms. However, it doesn't really make sense unless you know the basic definitions in abstract algebra. But, then, I don't think the abstract concept of a vector space really makes that much sense until then, anyway. Until then, I don't think you should worry too much about the exact definition. Better to have all the examples in mind and intuitively think that a vector space is just a place where you can add vectors and multiply them by scalars, plus, having a few examples in mind, particularly ℝ^n and ℂ^n.

Incidentally, it's weird to me that the word "closure" would be used at all in the definition. Closure presupposes there is an addition operation and a multiplication operation. Usually, the term closure is reserved for subspaces, subgroups, subrings, etc. Sub-stuff. Closure is the condition for a subset to be a vector space or some other algebraic gadget.

 

[Aziza]

The statement:
-f =(-1)*f
is not trivial and requires a proof. The axiom is not redundant, it provides the basis for proving the above statement.

But "closure under scalar multiplication" means you can multiply by negative numbers as well...
so are you saying that without this axiom, multiplication by negative numbers is not defined? So we know that (-1)*(f) exists and is in V, but we do not know it is -f without this axiom?

 

[homeomorphic]

Oh, and I forgot to say, few people really care about having the fewest possible axioms. Some of the usual axioms for a group are redundant. No one cares. Well, more or less no one.

 

[homeomorphic]

The statement:
-f =(-1)*f
is not trivial and requires a proof. The axiom is not redundant, it provides the basis for proving the above statement.

No, it is redundant. I worked it out. If you don't assume additive inverses for the vector addition, you get it for free, assuming all the other axioms. The OP is right. But again, no one cares.

 

[homeomorphic]

Okay, let's work it out.

Assume the following:

(1) (kl)v = k(lv)
(2) 1 v = v
(3) (k+l)v = kv + lv

where k and l are understood to be any scalars and v is any vector.

Assuming these, we have

v+ (-1)v = 1v + (-1)v by (2)

= (1+ -1)v by (3)

= 0 v

It remains to show that 0 v is the 0 vector.

0 v + v = 0 v + 1 v = (0+1)v = 1v = v

But identities are unique in a monoid, so 0 v is the zero vector.

Thus, -1 v is an additive inverse for v with respect to vector addition.

QED

No need to assume additive inverses exist. But you don't gain a lot by going through this argument, just to show that it's redundant, and that's why no one bothers with it.

 

[Aziza]

Okay, let's work it out.

Assume the following:

(1) (kl)v = k(lv)
(2) 1 v = v
(3) (k+l)v = kv + lv

where k and l are understood to be any scalars and v is any vector.

Assuming these, we have

v+ (-1)v = 1v + (-1)v by (2)

= (1+ -1)v by (3)

= 0 v

It remains to show that 0 v is the 0 vector.

0 v + v = 0 v + 1 v = (0+1)v = 1v = v

But identities are unique in a monoid, so 0 v is the zero vector.

Thus, -1 v is an additive inverse for v with respect to vector addition.

QED

No need to assume additive inverses exist. But you don't gain a lot by going through this argument, just to show that it's redundant, and that's why no one bothers with it.

haha well my teacher is saying it is not redundant, so I wanted to see who is right! :D

 

[Useful nucleus]

Very nice! However, I still believe that the axiom is not redundant. To clarify things, let me state the axiom again:
"For each vector v in the vector space V there is a "unique" vector -v in V such that
v+ (-v) = 0."

homeomorphic showed that:
-v = -1* (v)
but this does not show the uniquenss of the additive inverse. I will give it a try and see if I can prove the uniqueness of the additive inverse without the axiom.

 

[Useful nucleus]

Probably it is better to say that homeomorphic showed that there exists an additive inverse but did not show it is unique.

 

[Stephen Tashi]

and these two properties satisfy eight axioms

It isn't the "properties" that satisfy axioms, its the "operations" of multiplication and addition that may satisfy axioms. The technical details depend on what the axioms say.

 

[pwsnafu]

Very nice! However, I still believe that the axiom is not redundant. To clarify things, let me state the axiom again:
"For each vector v in the vector space V there is a "unique" vector -v in V such that
v+ (-v) = 0."

homeomorphic showed that:
-v = -1* (v)
but this does not show the uniquenss of the additive inverse. I will give it a try and see if I can prove the uniqueness of the additive inverse without the axiom.

Suppose w₁ and w₂ are additive inverses of v. Thus
$v+w_1=0$
$v+w_1+w_2=w_2$
but since w₂ is an additive inverse of v, we have
$w_1=w_2$
QED

 

[Klungo]

The way I learned it.

For all v in V, there exists w in V such that v+w=0.

Then you prove that w=-v.

 

[jostpuur]

It remains to show that 0 v is the 0 vector.

0 v + v = 0 v + 1 v = (0+1)v = 1v = v

But identities are unique in a monoid, so 0 v is the zero vector.

If you want to prove that 0v is 0, you should prove that 0v+x=x for all vectors x. How do you accomplish that?

 

[Erland]

0 v + v = 0 v + 1 v = (0+1)v = 1v = v

But identities are unique in a monoid, so 0 v is the zero vector.

Not true, according to the standard definition of a monoid. There are (nontrivial) monoids for which a + a = a for all a, for example. You need a cancellation law for vector addition ( a + b = a + c --> b = c ). This condition is weaker, for monoids, than the existence of inverses, but it is not stated among the vector space axioms.

This still doesn't prove that the axiom of inverses is not redundant, but my guess is that it is, and I will try to construct an example of structure not satisfying this axiom but all the other vector space axioms...

I'll be back...

 

[Erland]

This still doesn't prove that the axiom of inverses is not redundant, but my guess is that it is, and I will try to construct an example of structure not satisfying this axiom but all the other vector space axioms...

I'll be back...

It turns out to be easy to find such examples. Here is perhaps the simplest of them all:

Put V={0,1} and let F be an arbitrary field (for example, the field of real numbers, R). F will be our field of scalars.

We define an addition operation on V by 0+0=0 and 0+1=1+0=1+1=1. We use 0 as "zero vector" here, which makes sense since 0+0=0 and 1+0=1. Also, notice that x+x=x, for all x in V.

We define "multiplication of scalar and vector" simply by rx=x, for all r in F and x in V. Notice that in particular, 01=1.

Now, it is easy to verify that all the vector space axioms are satisfied by this structure, except the existence of inverse vectors, which is false since there is no x in V such that 1+x=0.

The axiom of existence of inverse vectors is therefore NOT redundant.

We could replace it with a cancellation law, which is apparently (but in this case not really) weaker, but this would be rather pointless.

I am not sure whether or not any of the other vector space axioms are redundant, though.

 

[Erland]

I am not sure whether or not any of the other vector space axioms are redundant, though.

It turns out that the axiom of commutativity of vector addition (x+y=y+x) is redundant.

The axioms involving multiplication with scalars are essential in proving this commutativity from the other axioms. It would not be redundant if we had the additive structure only, since there are many nonabelian groups.

 

[Erland]

The axiom of existence of inverse vectors is therefore NOT redundant.

We could replace it with a cancellation law, which is apparently (but in this case not really) weaker, but this would be rather pointless.

We could also replace it with the condition

0v=0, for all v in V,

because then, homeomorphic's attempted proof will work. This could perhaps be considered as a simpler condition than the existence of additive inverses.
On the other hand, this proof depends upon the axioms involving multiplication with scalars, while if we keep the additive inverses axiom, the axioms involving vector addition only are the axioms for an abelian group. Thus the addition axioms can be summarized by just stating that V is an abelian group under addition.

 

[jostpuur]

Demanding

0v=0

and

1v=v

as axioms would look pretty natural to me. :!)

 

[Erland]

Demanding

0v=0

and

1v=v

as axioms would look pretty natural to me. :!)

...and then we wouldn't need -v as a basic notion, but we can define it as -v=(-1)v, and then easily prove that v+(-v)=0, as Aziza did.

Since students usually learn linear algebra before abstract algebra, and therefore don't know what an Abelian group is. this axiomatization could actally be better for them than the standard one,