From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Q: regarding space
Date: 12 Jul 1999 16:53:36 GMT
Newsgroups: sci.math
Keywords: model theory, independence of vector space axioms
In article <7mc1n1$7kg$1@vixen.cso.uiuc.edu>,
research wrote:
>I wonder if there is any definition of the following 'space'.
>The space I'm considering is same as the vector space V
>except the axiom that:
>
>For a vector v (which is \in V),
>there is a vector notated as -v \in V, for which
>v + (-v) = 0
Attention desert-island mathematicians! This one's for you!
Recall that a (real) Vector Space is defined to be a set V on which two
binary operations "+" : V x V -> V and "*" : R x V -> V are defined,
meeting these axioms:
1. forall u,v,w in V: (u+v)+w = u+(v+w)
2. forall u,v in V: u+v = v+u
3. exists z in V forall v in V: v+z = v
4. forall v in V exists w in V: v+w = z
5. forall u,v in V forall r in R: r*(u+v) = r*u+r*v
6. forall r,s in R forall v in V: (r+s)*v = r*v+s*v
7. forall r,s in R forall v in V: (rs)*v = r*(s*v)
8. forall v in V: 1*v = v
It is proposed to drop axiom #4. It's a little hard to drop just that axiom,
since w = (-1)*v comes pretty close to satisfying what is needed:
(-1)*v + v = (-1)*v+1*v=(-1+1)*v= 0*v
So in the presence of the other axioms, #4 is equivalent to the condition
0*v = zero-vector z.
This is turn can be deduced from any of a number of other conditions,
e.g. the axiom forall v in V, v + v = v => v = z. But it appears to be
true that _some_ other condition is needed to show that 0*v = z,
otherwise all we know is that multiplication by 0 is a homomorphism
from the semigroup (V,+) to the sub-semigroup of idempotent elements
of V (a sub-semigroup on which R acts trivially: r*v=v).
Challenge: find a model (V,+,*) in which the other 7 axioms are
satisfied but in which there is a vector v in V for which 0*v is not
the zero vector.
SPOILER
Let V = R union {z}. Define "+" so that its restriction to real numbers
is ordinary addition, and then r+z=z+r = r for all real numbers r; only
z+z=z gives a sum not in R. Define "*" on R x R to be ordinary
multiplication, and define r*z = z for all reals r. Just for clarity
let me remark that 0*v is _never_ equal to z unless v=z, and v+w is
never equal to z unless v=w=z, so axiom 4 is rather aggressively violated!
Bonus challenge: for each of the other seven axioms find a model satisfy
all axioms except that one. This is a little unsatisfactory in the case
of deleting axiom #3, without which #4 has no meaning. Probably the
right replacement for #4 is the cancellation law:
4a. forall u,v,w in V : u+v = w+v => u=w
For in this case if you add axiom #3 you can recover axiom #4: simply
let w=(-1)*v and prove (v+w)+v = z+v, so that #4a => v+w=z.
dave (recent Linear Algebra teacher)
==============================================================================
From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Q: regarding space
Date: 15 Jul 1999 14:05:09 GMT
Newsgroups: sci.math
This follows up my article <7md6ig$j9i$1@gannett.math.niu.edu>, since
I see no one else took the bait.
A recent post prompted me to look up a family of examples I have used
to annoy my linear algebra students: examples showing the necessity of
each of the axioms for Vector Spaces. I summarized those axioms and asked
for examples which showed that each was independent of the others.
In detail, I first gave the list of axioms:
>Recall that a (real) Vector Space is defined to be a set V on which two
>binary operations "+" : V x V -> V and "*" : R x V -> V are defined,
>meeting these axioms:
>1. forall u,v,w in V: (u+v)+w = u+(v+w)
>2. forall u,v in V: u+v = v+u
>3. exists z in V forall v in V: v+z = v
>4. forall v in V exists w in V: v+w = z
>5. forall u,v in V forall r in R: r*(u+v) = r*u+r*v
>6. forall r,s in R forall v in V: (r+s)*v = r*v+s*v
>7. forall r,s in R forall v in V: (rs)*v = r*(s*v)
>8. forall v in V: 1*v = v
I then issued this challenge: for each of the eight axioms, find a model
satisfying all axioms except that one. In order to make sense of axiom 4 when
axiom 3 is removed, I replaced it with the cancellation law:
>4a. forall u,v,w in V : ( u+v = w+v ) => ( u=w )
This challenge was a little sneaky since in fact axiom #3 is _not_ independent
of the others; one can define a vector space using only axioms 1,2,4a, and 5-8!
It's also a little arbitrary since in some cases the axioms have to be
defined just-so in order to pull off the stunt of creating models satisfying
the other n-1 axioms, whereas the set of axioms for a vector space are
really very "natural" in their usual form. But here goes my set of
non-vector-spaces. Nicer examples welcome.
no-1: Let V=R^2, "*" = ordinary scaling, but define u"+"v to be
|cos(theta)| (u+v), where theta is the angle between u and v;
if either of u or v is zero, take theta to be zero.
no-2: Let V=R, u "+" v = u for any two vectors u and v, and (likewise)
r*u = u for any vector u and real number r.
Note that this only succeeds because I have set cancellation (4a)
and the zero law (3) in the appropriate order; of course
_if_ commutativity (2) is assumed, the order becomes irrelevant!
no-3: Can't happen! For any v and w we have
0*w + (0*w + 0*v) = (0*w + 0*w) + 0*v = (0+0)*w + 0*v = 0*w + 0*v
which by symmetry (and commutativity) is then equal also to
0*v + (0*w + 0*v). By cancellation we conclude 0*w=0*v.
So if we pick any v and define z = 0*v, then for any other
w in V we get w + z = w + 0*v = 1*w + 0*w = (1+0)*w = w.
Thus this z is a zero vector.
no-4: As in previous post, this model satisfies 1,2,3,5-8 but not 4 or 4a:
Let V = R union {z}. Define "+" so that its restriction to
real numbers is ordinary addition, and then r+z=z+r = r for all real
numbers r; only z+z=z gives a sum not in R. Define "*" on R x R to be
ordinary multiplication, and define r*z = z for all reals r. Just for
clarity let me remark that 0*v is _never_ equal to z unless v=z, and
v+w is never equal to z unless v=w=z, so axiom 4 is rather
aggressively violated!
no-5: This one is a little tricky. It's easier to violate in the complex case:
Let V = C x C with usual vector addition but to define scalar
multiplication, let sigma be any nontrivial automorphism of the
complex field (e.g. complex conjugation) and then let r*(x,y) =
(r.x, r.y) if x <> 0
(0, sigma(r).y) if x = 0.
We can use the same trick over any other field, except that R has
NO nontrivial automorphisms! And if we just restrict the previous
example to the reals inside C if sigma is complex conjugation,
then axiom #5 _will_ hold. But we can use the fact that C has
automorphisms which do not preserve R (not even setwise), and
use the above formulas to define a scalar multiplication on C x C
(or if you prefer, on R x C = R^3.)
Remark: r*(u+v) = r*u+r*v automatically holds (given the other
axioms) for all _rational_ r. So e.g. continuity would force #5, too.
no-6: Let V=R, "+" = usual addition, r*v = r^2 . v
no-7: Let V=R, "+" = usual addition, r*v = phi(r).v where phi : R -> Q
is any homomorphism of (additive) groups
no-8: Let V=R, "+" = usual addition, r*v = 0 for all r
dave
==============================================================================
From: Pertti Lounesto
Subject: Re: Q: regarding space
Date: Thu, 15 Jul 1999 18:01:24 +0200
Newsgroups: sci.math
Dave Rusin wrote:
> A recent post prompted me to look up a family of examples I have used
> to annoy my linear algebra students: examples showing the necessity of
> each of the axioms for Vector Spaces. I summarized those axioms and asked
> for examples which showed that each was independent of the others.
> In detail, I first gave the list of axioms:
>
> >Recall that a (real) Vector Space is defined to be a set V on which two
> >binary operations "+" : V x V -> V and "*" : R x V -> V are defined,
> >meeting these axioms:
> >1. forall u,v,w in V: (u+v)+w = u+(v+w)
> >2. forall u,v in V: u+v = v+u
> >3. exists z in V forall v in V: v+z = v
> >4. forall v in V exists w in V: v+w = z
> >5. forall u,v in V forall r in R: r*(u+v) = r*u+r*v
> >6. forall r,s in R forall v in V: (r+s)*v = r*v+s*v
> >7. forall r,s in R forall v in V: (rs)*v = r*(s*v)
> >8. forall v in V: 1*v = v
>
> I then issued this challenge: for each of the eight axioms, find a model
> satisfying all axioms except that one. In order to make sense of axiom 4 when
> axiom 3 is removed, I replaced 4 with the cancellation law:
> >4a. forall u,v,w in V : ( u+v = w+v ) => ( u=w ).
Your 4a resembles 3 and should replace 3 rather than 4.
Rename your 4a as 3a and replace your 4 by a new 4a:
4a. for all u in V exists v in V: (u+v)+u = u.