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m7ossny wrote:
> So what do you recommend to resolve this!!!!!!?
>
> So you say that a vector 'v' with distance 0 does not mean it is a zero
> vector.
Yes. But then you don't have a metric space; what you have in that case
is called a pseudometric space. See
http://mathworld.wolfram.com/Metric.html .
> Consider this example. The metric space (S, sigma). S is a set of
> sequence of observations.
> s1=[1, 2, 3, 3, 4, 3, 3, 3, 3, 4, 5, 6, 7, 8, 9] belongs to S.
> s2=[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] belongs to S too.
>
> s1+s2=[2, 3, 4, 4, 5, 4, 4, 4, 4, 5, 6, 7, 8, 9, 10] also belongs to S.
>
> sigma is the standard deviation of the observations.
Assuming that s1+s2 is defined componentwise, you have a vector space
but not a metric. Any sequence of the form [a,a,a,a,...,a] will have
"length" 0.
Now for some advanced math ... If you have a pseudometric, and let Z be
the set of all sequences with sigma(s) = 0, then you can "mod out" by Z
and get a vector space where sigma _is_ a metric.
--- Christopher Heckman
> Now,
>
> sigma(s1)=sigma1
> sigma(s2)=0
>
> sigma(s1+s2)=sigma(s1)
>
> sigma(s1+s2, s1)=sigma(s1)-sigma(s1+s2)=0
>
> 1. Does not this mean that s1+s2 equals s1 (according to the metric)!?
> 2. If this is true, doesnot this mean that s2 is a ZERO vector!?
> 3. Consider s3=[2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2],
> Doesnot s1+s3 equal s1 (according to the metric)!?
>
> Now we have two zeros s2, s3!? Am right or missing something!?
>
>
> Thanks alot for time, Bye.
>
>
>
> On Oct 31, 3:04 pm, Virgil <vir...@xxxxxxxxxxx> wrote:
> > In article <1162268294.384599.319...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> >
> >
> >
> > "m7ossny" <m7os...@xxxxxxxxx> wrote:
> > > Hi,
> >
> > > I am not an Algebra Guru and have some question. Consider a metric
> > > space of vectors 'V' with a distance function 'd'. The vector space 'V'
> > > has a primary operator '+'. The problem is that I have to define zero
> > > vector where
> >
> > > 1. d(ZERO)=0.
> > > 2. v + ZERO =ZERO + v = v (for all 'v' in 'V')
> >
> > > Some times the '+' operator and 'd' function are complex enough that
> > > there might be more than one zero vector (a subset of vectors that all
> > > has absolute distance equal to zero). Some other cases there is a zero
> > > vector for each vector.
> >
> > > Does this sound right!? Or am I missing something!?There is no more that
> > > one zero-vector per vector space, with or without
> > a metric.
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