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From: Gerry Myerson on 3 Sep 2009 19:13
In article
<62dff4d4-4cbb-4618-8967-ef3bcd4fd52a(a)e8g2000yqo.googlegroups.com>,
Tonico <Tonicopm(a)yahoo.com> wrote:

> On Sep 3, 10:15�pm, Mc Lauren Series <mclaurenser...(a)gmail.com> wrote:
> > On Sep 4, 12:06�am, Robert Israel
> >
> >
> >
> >
> >
> > <isr...(a)math.MyUniversitysInitials.ca> wrote:
> > > Mc Lauren Series <mclaurenser...(a)gmail.com> writes:
> >
> > > > In a set of vectors, there can be a maximum of two 2-D vectors which
> > > > are linearly independent. Any set of three or more 2-D vectors are
> > > > linearly dependent.
> >
> > > > Similalry, there can be a maximum of three 3-D vectors which are
> > > > linearly independent. Can this be generalized for N-D vectors that
> > > > there can be a maximum of N N-D vectors which are linearly
> > > > independent?
> >
> > > Look up the definition of "dimension".
> > > --
> > > Robert Israel � � � � � � �isr...(a)math.MyUniversitysInitials.ca
> > > Department of Mathematics � � � �http://www.math.ubc.ca/~israel
> > > University of British Columbia � � � � � �Vancouver, BC, Canada
> >
> > Here, I am using it in the sense of components. By 2-D vectors, I mean
> > vectors having two components. Sorry for the confusion. What is the
> > answer to my question?-
>
>
> The answer is yes, as long as you're talking of vectors with
> components from a field, like the reals of complex.
> The appropiate context though is what Robert I. already told you: look
> up "dimension of vector space", in linear algebra books (or in google,
> of course).
>
> Tonio

With all due respect, it's not quite as easy as looking up
the definition of dimension. A basis for R^n (where R^n is
is what OP would refer to as the set of all n-D vectors) is
a linearly independent set of vectors that spans R^n, and
there's a theorem that says that every basis of R^n has
exactly n elements, and the dimension of a vector space is
defined to be the number of elements in a basis. So
every basis for, say, R^7 has 7 elements; how does that imply
that any set of 8 vectors in R^7 is linearly dependent? Well,
Tonio, you know, and Robert knows, and I know (I just taught it
to a class last week, as it happens), but it does take a little work
to prove it.

--
Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Gerry Myerson on 3 Sep 2009 23:48
In article <rbisrael.20090903230235$68f6(a)news.acm.uiuc.edu>,
Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote:

> Mc Lauren Series <mclaurenseries(a)gmail.com> writes:
>
> > On Sep 4, 12:06=A0am, Robert Israel
> > <isr...(a)math.MyUniversitysInitials.ca> wrote:
> > > Mc Lauren Series <mclaurenser...(a)gmail.com> writes:
> > >
> > > > In a set of vectors, there can be a maximum of two 2-D vectors which
> > > > are linearly independent. Any set of three or more 2-D vectors are
> > > > linearly dependent.
> > >
> > > > Similalry, there can be a maximum of three 3-D vectors which are
> > > > linearly independent. Can this be generalized for N-D vectors that
> > > > there can be a maximum of N N-D vectors which are linearly
> > > > independent?
> > >
> > > Look up the definition of "dimension".
> > > --
> > > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0
> > > =A0isr...(a)math.MyUniversitysInitial=
> > s.ca
> > > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel
> > > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC,
> > > Cana=
> > da
> >
> > Here, I am using it in the sense of components. By 2-D vectors, I mean
> > vectors having two components. Sorry for the confusion. What is the
> > answer to my question?
>
> I mean dimension in the sense of vector spaces. Once you understand
> that, and see how to find the dimension of R^N, you will have the answer
> to your question.

Maybe so. But also one can answer the question without knowing
what a vector space is, nor how dimension is defined in one. One
can prove that n + 1 elements of R^n must be linearly dependent
using little more than the basics of elementary row operations on
matrices.

--
Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: David C. Ullrich on 4 Sep 2009 10:00
On Fri, 04 Sep 2009 13:48:18 +1000, Gerry Myerson
<gerry(a)maths.mq.edi.ai.i2u4email> wrote:

>In article <rbisrael.20090903230235$68f6(a)news.acm.uiuc.edu>,
> Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote:
>
>> Mc Lauren Series <mclaurenseries(a)gmail.com> writes:
>>
>> > On Sep 4, 12:06=A0am, Robert Israel
>> > <isr...(a)math.MyUniversitysInitials.ca> wrote:
>> > > Mc Lauren Series <mclaurenser...(a)gmail.com> writes:
>> > >
>> > > > In a set of vectors, there can be a maximum of two 2-D vectors which
>> > > > are linearly independent. Any set of three or more 2-D vectors are
>> > > > linearly dependent.
>> > >
>> > > > Similalry, there can be a maximum of three 3-D vectors which are
>> > > > linearly independent. Can this be generalized for N-D vectors that
>> > > > there can be a maximum of N N-D vectors which are linearly
>> > > > independent?
>> > >
>> > > Look up the definition of "dimension".
>> > > --
>> > > Robert Israel =A0 =A0 =A0 =A0 =A0 =A0
>> > > =A0isr...(a)math.MyUniversitysInitial=
>> > s.ca
>> > > Department of Mathematics =A0 =A0 =A0 =A0http://www.math.ubc.ca/~israel
>> > > University of British Columbia =A0 =A0 =A0 =A0 =A0 =A0Vancouver, BC,
>> > > Cana=
>> > da
>> >
>> > Here, I am using it in the sense of components. By 2-D vectors, I mean
>> > vectors having two components. Sorry for the confusion. What is the
>> > answer to my question?
>>
>> I mean dimension in the sense of vector spaces. Once you understand
>> that, and see how to find the dimension of R^N, you will have the answer
>> to your question.
>
>Maybe so. But also one can answer the question without knowing
>what a vector space is, nor how dimension is defined in one. One
>can prove that n + 1 elements of R^n must be linearly dependent
>using little more than the basics of elementary row operations on
>matrices.

And proving that is precisely the same as proving that R^n
has dimension n.



David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
From: Gerry on 4 Sep 2009 18:46
On Sep 5, 12:00 am, David C. Ullrich <dullr...(a)sprynet.com> wrote:
> On Fri, 04 Sep 2009 13:48:18 +1000, Gerry Myerson
> <ge...(a)maths.mq.edi.ai.i2u4email> wrote:

> >Maybe so. But also one can answer the question without knowing
> >what a vector space is, nor how dimension is defined in one. One
> >can prove that n + 1 elements of R^n must be linearly dependent
> >using little more than the basics of elementary row operations on
> >matrices.
>
> And proving that is precisely the same as proving that R^n
> has dimension n.

I don't think so. You can prove R^n has dimension n by pointing
out that the set {(1, 0, ..., 0), ..., (0, ..., 0, 1)} has
n elements, is linearly independent, and spans n.
--
GM
From: David C. Ullrich on 5 Sep 2009 08:42
On Fri, 4 Sep 2009 15:46:06 -0700 (PDT), Gerry <gerry(a)math.mq.edu.au>
wrote:

>On Sep 5, 12:00�am, David C. Ullrich <dullr...(a)sprynet.com> wrote:
>> On Fri, 04 Sep 2009 13:48:18 +1000, Gerry Myerson
>> <ge...(a)maths.mq.edi.ai.i2u4email> wrote:
>
>> >Maybe so. But also one can answer the question without knowing
>> >what a vector space is, nor how dimension is defined in one. One
>> >can prove that n + 1 elements of R^n must be linearly dependent
>> >using little more than the basics of elementary row operations on
>> >matrices.
>>
>> And proving that is precisely the same as proving that R^n
>> has dimension n.
>
>I don't think so. You can prove R^n has dimension n by pointing
>out that the set {(1, 0, ..., 0), ..., (0, ..., 0, 1)} has
>n elements, is linearly independent, and spans n.

Erm, right.

That proves R^n has dimension n _if_ we have _already_
proved that the dimension is well-defined by "the dimension
is the number of elements in a basis". Proving that that
set is a basis does not show that R^n does not also have
dimension n+1 in addition to having dimension n.

Let's start over. Robert said that the OP should look up
the definition of "dimension" for a proof that n+1
vectors cannot be independent. You suggested that
that was not needed, all we needed was an argument
about row operations.

Nobody suggested that the definition per se proved
anything. A definition obviously cannot prove anything
like the fact in question. Along with learning the
definition of "dimension" one _also_ learns that
the definition "the dimension is the number of elements
in a basis" _makes sense_. Which is to say that one
also learns that any two bases have the same number
of elements. And that argument you suggested _is_
exactly how one proves that _any_ basis for R^n
has dimension n. The definition of dimension is
very relevant.

**************

A shorter pithier and less comprehensible version
of what I just said:

Fine. Ok, let's replace

(i) That's exactly how you prove that R^n has dimension n

with

(ii) That's exactly how you prove that R^n does not have
dimension other than n.



David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
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