About the proof of the linear correlation for n +1 n-dimensional vectors.

[Hongyi Zhao]

Hi all,

It's well known that n+1 n-dimensional vectors are linearly correlative.

But for the proof of this proposition, the common methods are based rank
of matrix, theory of determinant, theory of linear equations, and so on.

I want to prove it only based on the definition of linear (in)dependence,
but failed to achieve my aim. Could someone please give me some hints?

Best regards
--
.: Hongyi Zhao [ hongyi.zhao AT gmail.com ] Free as in Freedom :.

[Tim Little]

On 2012-02-12, Hongyi Zhao <hongy...@gmail.com> wrote:
> It's well known that n+1 n-dimensional vectors are linearly correlative.

[...]

> I want to prove it only based on the definition of linear
> (in)dependence, but failed to achieve my aim. Could someone please
> give me some hints?

Think about the definition of "dimension" for vector spaces.


--
Tim

[Hongyi Zhao]

On Sun, 12 Feb 2012 11:09:24 +0000, Tim Little wrote:

> Think about the definition of "dimension" for vector spaces.

Do you mean something like "Dimension theorem for vector spaces"?

[hagman]

Not even theorem, merely definition.
The dimension of a vector space is the cardinality of a basis.
A basis is a maximal linear independent family.
Hence no linear independent family of vectors can have a larger
cardinality than the dimension of the vector space
(Of course, it takes a theorem, not just a definition, to show that
dimension is a well-defined concept; depending on your original setup
you may also need to show that dim K^n = n)

Hagman

[Shmuel Metz]

In <jh80p1$1d6$1...@aspen.stu.neva.ru>, on 02/12/2012

at 09:29 AM, Hongyi Zhao <hongy...@gmail.com> said:

>Could someone please give me some hints?

Yes; read the definition of "n-dimensional".

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to spam...@library.lspace.org

[Tim Little]

On 2012-02-12, Hongyi Zhao <hongy...@gmail.com> wrote:

> On Sun, 12 Feb 2012 11:09:24 +0000, Tim Little wrote:
>> Think about the definition of "dimension" for vector spaces.
>
> Do you mean something like "Dimension theorem for vector spaces"?

No, the definition suffices. The dimension theorem proves that the
space has a well-defined dimension, but it's not needed here: one of
your premises was that the space has dimension n.

You might use the dimension theorem if you had a formally weaker
premise such as "there exists a basis of size n for the space",
leaving open the possibility of having bases of other sizes as well.


--
Tim

[David C. Ullrich]

On 13 Feb 2012 03:13:11 GMT, Tim Little <t...@little-possums.net>
wrote:

>On 2012-02-12, Hongyi Zhao <hongy...@gmail.com> wrote:
>> On Sun, 12 Feb 2012 11:09:24 +0000, Tim Little wrote:
>>> Think about the definition of "dimension" for vector spaces.
>>
>> Do you mean something like "Dimension theorem for vector spaces"?
>
>No, the definition suffices. The dimension theorem proves that the
>space has a well-defined dimension, but it's not needed here: one of
>your premises was that the space has dimension n.

Oh come now!

First, the definition of the dimension of the vector space V
is "the number of elements in a basis for V", which makes no
sense until we've proved some not-quite-trivial theorem.
The definition is _not_ "the number of elements of V,
if V happens to have the property that any two bases
have the same cardinality".

Second, even if we took the second version of the definition,
the utterly trivial theorem that results is of no use whatever,
for example it doesn't apply to the case of n+1 vectors in
R^n, until we've proved something. Note as well that the
OP said he tried to prove this result but couldn't - it
follows that he must not be referring to the utterly
trivial and totally useless result that you correctly
point out follows just from the definition.

I mean really.

[David C. Ullrich]

On Sun, 12 Feb 2012 11:24:00 -0500, Shmuel (Seymour J.) Metz
<spam...@library.lspace.org.invalid> wrote:

>In <jh80p1$1d6$1...@aspen.stu.neva.ru>, on 02/12/2012
> at 09:29 AM, Hongyi Zhao <hongy...@gmail.com> said:
>
>>Could someone please give me some hints?
>
>Yes; read the definition of "n-dimensional".

That definition makes no sense until _after_ we've
proved a non-trivial theorem. That theorem is exactly
what he's askng about.

[David C. Ullrich]

On Sun, 12 Feb 2012 07:56:57 -0800 (PST), hagman
<goo...@von-eitzen.de> wrote:

>On 12 Feb., 16:26, Hongyi Zhao <hongyi.z...@gmail.com> wrote:
>> On Sun, 12 Feb 2012 11:09:24 +0000, Tim Little wrote:
>> > Think about the definition of "dimension" for vector spaces.
>>
>> Do you mean something like "Dimension theorem for vector spaces"?
>>
>> Best regards
>> --
>> .: Hongyi Zhao [ hongyi.zhao AT gmail.com ] Free as in Freedom :.
>
>Not even theorem, merely definition.
>The dimension of a vector space is the cardinality of a basis.

Which makes no sense until we've proved that any two bases
have the same cardinality. _If_ we know that then yes, his
question is trivial.

>A basis is a maximal linear independent family.
>Hence no linear independent family of vectors can have a larger
>cardinality than the dimension of the vector space
>(Of course, it takes a theorem, not just a definition, to show that
>dimension is a well-defined concept; depending on your original setup
>you may also need to show that dim K^n = n)

Ah, so you realize all that. How can you possibly think that
what he was really interested in was an answer to his question
that didn't even apply to the case of n+1 vectors in K^n?

>
>Hagman

[David C. Ullrich]

On Sun, 12 Feb 2012 09:29:06 +0000 (UTC), Hongyi Zhao
<hongy...@gmail.com> wrote:

>Hi all,
>
>It's well known that n+1 n-dimensional vectors are linearly correlative.

That's "linearly independent" in English, by the way.

>
>But for the proof of this proposition, the common methods are based rank
>of matrix, theory of determinant, theory of linear equations, and so on.
>
>I want to prove it only based on the definition of linear (in)dependence,
>but failed to achieve my aim. Could someone please give me some hints?

Pay no attention to the silly replies you've got so far.

Basically you need to prove this: Assuming K is the scalar field:

Thm. n-1 vectors cannot span K^n.

The proof is by induction on n. Start with n = 2: Suppose
that (a,b) spans K^2. Then (1,0) = c(a,b). Hence c <> 0,
and hence b = 0. Similarly a = 0, contradiction.

Now suppose that we know that n-2 vectors cannot span
K^{n-1}. It follows that if v_1, ... v+{n-1} span K^{n-1}
then v_1, ... , v_{n-1} must be independent.
(If not then one of v_1, ... v_{n-1} is a linear combination
of the others, and hence "the others" give n-2 vectors
spanning K^{n-1}, contradiction.)

Now suppose that v_1, ... , v_{n-1} span K^n. Choose
scalars c_j so that

(0,0,...,0,1) = c_1 v_1 + ... + c_{n-1} v_{n-1}.

Define P : K^n -> K^{n-1} by

Px = (x_1,...,x_{n-1}).

Now Pv_1, ... Pv_{n-1} certainly span K^{n-1},
and hence the induction hypothesis shows that
Pv_1,...,Pv_{n-1} are independent (see the
comment above).

But

c_1 Pv_1 + ... + c_{n-1} P v_{n-1}

= P(0,0,...,0,1) = 0,

so c_1, ... c_{n-1} = 0, contradiction. QED.



>
>Best regards

[David C. Ullrich]

On Mon, 13 Feb 2012 07:39:17 -0600, David C. Ullrich
<ull...@math.okstate.edu> wrote:

>On Sun, 12 Feb 2012 09:29:06 +0000 (UTC), Hongyi Zhao
><hongy...@gmail.com> wrote:
>
>>Hi all,
>>
>>It's well known that n+1 n-dimensional vectors are linearly correlative.
>
>That's "linearly independent" in English, by the way.
>
>>
>>But for the proof of this proposition, the common methods are based rank
>>of matrix, theory of determinant, theory of linear equations, and so on.
>>
>>I want to prove it only based on the definition of linear (in)dependence,
>>but failed to achieve my aim. Could someone please give me some hints?
>
>Pay no attention to the silly replies you've got so far.
>
>Basically you need to prove this: Assuming K is the scalar field:
>
>Thm. n-1 vectors cannot span K^n.

Just to dot all the i's:

Cor: n+1 vectors in K^n cannot be independent.

Lemma 1. Any maximal independent set in a vector space must span.

Trivial from the definitions.

Lemma 2. Any independent set in K^n can be extended to a
_finite_ maximal independent set.

The reason I'm posting this is I just realized that there exist
simple proofs of something just like Lemma 2, except that
they don't give a _finite_ maximal independent set. Took
me a second:

Proof of Lemma 2: Say v_1, ... v_k in K^n are independent.
Let e_1,...,e_n be the standard basis for K^n.

Note that the set v_1,...v_k,e_1,...,e_n is clearly not
independent. Hence there exists a set

S subset {e_1,...e_n}

such that

B = {v_1,...,v_k} union S

is a maximal independent subset of K^n. QED.

Proof of the cororllary: Say v_1,...,v_{n+1} in K^n
are independent. By Lemma 2 this set can be extended
to a finite maximal independent set B, so that B has
m elements, m > n.

Now B is a basis for K^n, by Lemma 1. But since B is
indepdendent, B is isomorphic to K^m. Hence the
inverse image of the standard basis for K^n under
this isomoprhism is a set of n vectors which span K^m;
since n < m this contradicts the theorem. QED.

[Tonico]

*** After all you said to the first repliers to the OP, and with which
I mostly agree unless the way it was said is not the nicest out there,
the above "proof" of lemma 2 is pretty peculiar, to say the least.

First, you assume you already know K^n has dimension n (or, if you
haven't yet defined dimension, the you pressume there exists a basis
with n vectors in K^n) .

Second, you assume you already know that a basis is a maximal
independent set (otherwise how the set v_1,...,v_k, e_1,...,e_n is
CLEARLY lin. dependent?).

But if we already know all the things you assume, how come we haven't
yet come across lemma 2?
I, for one, don't know of any proof of well definition of dimension
that doesn't use lemma 2.

Now, please do pay attention: I am NOT saying that "dimension", "basis
is a maximal lin. indep. set" and all the things you assume cannot be
proved WITHOUT using lemma 2; what I'm saying is how can YOU know this
is what the OP wanted without first asking him for some further info
about his background and EXACT wish?

I think the OP was a little odd: does he want to prove ""directly""
that if v_1,..,v_{n+1}

are vectors in K^n then they are lin. indep. DIRECTLY? Then he must
prove that if

a_1*v_1 +...+ a_{n+1}*v_{n+1} = 0, for some scalars a_1,..,a_{n+1} in
K (well, this is yet

ANOTHER assumption which seems to be pretty obvious from the question:
we're talking of

K^n as vector space over K), then a_1 = ...= a_{n+1} = 0...

How can this be obtained? Well, pretty easily using (homogeneous)
linear systems of

equations over a field, or forming a matrix and saying something about
its reduced form

and rank and...well, and the things the OP did NOT want to take into
consideration.

Then what EXACTLY did he mean? I'm not sure and I can't understand how
you can be.

Tonio ****

> >>Best regards-

[David C. Ullrich]

On Mon, 13 Feb 2012 06:52:56 -0800 (PST), Tonico <Toni...@yahoo.com>
wrote:

The reason the way it was said was not the nicest out there
is that the people who wrote those posts understood very
well what the OP was really after.

>the above "proof" of lemma 2 is pretty peculiar, to say the least.
>
>First, you assume you already know K^n has dimension n

No, I don't.

> (or, if you
>haven't yet defined dimension, the you pressume there exists a basis
>with n vectors in K^n) .

Yes I do assume that. That's obvious: the standard basis for
K^n is a basis with n vectors. It's obvious that every vector
in K^n is a linear combination of e_1,...,e_n, and it's obvious
from the definition that e_1,...,e_n are independent

>Second, you assume you already know that a basis is a maximal
>independent set

This is trivial from the definitions. A basis is independent from
the definition of "basis". And if B is a basis and v is in V
then v is a linear combination of the elements of B,
hence B union {v} is not independent. So B is a maximal
independent set -- B is independent, and no proper
superset of B is independent, which is exactly what
"maximal independent set" means.

And it's also obvious that a maximal independent set
is a basis. Say B is a maximal independent set.
Then B is independent. And B must span V:
If v is in V then maximality of B shows that
B union {v} is not independent, and hence,
since B is independent, v must be a linear
combinattion of the elements of B.

Hmm. Maybe you just misread all this. When I say
"maximal independent set" I mean exactly that.
I _don't_ mean "independent set with maximal
cardinality" - a priori there could be two different
maximal independent sets with different cardinalities.

Or maybe you're not misreading it that way - that's
just the only way I'm able to make sense of your
comments, saying I'm assuming all those things
that I'm not assuming...

>(otherwise how the set v_1,...,v_k, e_1,...,e_n is
>CLEARLY lin. dependent?).

That has nothing to do with the fact that a basis is a maximal
indepedent set. It's _obvious_. Because it's obvious that
e_1,...,e_n span K^n, hence it's obvious that v_1 is
a linear combination of e_1,...,e_n, which says precisely
that v_1,e_1,...,e_n is not independent.

>But if we already know all the things you assume, how come we haven't
>yet come across lemma 2?

That would be a good question, if in fact I assumed the things
you say I assume.

>I, for one, don't know of any proof of well definition of dimension
>that doesn't use lemma 2.
>
>Now, please do pay attention: I am NOT saying that "dimension", "basis
>is a maximal lin. indep. set" and all the things you assume cannot be
>proved WITHOUT using lemma 2; what I'm saying is how can YOU know this
>is what the OP wanted without first asking him for some further info
>about his background and EXACT wish?

Because he said he'd tried to prove all this and failed,
and based on previous posts I have the impression
that he's able to prove triivialities.

>
>I think the OP was a little odd: does he want to prove ""directly""
>that if v_1,..,v_{n+1}
>
>are vectors in K^n then they are lin. indep. DIRECTLY? Then he must
>prove that if
>
>a_1*v_1 +...+ a_{n+1}*v_{n+1} = 0, for some scalars a_1,..,a_{n+1} in
>K (well, this is yet
>
>ANOTHER assumption which seems to be pretty obvious from the question:
>we're talking of
>
>K^n as vector space over K), then a_1 = ...= a_{n+1} = 0...
>
>How can this be obtained? Well, pretty easily using (homogeneous)
>linear systems of
>
>equations over a field, or forming a matrix and saying something about
>its reduced form
>
>and rank and...well, and the things the OP did NOT want to take into
>consideration.
>
>Then what EXACTLY did he mean? I'm not sure and I can't understand how
>you can be.

I'm not sure EXACTLY what he meant. But he asked about a non-trivial
theorem that he said he didn't know how to prove without various
machinery - I showed him a proof that doesn't use any of those things.

Read it again, and let me know what gaps exist that are not trivial to
fill.

[Tonico]

On Feb 13, 5:19 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:
> On Mon, 13 Feb 2012 06:52:56 -0800 (PST), Tonico <Tonic...@yahoo.com>

++++ I think the only "obvious" thing here is we both have pretty
different views about what preceeds what and how in basic linear
algebra.

To begin with I don't think it is obvious there are maximal
independent sets in K^n unless we already proved some other stuff
beforehand, say like K^n is finitely generated. Of course, that a
maximal ind. set is a basis seems to naturally come AFTER we've
already defined basis, AFTER we've already shown such a notion is
sound.

You also write "And if B is a basis and v is in V then v is a linear

combination of the elements of B, hence B union {v} is not

independent." is, in my book, part of a rather important, basic
theorem: if a given set of vectors B in a v.s. V is lin. ind. and v is
in V, then v is a lin. combination of elements in B iff B U {v} is
lin. dependent.
Perhaps for you the above is "obvious", but then everything is after
one understands this stuff, something which I'm not sure the O.P. has
attained. ++++

> Hmm. Maybe you just misread all this. When I say
> "maximal independent set" I mean exactly that.
> I _don't_ mean "independent set with maximal
> cardinality" - a priori there could be two different
> maximal independent sets with different cardinalities.

++++ Oh, I did understand what you meant. The question here is whether
the OP did... ++++

>
> Or maybe you're not misreading it that way - that's
> just the only way I'm able to make sense of your
> comments, saying I'm assuming all those things
> that I'm not assuming...
>
> >(otherwise how the set v_1,...,v_k, e_1,...,e_n is
> >CLEARLY lin. dependent?).
>
> That has nothing to do with the fact that a basis is a maximal
> indepedent set. It's _obvious_.

++++ Once again, for you lots of things are obvious in a way I'm sure
most beginning students wouldn't quite understand, or at least that's
the way I see things. Anyway, "it is obvious" that your way of
presenting things could or couldn't be clearer or more accurate to the
OP than the way the first answerers chose, and I can't see how can we
know this without first knowing from the OP her/his opinion. ++++

Because it's obvious that
> e_1,...,e_n span K^n, hence it's obvious that v_1 is
> a linear combination of e_1,...,e_n, which says precisely
> that v_1,e_1,...,e_n is not independent.

++++ Err...yes, I'm aware of this fact. The point isn't whether this
is true or not but whether this is clear to the OP. +++

>
> >But if we already know all the things you assume, how come we haven't
> >yet come across lemma 2?
>
> That would be a good question, if in fact I assumed the things
> you say I assume.
>
> >I, for one, don't know of any proof of well definition of dimension
> >that doesn't use lemma 2.
>
> >Now, please do pay attention: I am NOT saying that "dimension", "basis
> >is a maximal lin. indep. set" and all the things you assume cannot be
> >proved WITHOUT using lemma 2; what I'm saying is how can YOU know this
> >is what the OP wanted without first asking him for some further info
> >about his background and EXACT wish?
>
> Because he said he'd tried to prove all this and failed,

++++ I beg to differ: what the OP said is that he knew how to prove
that with linear systems of eq's, matrices and stuff, and he wanted a
proof "based on the definition of linear independence". Here, perhaps,
would be a good idea to ask the OP what EXACTLY does he mean by that,
since the standard proof I can recall right now use the very
definition of lin. ind., deducing what's needed using system of eq's
and/or matrices, which are usually learnt before.

> and based on previous posts I have the impression
> that he's able to prove triivialities.
>
>
>
>
>
>
>
> >I think the OP was a little odd: does he want to prove ""directly""
> >that if v_1,..,v_{n+1}
>
> >are vectors in K^n then they are lin. indep. DIRECTLY? Then he must
> >prove that if
>
> >a_1*v_1 +...+ a_{n+1}*v_{n+1} = 0, for some scalars a_1,..,a_{n+1} in
> >K (well, this is yet
>
> >ANOTHER assumption which seems to be pretty obvious from the question:
> >we're talking of
>
> >K^n as vector space over K), then a_1 = ...= a_{n+1} = 0...
>
> >How can this be obtained? Well, pretty easily using (homogeneous)
> >linear systems of
>
> >equations over a field, or forming a matrix and saying something about
> >its reduced form
>
> >and rank and...well, and the things the OP did NOT want to take into
> >consideration.
>
> >Then what EXACTLY did he mean? I'm not sure and I can't understand how
> >you can be.
>
> I'm not sure EXACTLY what he meant. But he asked about a non-trivial
> theorem that he said he didn't know how to prove without various
> machinery - I showed him a proof that doesn't use any of those things.

++++ I think you should read again the OP's original two (only) posts:
nowhere there mentions he anything about non-trivial, theorems or
stuff, and now it seems to me clearer why you did what you did in your
responses, which seemed to be pretty remote from what the OP actually
wanted.

Of course, I could be wrong and the OP DID actually want what you
wrote, but again: how could you know this so as to disqualify what the
others wrote?

Tonio +++

> >> >>Best regards-- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -

[David C. Ullrich]

On Mon, 13 Feb 2012 12:18:55 -0800 (PST), Tonico <Toni...@yahoo.com>
wrote:

For heaven's sake. Which of the following does not count
as "obvious" to you?

(i) The "standard basis" is in fact a basis for K^n.

(ii) If v is a linear combination of v_1,...,v_n,
then v, v_1, ... , v_n are not independent.

>Of course, that a
>maximal ind. set is a basis seems to naturally come AFTER we've
>already defined basis, AFTER we've already shown such a notion is
>sound.
>
>You also write "And if B is a basis and v is in V then v is a linear
>combination of the elements of B, hence B union {v} is not
>independent." is, in my book, part of a rather important, basic
>theorem: if a given set of vectors B in a v.s. V is lin. ind. and v is
>in V, then v is a lin. combination of elements in B iff B U {v} is
>lin. dependent.
>Perhaps for you the above is "obvious",

??? Look. Say B = b_1, ... b_n is a basis for V and v is in
V. Then there exist c_1, ..., c_n such that

v = c_1 v_1 + ... + c_n v_n.

Hence

-v + c_1 v_1 + ... + c_n v_n = 0,

hence v, v_1, ... , v_n is not independent.

What part of that is not obvious?

>but then everything is after
>one understands this stuff,

If you say so. It was not at all obvious to me how
to prove that n+1 vectors in K^n cannot be independent.

You shouldn't use quote marks unless it's a precise quote.
Here's a _quote_ from the OP:

"I want to prove it only based on the definition of linear
(in)dependence,
but failed to achieve my aim."

He says he wants a proof based _only_ on the definition of
linear independence, which is what I provided.

>Here, perhaps,
>would be a good idea to ask the OP what EXACTLY does he mean by that,
>since the standard proof I can recall right now use the very
>definition of lin. ind., deducing what's needed using system of eq's
>and/or matrices, which are usually learnt before.

And which are exacty what he said he wanted to avoid. Another
_quote_:

"But for the proof of this proposition, the common methods are
based rank of matrix, theory of determinant, theory of linear
equations, and so on.

I want to prove it only based on the definition of linear
(in)dependence, but failed to achieve my aim."

That makes it very clear that the standard proof you recall
is exactly what he doesn't want. He knows you can prove
it that way.

>
>
>> and based on previous posts I have the impression
>> that he's able to prove triivialities.
>>
>>
>>
>>
>>
>>
>>
>> >I think the OP was a little odd: does he want to prove ""directly""
>> >that if v_1,..,v_{n+1}
>>
>> >are vectors in K^n then they are lin. indep. DIRECTLY? Then he must
>> >prove that if
>>
>> >a_1*v_1 +...+ a_{n+1}*v_{n+1} = 0, for some scalars a_1,..,a_{n+1} in
>> >K (well, this is yet
>>
>> >ANOTHER assumption which seems to be pretty obvious from the question:
>> >we're talking of
>>
>> >K^n as vector space over K), then a_1 = ...= a_{n+1} = 0...

Didn't read that paragraph carefully til just now. The result you
mention
here cannot be obtained by any method whatever.

>> >How can this be obtained? Well, pretty easily using (homogeneous)
>> >linear systems of
>>
>> >equations over a field, or forming a matrix and saying something about
>> >its reduced form
>>
>> >and rank and...well, and the things the OP did NOT want to take into
>> >consideration.
>>
>> >Then what EXACTLY did he mean? I'm not sure and I can't understand how
>> >you can be.
>>
>> I'm not sure EXACTLY what he meant. But he asked about a non-trivial
>> theorem that he said he didn't know how to prove without various
>> machinery - I showed him a proof that doesn't use any of those things.
>
>
>++++ I think you should read again the OP's original two (only) posts:
>nowhere there mentions he anything about non-trivial, theorems or
>stuff,

I didn't say he mentioned the word "non-trivial". I said he wanted
a proof of a non-trivial theorem. He did - the theorem he asked
about is non-trvial.

>and now it seems to me clearer why you did what you did in your
>responses, which seemed to be pretty remote from what the OP actually
>wanted.
>
>Of course, I could be wrong and the OP DID actually want what you
>wrote, but again: how could you know this so as to disqualify what the
>others wrote?

What the others wrote, saying that it's clear from the definition
of linear independence, is just silly. Because the definition of
linear independence cannot be given until we know a certain
non-trivial fact. That non-trivial fact can be stated in various
obviously equivalent ways: (i) n+1 vectors in K^n cannot be
independent (ii) n-1 vectors in K^n cannot span K^n
(iii) In a finitely generated vector space the cardinality
of any spanning set is >= the cardinality of any independent
set (iv) any two bases have the same cardinality.

He asked about (i) (comment on that below). (i) is
equivalent to (iv). And we cannot state the definition
of "dimension" until after we've proved (iv). So saying
that (i) is immediate from the definition of dimension
is just silly.

To put it another way: Do you _really_ think that what
he wanted was an answer to his original question that
does not even suffice to show that n+1 vectors in
R^n must be dependent?

What he actually asked about was n+1 "n-dimensional"
vectors. There is no such thing as an "n-dimensional
vector". What he meant by "n-dimensional vector"
was "element of K^n" (or maybe just "element of R^n").

How do I know? Years of experience. Also there's the
fact that "n-dimensional vector" makes more sense meaning
"element of K^n". The vector (1,2,3) looks like a "3-d"
vector.

[hagman]

And once again everybody seems to treat a basis as a *set* of vectors
(or at least the non-commented use of "union" suggests so).
This way the quoted statement

if a given set of vectors B in a v.s. V is lin. ind. and v is
in V, then v is a lin. combination of elements in B iff B U {v} is
lin. dependent.

is "obviously" wrong:
Let V = R^1. Then V is a vector space and (1) is a vector in V,
hence B = {(1)} is a set of vectors in V.
B is linear independent.
Let v = (1). Then v is in v.
Now check what the quote joins with an "iff":
a) v is a linear combination of elements in B: true as v = 1*(1)
b) B U {v} is linear dependent: false as B U {v} = {(1)}

I just wanted to add this remark to underline that "obvios"
is often not that obvious.

hagman

[David C. Ullrich]

Yes. It's a standard abuse of language, at least in elementary
linear algebra books - bases are really sequences, not sets,
but they're referred to as sets. Sorry, I assumed we understood
that that was the convention.

Proof that those books abuse the language in exactly the same
way as I did: The standard definition is this:

"Vectors x_1, ... x_n are dependent if there exist scalars
c_1, ... c_n, not all 0, such that c_1 x_1 + ... + c_n x_n = 0".

That's the way the definition is stated in every elementary
linear algebra book I've ever seen. Those books nonetheless
talk about independent "sets", but if they really mean "set"
then the definition above is wrong (unless we add the condition
that x_1, ... x_n are distinct).

For example, if x <> 0, x_1 = x and x_2 = x then the
definition above says that the "set" x_1, x_2 is dependent,
because x_1 - x_2 = 0. On the other hand, if they really
were talking about sets then {x_1, x_2} = {x_1},
an independent set of vectors.

Sorry. Meant to acknowledge this, got distracted by something.
But it's not just me, it really is a standard wrongness.

[Tonico]

> hagman-


Agreed and point taken. I think we all can be careless sometimes when
talking about something we "know very well". For example, my mistake
in not pointing out that v should be different from all v_1,...,v_n in
that theorem I mention (or changing the wording altogether) is more
serious, imho, than others' when they refer to "sets" instead to
sequences or "ordered sets".

Tonio

[Tonico]

On Feb 14, 12:55 am, David C. Ullrich <ullr...@math.okstate.edu>
wrote:
> On Mon, 13 Feb 2012 12:18:55 -0800 (PST), Tonico <Tonic...@yahoo.com>

**** For sake's heaven: none, UNLESS we've already, of course, learnt
what basis and lin. (in)dependency is. This is so trivial that it is
boring, yet I think this is what the talk was about. ****

> >Of course, that a
> >maximal ind. set is a basis seems to naturally come AFTER we've
> >already defined basis, AFTER we've already shown such a notion is
> >sound.
>
> >You also write "And if B is a basis and v is in V then v is a linear
> >combination of the elements of B, hence B union {v} is not
> >independent." is, in my book, part of a rather important, basic
> >theorem: if a given set of vectors B in a v.s. V is lin. ind. and v is
> >in V, then v is a lin. combination of elements in B iff B U {v} is
> >lin. dependent.
> >Perhaps for you the above is "obvious",
>
> ??? Look. Say B = b_1, ... b_n is a basis for V and v is in
> V. Then there exist c_1, ..., c_n such that
>
>   v = c_1 v_1 + ... + c_n v_n.
>
> Hence
>
>    -v + c_1 v_1 + ... + c_n v_n = 0,
>
> hence v, v_1, ... , v_n is not independent.
>
> What part of that is not obvious?

**** Either you're being sternly stubborn or else you've completely
misread/dismissed what I wrote before. ***

> >but then everything is after
> >one understands this stuff,
>
> If you say so. It was not at all obvious to me how
> to prove that n+1 vectors in K^n cannot be independent.
>

**** It wasn't for me, either, and this is precisely the point: what
did the OP mean? I can't say for sure, yet YOU seemed to know exactly
and, for this, you chose to dismiss and even to belittle what others
wrote back, and I still can't see how your way is better or clearly
preferable to others'. ****

**** I know and knew that from the beginning, yet this is not the
point we were dealing with but HOW to know precisely what'd a proof of
the wanted fact can be attained NOT by the basic methods the OP ruled
out.

I think we've exhausted the subject here. I want to point out that I
wasn't trying to prove you're wrong, mostly because of the simple
reason you weren't, but only trying to find out why you, an
experimented and well known contributor here, felt like belittling
others' contributions without being, imho, sure what the OP actually
wanted.

Tonio

>
>
>
>
>
>
>
> >> and based on previous posts I have the impression
> >> that he's able to prove triivialities.
>
> >> >I think the OP was a little odd: does he want to prove ""directly""
> >> >that if v_1,..,v_{n+1}
>
> >> >are vectors in K^n then they are lin. indep. DIRECTLY? Then he must
> >> >prove that if
>
> >> >a_1*v_1 +...+ a_{n+1}*v_{n+1} = 0, for some scalars a_1,..,a_{n+1} in
> >> >K (well, this is yet
>
> >> >ANOTHER assumption which seems to be pretty obvious from the question:
> >> >we're talking of
>
> >> >K^n as vector space over K), then a_1 = ...= a_{n+1} = 0...
>
> Didn't read that paragraph carefully til just now. The result you
> mention
> here cannot be obtained by any method whatever.
>
>
> >> >How can this be obtained? Well, pretty easily using (homogeneous)
> >> >linear systems of
>
> >> >equations over a field, or forming a matrix and saying something about

> >> >its-

[Kaba]

David C. Ullrich wrote:
> Yes. It's a standard abuse of language, at least in elementary
> linear algebra books - bases are really sequences, not sets,
> but they're referred to as sets. Sorry, I assumed we understood
> that that was the convention.

--8x--

> Proof that those books abuse the language in exactly the same
> way as I did: The standard definition is this:
>
> "Vectors x_1, ... x_n are dependent if there exist scalars
> c_1, ... c_n, not all 0, such that c_1 x_1 + ... + c_n x_n = 0".
>
> That's the way the definition is stated in every elementary
> linear algebra book I've ever seen. Those books nonetheless
> talk about independent "sets", but if they really mean "set"
> then the definition above is wrong (unless we add the condition
> that x_1, ... x_n are distinct).

Actually, the notation {x_1, ... x_n}, or {x_i}_{i in S} in general,
stands for an injective family of sets (an injective sequence in case
S subset NN), i.e. an indexed _set_, or an injective function from S to
V. A general family of sets, or a function from S to V, is denoted by
(x_1, ..., x_n), or (x_i)_{i in S} in general (a sequence in case S
subset NN).

It is ok to write {1, 2, 2, 3} = {1, 2, 3}. The convention only applies
after an indexing is imposed on a set.

> For example, if x <> 0, x_1 = x and x_2 = x then the
> definition above says that the "set" x_1, x_2 is dependent,
> because x_1 - x_2 = 0. On the other hand, if they really
> were talking about sets then {x_1, x_2} = {x_1},
> an independent set of vectors.

By the previous convention, this notation breaks the injectivity of
indexing.

--
http://kaba.hilvi.org

[David C. Ullrich]

On Mon, 13 Feb 2012 19:51:28 -0600, David C. Ullrich
<ull...@math.okstate.edu> wrote:

>On Mon, 13 Feb 2012 15:15:37 -0800 (PST), hagman
><goo...@von-eitzen.de> wrote:
>

>[...]

Two comments on this, one to clear up a possible misunderstanding:

I'm pleading guilty to a standard abuse of language, not agreeing
that anything I said was actually wrong. _If_ we bear in mind
that so-called "sets" of vectors are really sequences of
vectors (and then interpret things like "union" in the obvious
way) then (as far as I know) everything I said here is correct.

A more striking bit of evidence that this abuse of language
really is perfectly standard: In spite of the fact that in
all those books one reads about independent "sets" of
vectors and sees no mention of independent sequences
of vectors, one also sees the fundamentally important
theorem that a square matrix is invertible if and only
if the rows are independent. If we're actually talking
about the _set_ of the rows then this theorem is
false; consider the matrix [[1,1], [1,1]].

[Kaba]

David C. Ullrich wrote:
> A more striking bit of evidence that this abuse of language
> really is perfectly standard: In spite of the fact that in
> all those books one reads about independent "sets" of
> vectors and sees no mention of independent sequences
> of vectors, one also sees the fundamentally important
> theorem that a square matrix is invertible if and only
> if the rows are independent. If we're actually talking
> about the _set_ of the rows then this theorem is
> false; consider the matrix [[1,1], [1,1]].

Similarly, it is standard to say "RR^n is a vector space", which is
strictly speaking false. Explicitly speaking, one should say something
like (RR^n, +, *) is a vector space. I tried for some time to go with
this more stricter style, but it quickly showed itself too bothersome.
For example, if by the explicit style I say "V is a vector space", then
I would have problems referring to the elements of the vector space (one
would need something like x in ElementsOf(V)).

Taking the ZFC as a basis, every mathematical object is a set. Thus,
whatever object one is talking about, there should be a corresponding
set. In your example the confusion rises not on taking the set of rows.
It is about a wrong definitions of rows. Explicitly speaking again, the
rows should be associated with a row number (for example), or some other
unique identifier. Then the matrix inversion theorem becomes true again.

After the previous is understood, that we can always make things
precise, we can again make things implicit. I wouldn't call it an abuse
of language.

--
http://kaba.hilvi.org

[Shmuel Metz]

In <0g9ij797vhobh8pdj...@4ax.com>, on 02/13/2012

at 09:19 AM, David C. Ullrich <ull...@math.okstate.edu> said:

>The reason the way it was said was not the nicest out there is that
>the people who wrote those posts understood very well what the OP was
>really after.

Nonsense. Guessing what an unclear poster is really after is a black
art with a high error rate.

[Shmuel Metz]

In <h02ij7hec3r2ck854...@4ax.com>, on 02/13/2012

at 06:55 AM, David C. Ullrich <ull...@math.okstate.edu> said:

>That definition makes no sense until _after_ we've
>proved a non-trivial theorem. That theorem is exactly
>what he's askng about.

No. He wrote "It's well known that n+1 n-dimensional vectors are
linearly correlative." So he's not asking how to prove that the vector
space is n-dimensional, he's taking it as a given. Also, the theorem
that two bases for a finite-dimensional vector space have the same
cardinality is trivial.

[David C. Ullrich]

On Mon, 13 Feb 2012 12:51:11 -0500, Shmuel (Seymour J.) Metz
<spam...@library.lspace.org.invalid> wrote:

>In <h02ij7hec3r2ck854...@4ax.com>, on 02/13/2012
> at 06:55 AM, David C. Ullrich <ull...@math.okstate.edu> said:
>
>>That definition makes no sense until _after_ we've
>>proved a non-trivial theorem. That theorem is exactly
>>what he's askng about.
>
>No. He wrote "It's well known that n+1 n-dimensional vectors are
>linearly correlative." So he's not asking how to prove that the vector
>space is n-dimensional, he's taking it as a given.

I doubt it. Strictly speaking there's no such thing as an
"n-dimensional vector", so again we have to guess. It seems
very likely to me that what he meant by that phrase is simply
"element of K^n" or maybe just R^n.

Why do I think that? One, because of experience with the way
students speak. Two, because, for example, it makes sense
to me to refer to (1,2,3) as a "3-dimensional vector"
even without knowing anything about the definition of dimension.
(And somehow the fact that he said "You mean the dimension
theorem?" when someone said that the definition of dimension
was all that was needed points me in this direction as well.)

>Also, the theorem
>that two bases for a finite-dimensional vector space have the same
>cardinality is trivial.

Really? That's news to me. Or not, or course, depending
on what you mean by "trivial". To clarify what you mean
by that word, could you show us the trivial proof?

I mean, I know how to prove it, but it doesn't seem trivial
to me (and btw Tonico has been complaining about me
claiming things are obvious that he insists are not obvious).

Oh. Yes, the theorem is trivial _after_ we've done a certain
amount of work regarding things like row-reduced
forms for matrices, etc. He specified he wanted a proof
just from the definitions, without all that machinery.
A proof that doesn't use that machinery, but which
uses things that do use that machinery, doesn't really
count. So we need a trivial proof that any two bases
for K^n have cardinality n, _without_ that stuff.

[Hongyi Zhao]

On Mon, 13 Feb 2012 12:18:55 -0800, Tonico wrote:

> ++++ I beg to differ: what the OP said is that he knew how to prove that
> with linear systems of eq's, matrices and stuff, and he wanted a proof
> "based on the definition of linear independence". Here, perhaps,
> would be a good idea to ask the OP what EXACTLY does he mean by that,

Thanks for all of the helps here, the EXACT thing that I want to said is:

Suppose we have x_i belongs to C ^ n , i =1, 2, ..., n+1.

Then for the following summation:

sum (a_i * x_i ) = 0 , we can find a set of a_i, in which, not all of
them are equal to 0, aha, this is just the def. of lin. indep.

Just as I have said, I want to prove the above thing without trouble the
rank of matrix, and the theory of lin. equs., which are all well stated
in the usual textbooks ;-)

The reason for making this issue is that, I think this issue is a more
basic thing then the rank of matrix, and the theory of lin. equs. and so
on. So, we shouldn't use the more advanced concept / methods to prove
it.

[Tonico]

Oh, finally...thanx for clarifying! This is just one of the options I
wrote about, though I suppose you meant to say that given n+1 elements
{x_1,...,x_{n+1}} in C^n over C then you can ALWAYS find n+1 scalars
a_1,...,a_{n+1} in C, NOT ALL OF THEM zero, s.t. we get the trivial
combination SUM {i,1, n+1} a_i*x_i = 0 .

Now, just as I wrote before, I don't know how to achieve this without
systems of linear equations or matrices, both things which, imho, are
LESS advanced concepts than linear independence in vector spaces.

Tonio

[David C. Ullrich]

In case you missed it in all the noise, what you say you wanted to
prove is exactly what I thought you wanted to prove, and in fact
I've posted a proof.

>
>Best regards

[David C. Ullrich]

On Wed, 15 Feb 2012 04:20:23 -0800 (PST), Tonico <Toni...@yahoo.com>
wrote:

You suppose he "meant to say" this? This is exactly what he said.

He didn't include the word ALWAYS. When one says "Suppose A.
Then B" that _means_ "If A is true then B is ALWAYS true".

>Now, just as I wrote before, I don't know how to achieve this without
>systems of linear equations or matrices,

Why do you keep saying this? I posted a proof - if you read it then
you _will_ know how to prove it without systems of linear equations
and matrices.

[Timothy Murphy]

Tonico wrote:

>> Suppose we have x_i belongs to C ^ n , i =1, 2, ..., n+1.
>>
>> Then for the following summation:
>>
>> sum (a_i * x_i ) = 0 , we can find a set of a_i, in which, not all of
>> them are equal to 0, aha, this is just the def. of lin. indep.

> Now, just as I wrote before, I don't know how to achieve this without
> systems of linear equations or matrices, both things which, imho, are
> LESS advanced concepts than linear independence in vector spaces.

I didn't see David Ullrich's proof,
but surely it is easy enough to prove by induction on n?

If the n'th coefficients of x_1,...,x_{n+1} are all 0,
one is already in C^{n-1}.
Otherwise choose one, say x_{n+1}, and subtract multiples of this
from x_1,...,x_n to get n vectors y_i = x_i - c_i x_{n+1} (i = 1,2,...,n)
with last coefficient 0.
Then apply the inductive hypothesis.


--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College Dublin

[David C. Ullrich]

On Tue, 14 Feb 2012 10:50:30 -0600, David C. Ullrich
<ull...@math.okstate.edu> wrote:

>On Mon, 13 Feb 2012 12:51:11 -0500, Shmuel (Seymour J.) Metz
><spam...@library.lspace.org.invalid> wrote:
>
>>In <h02ij7hec3r2ck854...@4ax.com>, on 02/13/2012
>> at 06:55 AM, David C. Ullrich <ull...@math.okstate.edu> said:
>>
>>>That definition makes no sense until _after_ we've
>>>proved a non-trivial theorem. That theorem is exactly
>>>what he's askng about.
>>
>>No. He wrote "It's well known that n+1 n-dimensional vectors are
>>linearly correlative." So he's not asking how to prove that the vector
>>space is n-dimensional, he's taking it as a given.
>
>I doubt it. Strictly speaking there's no such thing as an
>"n-dimensional vector", so again we have to guess. It seems
>very likely to me that what he meant by that phrase is simply
>"element of K^n" or maybe just R^n.
>
>Why do I think that? One, because of experience with the way
>students speak. Two, because, for example, it makes sense
>to me to refer to (1,2,3) as a "3-dimensional vector"
>even without knowing anything about the definition of dimension.
>(And somehow the fact that he said "You mean the dimension
>theorem?" when someone said that the definition of dimension
>was all that was needed points me in this direction as well.)

Heh. Finally, in case you missed it, he just made a post saying
that that was what he meant.

[Shmuel Metz]

In <v93jj7d1beb4dlcjp...@4ax.com>, on 02/13/2012

at 04:55 PM, David C. Ullrich <ull...@math.okstate.edu> said:

>the theorem he asked about is non-trvial.

It's a trivial application of induction.

>What the others wrote, saying that it's clear from the definition of
>linear independence, is just silly.

No: what's silly is confusing "linear independence" with
"n-dimensional" and pretending not only that you can read minds but
that others can as well.

>Because the definition of linear independence cannot be given
>until we know a certain non-trivial fact. That non-trivial fact
>can be stated in various obviously equivalent ways:
>(i) n+1 vectors in K^n cannot be independent (ii) n-1 vectors in
>K^n cannot span K^n (iii) In a finitely generated vector space
>the cardinality of any spanning set is >= the cardinality of any
>independent set (iv) any two bases have
>the same cardinality.

Balderdash. You are again confusing the definition of "linear
independence" with that of "n dimensional". That allegedly non-trivial
fact is needed to show that the dimensionality is well defined; it is
not needed to define linear independence.

>What he meant by "n-dimensional vector"
>was "element of K^n" (or maybe just "element of R^n").

You know that how, Uri Geller?

>How do I know? Years of experience.

Really? Assuming that classroom experience is enough to accurately
guess the intent of an unclear poster, that still wouldn't make it
silly for someone without the teaching experience to make a different,
equally plausible, guess.

>Also there's the fact that "n-dimensional vector" makes more
>sense meaning "element of K^n".

That's an opinion, not a fact.

[Shmuel Metz]

In <bo3lj7dkfk9prjkf5...@4ax.com>, on 02/14/2012

at 10:50 AM, David C. Ullrich <ull...@math.okstate.edu> said:

>Really? That's news to me. Or not, or course, depending
>on what you mean by "trivial". To clarify what you mean
>by that word, could you show us the trivial proof?

You use induction to show that you can replace each element of the
first basis with an element of the second. In the process you get a
1-1 mapping between the two bases.

>Oh. Yes, the theorem is trivial _after_ we've done a certain amount
>of work regarding things like row-reduced forms for matrices, etc.

I normally prefer to do things in a coordinate free fashion, so I
wasn't thinking in terms of matrices.

[David C. Ullrich]

On Tue, 14 Feb 2012 09:53:30 -0500, Shmuel (Seymour J.) Metz
<spam...@library.lspace.org.invalid> wrote:

>In <v93jj7d1beb4dlcjp...@4ax.com>, on 02/13/2012
> at 04:55 PM, David C. Ullrich <ull...@math.okstate.edu> said:
>
>>the theorem he asked about is non-trvial.
>
>It's a trivial application of induction.

Once again, you should really explain this in more detail,
just so we can see what you mean by "trivial".

Like in that noisy subthread - every time I claimed something
was obvious and Tonico objected, I explained very clearly
and completely just why it was obvious.

Lest someone be tempted to make some crack about how I
should figure it out for myself, we should note that I posted
a proof of this fact in this very thread, in the two posts in
the thread with actual mathematical content. And of course
the proof was by induction. But the proof I posted didn't
strike me as trivial - it took me a half hour to find it.

(I didn't _say_ that I'd proved that any basis for K^n has
cardinality n. But I posted a proof that n-1 vectors cannot
span, and then made a post showing how it follows that
n+1 vectors cannot be independent.)

>
>>What the others wrote, saying that it's clear from the definition of
>>linear independence, is just silly.
>
>No: what's silly is confusing "linear independence" with
>"n-dimensional"

Huh?

> and pretending not only that you can read minds but
>that others can as well.
>
>>Because the definition of linear independence cannot be given

Oh. Sorry. That was a typo. Of course this sentence is nonsense,
what I meant was that the definition of dimension cannot be
given until...

Sorry.

But you really could have guessed that it was a typo from the
fact that it was so utterly ridiculous. (This has something to do
wiith how I was able to divine what he meant from what
he wrote, by the way: With my interpretation it's a much
more reasonable question.)

Before replying that you couldn't guess anything based on
that because I've been saying ridiculous things in this
whole thread: People have contested statments I've made
about what is and what is not silly, and what the OP meant.
You can't point to anything I've said here that's
mathematically ridiculous.

>>until we know a certain non-trivial fact. That non-trivial fact
>>can be stated in various obviously equivalent ways:
>>(i) n+1 vectors in K^n cannot be independent

Of course, even if a person couldn't guess it was a typo because
it was so ridiculous, a person should really guess that at this point:
The word "independent" appears in one of the things I say we
need to prove before we can define "independent"?

>(ii) n-1 vectors in
>>K^n cannot span K^n (iii) In a finitely generated vector space
>>the cardinality of any spanning set is >= the cardinality of any
>>independent set (iv) any two bases have
>>the same cardinality.
>
>Balderdash. You are again confusing the definition of "linear
>independence" with that of "n dimensional". That allegedly non-trivial
>fact is needed to show that the dimensionality is well defined; it is
>not needed to define linear independence.


>>What he meant by "n-dimensional vector"
>>was "element of K^n" (or maybe just "element of R^n").
>
>You know that how, Uri Geller?

I hope you didn't miss the fact that it turns out
that this is exactly what he did mean.

>>How do I know? Years of experience.
>
>Really? Assuming that classroom experience is enough to accurately
>guess the intent of an unclear poster, that still wouldn't make it
>silly for someone without the teaching experience to make a different,
>equally plausible, guess.

Regardless of what we guess about the OP, the replies
that I've called silly have the property that without
a certain non-trivial result they don't even suffice
to show that n+1 elements in K^n must be dependent.
That seems silly to me.

And whether that result is trivial or not, the typical
proofs you see in books _do_ use the machinery that
the OP stated he wanted to avoid. More evidence
that this simply cannot be a helpful reply to whatever
question he meant to ask. (Assuming for the sake
of argument that his original post was not clear - seemed
perfectly clear to me.)

Suppose that X asks how to prove that every complex
polynomial has a root, and someone replies that this
is just the Fundmental Theorem of Algebra. Is that
silly? Depends on what we assume about X.

For "it follows from the definition of dimension" to
be a non-silly/helpful reply to the OP we have to
assume (at least) two things:

(i) The OP intended for "n-dimensional vector" to
mean "element of a vector space of dimension n"

(ii) The OP didn't see how to show that if V has
dimension n then n+1 vectors must be dependent.

The conjunction of (i) and (ii) is simply incongruous.
The typical student in an undergrad shut-up-and-
calculate linear algebra class isn't going to be asking
questions about how to prove things in the first
place. The set of people who even _know_ the
definition of "dimension", and who are even aware
of the fact that the definition requires that we first
prove something (as we saw when he said
"you mean the dimension theoorem?" in reply
to one reply) but who couldn't prove (ii) with
no trouble is very small.

On the other hand, if we assume, as seemed
natural to me, that he meant what it turns out he
did mean then it's a very reasonable question.
A person could know all the basic facts about linear
algebra and be fairly good at proving things without
knowing the answer to the question he asked
(I know because that description would apply to
me a few days ago.)

[David C. Ullrich]

On Wed, 15 Feb 2012 13:28:53 +0000, Timothy Murphy
<gayl...@eircom.net> wrote:

>Tonico wrote:
>
>>> Suppose we have x_i belongs to C ^ n , i =1, 2, ..., n+1.
>>>
>>> Then for the following summation:
>>>
>>> sum (a_i * x_i ) = 0 , we can find a set of a_i, in which, not all of
>>> them are equal to 0, aha, this is just the def. of lin. indep.
>
>> Now, just as I wrote before, I don't know how to achieve this without
>> systems of linear equations or matrices, both things which, imho, are
>> LESS advanced concepts than linear independence in vector spaces.
>
>I didn't see David Ullrich's proof,

What I did was needlessly roundabout, because I
wasn't paying attention: The first thing I saw how to prove
was that n-1 vectors cannot span, then I made another

post showing how it follows that n+1 vectors cannot be
independent.

>but surely it is easy enough to prove by induction on n?
>
>If the n'th coefficients of x_1,...,x_{n+1} are all 0,
>one is already in C^{n-1}.
>Otherwise choose one, say x_{n+1}, and subtract multiples of this
>from x_1,...,x_n to get n vectors y_i = x_i - c_i x_{n+1} (i = 1,2,...,n)
>with last coefficient 0.
>Then apply the inductive hypothesis.

One might complain that this is just the first few steps
in row-reducing the matrix having the x_j as rows.

Of course one could probably make a similar complaint
about the argument I posted - I think in what I wrote the
row-reduction was maybe just better hidden.

An honestly coordinate-free proof would be interesting;
it's not clear to me whether there is such a thing.

[Tonico]

On Feb 15, 3:21 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:
> On Wed, 15 Feb 2012 04:20:23 -0800 (PST), Tonico <Tonic...@yahoo.com>

**** No, it's not. The following is a copy of what he wrote:

"the EXACT thing that I want to said is:

Suppose we have x_i belongs to C ^ n , i =1, 2, ..., n+1.

Then for the following summation:

sum (a_i * x_i ) = 0 , we can find a set of a_i, in which, not all of
them are equal to 0, aha, this is just the def. of lin. indep."

For example, having the summation above with all a_i = 0 makes it
impossible to choose aset of a_i not all of them zero.

Perhaps what I wrote is EXACTLy what he meant, but I tend to be as
accurate as I can (with variable success) when writing mathematics, in
order to avoid confusions.

Tonio

[Tonico]

On Feb 15, 3:21 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:
> On Wed, 15 Feb 2012 04:20:23 -0800 (PST), Tonico <Tonic...@yahoo.com>

I really can't understand why you continue to believe that what you
post is the same as what the OP wants, but I suppose this is clear to
you...

Tonio

[Hongyi Zhao]

On Wed, 15 Feb 2012 07:12:00 -0600, David C. Ullrich wrote:

> In case you missed it in all the noise, what you say you wanted to prove
> is exactly what I thought you wanted to prove, and in fact I've posted a
> proof.

Dear David, thanks a lot for your wonderful proof ;-)

Regards

[David C. Ullrich]

On Wed, 15 Feb 2012 08:35:25 -0800 (PST), Tonico <Toni...@yahoo.com>
wrote:

This is becoming simply biizarre. You still can't understand why I
believe that what he posted was what he wanted, _after_
reading the post above in which he _says_ that that's what
he wanted?

>
>Tonio

[David C. Ullrich]

On Wed, 15 Feb 2012 10:22:57 -0500, Shmuel (Seymour J.) Metz
<spam...@library.lspace.org.invalid> wrote:

>In <bo3lj7dkfk9prjkf5...@4ax.com>, on 02/14/2012
> at 10:50 AM, David C. Ullrich <ull...@math.okstate.edu> said:
>
>>Really? That's news to me. Or not, or course, depending
>>on what you mean by "trivial". To clarify what you mean
>>by that word, could you show us the trivial proof?
>
>You use induction to show that you can replace each element of the
>first basis with an element of the second.

Meaning that if B is a basis and v_1, ... v_k is a basis then there
exists b in B such that b, v_2, ... , v_k is a basis?

Leaving aside the question of whether thinking about doing
that is "trivial": Yes, if you can do that you're done.
After about a minute's thought I don't see how to
do this - that doesn't prove it's not trivial, of course.

[Tonico]

On Feb 16, 4:46 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:
> On Wed, 15 Feb 2012 08:35:25 -0800 (PST), Tonico <Tonic...@yahoo.com>

It was bizarre even before, and you seem to be trying to make it even
more, but I don't.

Tonio

[Shmuel Metz]

In <rdmnj7pj4mgpe9mk0...@4ax.com>, on 02/15/2012

at 10:13 AM, David C. Ullrich <ull...@math.okstate.edu> said:

>Once again, you should really explain this in more detail, just so we
>can see what you mean by "trivial".

Basically Gaussian elimination, but without expressing anything in
matrix terms. If (x_1,...,x_m) and (y_1,...,y_N) are each linearly
independent and span, then you can replace one of the x_i with a y_j.
If you have exhausted one set without exhausting the other then you
can show a contradiction. It's not a one-liner when you flesh it out,
but it's still reasonably short and I consider it trivial.

>every time I claimed something was obvious and Tonico objected,

It's possible that Tonico and I do not have the same definition of
trivial, but I'd be surprised if anybody capable of reading, e.g.,
Halmos's Finite Dimensional Vector Spaces would find it difficult. Of
course, I lack your teacing experience.

[Shmuel Metz]

In <k76qj795hfh1213kp...@4ax.com>, on 02/16/2012

at 08:58 AM, David C. Ullrich <ull...@math.okstate.edu> said:

>Meaning that if B is a basis and v_1, ... v_k is a basis then
>there exists b in B such that b, v_2, ... , v_k is a basis?

Yes.

>Leaving aside the question of whether thinking about doing that is
>"trivial": Yes, if you can do that you're done. After about a
>minute's thought I don't see how to do this

Since v_1, ... v_k is maximal, for each b in B v_1, ... v_k, b is
linearly dependent and there exist scalars a_i, for i = 0..k, for
which a_0*b + Sigma_i=1^k a_i*v_i = 0. Since B is a basis, there must
be a b for which a_1 is nonzero. Then b, v_2, ... , v_k is a basis.

As I said, it's basically Gausian elimination, although I'm avoiding
matix notation and am not interested in computing an inverse.

[David C. Ullrich]

On Thu, 16 Feb 2012 12:30:44 -0500, Shmuel (Seymour J.) Metz
<spam...@library.lspace.org.invalid> wrote:

>In <k76qj795hfh1213kp...@4ax.com>, on 02/16/2012
> at 08:58 AM, David C. Ullrich <ull...@math.okstate.edu> said:
>
>>Meaning that if B is a basis and v_1, ... v_k is a basis then
>>there exists b in B such that b, v_2, ... , v_k is a basis?
>
>Yes.
>
>>Leaving aside the question of whether thinking about doing that is
>>"trivial": Yes, if you can do that you're done. After about a
>>minute's thought I don't see how to do this
>
>Since v_1, ... v_k is maximal, for each b in B v_1, ... v_k, b is
>linearly dependent and there exist scalars a_i, for i = 0..k, for
>which a_0*b + Sigma_i=1^k a_i*v_i = 0.

In fact since the v_j span V we can take a_0 = 1.

>Since B is a basis, there must
>be a b for which a_1 is nonzero. Then b, v_2, ... , v_k is a basis.

Because otherwise everything in B is a linear combination of
v_2, ... v_k, hence v_2,...v_k span V, since B spans V.

>Then b, v_2, ... , v_k is a basis.

The span of those vectors contains v_1, hence they span V.
And they're independent because, um, because...
oh. Because v_2,...,v_k are independent and b is not
a linear combination of v_2,..,v_k.

Heh. I like this. Not clear to me whether it's really "trivial",
but it does seem to me to be the most coordinate-free
argument I've seen.

[Shmuel Metz]

In <1gpsj7p0pe19j6eem...@4ax.com>, on 02/17/2012

at 08:49 AM, David C. Ullrich <ull...@math.okstate.edu> said:

>>Since v_1, ... v_k is maximal, for each b in B v_1, ... v_k, b is
>>linearly dependent and there exist scalars a_i, for i = 0..k, for
>>which a_0*b + Sigma_i=1^k a_i*v_i = 0.
>In fact since the v_j span V we can take a_0 = 1.

Don't need it for that step; since the v_i are linearly independent,
a_0 can't be 0 and we can renomalize a'_i = a_i/a_0.

>Heh. I like this.

I vaguely recall that it's the proof that Halmos used, but it's been a
few decades, so don't trust my recollection.