basis of dual space
Siu Lok Shun
dimensional space, for example F^\mathbb{N}?
--
Mike Mccarty Sr
Siu Lok Shun <ls...@hkusua.hku.hk> wrote:
)Who can write down explicitly a basis for the dual space of a infinitely
)dimensional space, for example F^\mathbb{N}?
Consider the analytic functions over the reals. This forms an infinitely
dimensional vector space. It has a basis {1,x,x^2,...}. What is the dual
of this space? Can you write a basis for it?
Mike
--
char *p="char *p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);}
This message made from 100% recycled bits.
I can explain it for you, but I can't understand it for you.
I don't speak for Alcatel <- They make me say that.
Lee Rudolph
>In article <8lanu7$sq...@hkunae.hku.hk>,
>Siu Lok Shun <ls...@hkusua.hku.hk> wrote:
>)Who can write down explicitly a basis for the dual space of a infinitely
>)dimensional space, for example F^\mathbb{N}?
>
>Consider the analytic functions over the reals. This forms an infinitely
>dimensional vector space. It has a basis {1,x,x^2,...}.
It most certainly has *not*, on (a) the reasonable interpretation of
your ellipsis, and (b) my guess as to what Siu Lok Shun means by "basis",
namely, "algebraic basis" (no questions of topology, no "infinite
sums", just pure linear algebra). Specifically, I will address this
test question to Siu Lok Sun: "is the countable set { x^n : n is a
non-negative integer} a basis, in your sense, of the uncountable-
dimensional real vectorspace of real-analytic functions from R to R?"
Please note that exp(x) is not a linear combination of elements of
this proposed basis (again, on the definition of "linear combination"
appropriate in linear algebra).
>What is the dual
>of this space? Can you write a basis for it?
However, I am happy to recycle these last two questions, in
a new context. "Consider the real vectorspace R^\infty of all
sequences {a_0,a_1,...,a_n,...} with only finitely many non-zero
entries. What is the dual of this space? Can you write a basis
for it?" (Notice that this space can be identified with the
real *polynomial* functions over the reals, a subspace of
the real-analytic functions over the reals.)
Lee Rudolph
David C. Ullrich
>Who can write down explicitly a basis for the dual space of a infinitely
>dimensional space, for example F^\mathbb{N}?
People are confused about exactly what you mean by various
things. Are you talking about topological vector spaces (eg Banach
spaces) or just vector spaces? (In the topological-vector-space
setting the word "basis" also means various things...)
Presumably since you mention F^N you're talking about
just algebraic vector spaces and you want the algebraic dual
(as opposed to the space of _continuous_ linear functionals,
which is what "dual space" means in topological vector spaces).
But I'm not certain what you mean by F^N: This could be
V1: the space of all sequences of elements of F
or
V2: the space of all sequences which vanish except for
finitely many terms.
These spaces have different duals. In fact as it happens
the dual of V2 is V1 and the dual of V1 is V2. The "standard basis"
is a basis for V2, but I don't believe that there _is_ a basis for
V1 that can be "written down".
So if all my assumptions about what you mean are
correct I believe the answer to your question is "I can" or
"nobody can", depending on what you mean by F^N .
Panh
David C. Ullrich <ull...@math.okstate.edu> a écrit dans le message :
3979a8cb...@nntp.sprynet.com...
> V1: the space of all sequences of elements of F
> or
>
> V2: the space of all sequences which vanish except for
> finitely many terms.
>
> These spaces have different duals. In fact as it happens
> the dual of V2 is V1 and the dual of V1 is V2. The "standard basis"
> is a basis for V2, but I don't believe that there _is_ a basis for
> V1 that can be "written down".
The dual of V1 is V2?
G. A. Edgar
<lrud...@panix.com> wrote:
> However, I am happy to recycle these last two questions, in
> a new context. "Consider the real vectorspace R^\infty of all
> sequences {a_0,a_1,...,a_n,...} with only finitely many non-zero
> entries. What is the dual of this space?
The dual is the space of all sequences (with no restriction
that only finitely many be non-zero).
> Can you write a basis for it?"
No, that requires the Axiom of Choice.
> (Notice that this space can be identified with the
> real *polynomial* functions over the reals, a subspace of
> the real-analytic functions over the reals.)
--
Gerald A. Edgar ed...@math.ohio-state.edu
G. A. Edgar
<ull...@math.okstate.edu> wrote:
> V1: the space of all sequences of elements of F
>
> or
>
> V2: the space of all sequences which vanish except for
> finitely many terms.
>
> These spaces have different duals. In fact as it happens
> the dual of V2 is V1
yes
> and the dual of V1 is V2.
no
> The "standard basis"
> is a basis for V2, but I don't believe that there _is_ a basis for
> V1 that can be "written down".
correct
david_...@my-deja.com
"G. A. Edgar" <ed...@math.ohio-state.edu.nospam> wrote:
> In article <3979a8cb...@nntp.sprynet.com>, David C. Ullrich
> <ull...@math.okstate.edu> wrote:
>
> > V1: the space of all sequences of elements of F
> >
> > or
> >
> > V2: the space of all sequences which vanish except for
> > finitely many terms.
> >
> > These spaces have different duals. In fact as it happens
> > the dual of V2 is V1
>
> yes
>
> > and the dual of V1 is V2.
>
> no
Aargh, no of course it isn't. It was clear that the
dual of V2 was V1. This morning it was not quite so clear
to me how to show that the dual of V1 was V2. Then I
"recalled" that if we're talking about the purely
algebraic dual of non-topological vector spaces then
X** is always X. (Yes, that's nonsense.)
As penance I should say (in some sense) what the
dual of V1 actually is: If B is a basis for V1 then
the dual of V1 is (ie is naturally identified with)
the class of all functions on B.
Otoh since V1 _is_ the dual of V2 there is a natural
embedding of V2 in the dual of V1 (X "is" a subset of
X** in general). If we take the basis B in the previous
paragraph to contain the "standard basis vectors",
regarding N as a subset of B, then we can see what
the image of V2 is in this description of V1*:
V2 corresponds to the functions on B which vanish
on B\N and also vanish except at finitely many points
of N. Making it clear that V2 is just a _tiny_ subset
of V1*.
> > The "standard basis"
> > is a basis for V2, but I don't believe that there _is_ a basis for
> > V1 that can be "written down".
>
> correct
>
> --
> Gerald A. Edgar ed...@math.ohio-state.edu
>
Sent via Deja.com http://www.deja.com/
Before you buy.
david_...@my-deja.com
"Panh" <pa...@club-internet.fr> wrote:
>
> David C. Ullrich <ull...@math.okstate.edu> a écrit dans le message :
> 3979a8cb...@nntp.sprynet.com...
> > or
> >
> > V2: the space of all sequences which vanish except for
> > finitely many terms.
> >
> > These spaces have different duals. In fact as it happens
> > is a basis for V2, but I don't believe that there _is_ a basis for
> > V1 that can be "written down".
>
No of course it is not, sorry. (It was early in the morning -
see my reply to Edgar for more details.)
Siu Lok Shun
--
I mean the (algebraic, not topological, basis for the dual space of a the
infinitely dimensional space over a field F, for example the ring of
polynomials over F. I know that generally the cardinality of the basis of
the dual space is equal to that of ths space if and only if the space is
finitely dimensional.
David C. Ullrich
Well then (in case you missed it, being distracted by my
momentary idiocy the other day) the answer is no you cannot
"write down" a basis for the dual. The dual of the space of
polynomials (~ sequences eventually 0) _is_ the space
of all sequences from F, (because in general if B is a basis
then the dual is the space of all functions from B to F)
and there's no way to "write down" a basis for the set of
all F-valued sequences.
We all know people who, when they make an error,
try to change the question to show they were right after all.
I hope nobody takes the following that way - that's not the
way it's intended. I was wrong wrong wrong, the following
is a _different_ question:
I wonder whether in some sense the set of all
linear functionals on V1 which can be "written down" is
V2. That's a little vague - one precise version would be to
ask whether it's consistent with ZF (minus AC) that the
dual of V1 is V2. (Or that in general if V is the space
of all F-valued functions on a set B then the dual is
the space of all functions on B vanishing except at
finitely many points.)
Siu Lok Shun
But by the axiom of choice, such a basis exists, although its cardinality
is "greater" than that of the space,and I remember that there is a
formula for the cardinality of the dual space in Jacobson's book .
> We all know people who, when they make an error,
> try to change the question to show they were right after all.
> I hope nobody takes the following that way - that's not the
> way it's intended. I was wrong wrong wrong, the following
> is a _different_ question:
> I wonder whether in some sense the set of all
> linear functionals on V1 which can be "written down" is
> V2. That's a little vague - one precise version would be to
> ask whether it's consistent with ZF (minus AC) that the
> dual of V1 is V2. (Or that in general if V is the space
> of all F-valued functions on a set B then the dual is
> the space of all functions on B vanishing except at
> finitely many points.)
--
Dave L. Renfro
[sci.math 21 Jul 2000 23:55:51 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/dwunshalclo>
wrote
> Who can write down explicitly a basis for the dual space of a
> infinitely dimensional space, for example F^\mathbb{N}?
[The following is formatted for fixed font size.]
I assume you mean "Hamel basis" (vector spaces), and not
"Schauder basis" (normed spaces).
I'll denote the cardinality of a set E by |E|. By 'c', I mean
the cardinal number 2^{Aleph_0}. R denotes the set (or field,
depending on context) of real numbers and N denotes the set
of positive integers.
THEOREM 1: If V is a vector space over a field F and dim(V) is
infinite, then dim(V*) = |F|^[dim(V)].
This is proved on page 247 of Nathan Jacobson's book "Lectures
in Abstract Algebra: Volume II--Linear Algebra", Van Nostrand,
1953 [QA266 .J158L] [This was reprinted as Springer-Verlag's
Graduate Texts in Math. #31, QA162 .J3 1975.]
In particular, if V is infinite dimensional, then dim(V*) > dim(V)
(indeed, dim(V*) will have cardinality at least c), which implies
that V* and V are not isomorphic. [Recall that any field has
cardinality at least 2.]
I don't know if by F^N you mean the direct sum (also called the
subdirect product) of Aleph_0 many copies of F or the direct
product of Aleph_0 many copies of F. The latter can be realized
as the set of all infinite sequences of elements from F (with
component-wise operations), whereas the former can be realized
as the set of all infinite sequences of elements from F which
contain only finitely many nonzero terms. You probably mean direct
product, since one commonly writes F^(N) if a direct sum is
intended. However, since the dual of F^(N) is the direct product
of Aleph_0 copies of F [This follows from the more general fact
that Hom[SUM(V_i), W] is isomorphic to Hom[PRODUCT(V_i), W],
where the V_i's (not necessarily a countable collection) and W
are vector spaces and 'SUM', 'PRODUCT' represent direct sum and
direct product.], just finding a basis for F^N (direct product)
is essentially equivalent to finding a basis for the dual space
of F^(N) (direct sum). I'll pass on exhibiting a basis for the
dual space of F^N (direct product), and focus on finding a basis
for the dual space of F^(N) (direct sum). In fact, I'll only
consider the problem of exhibiting a linearly independent set
in F^(N) (direct sum) having cardinality c.
There is a short argument given in Example 6.4 on page 46 of
William C. Brown's book "A Second Course in Linear Algebra",
John Wiley & Sons, 1988 [QA184 .B765 1987], for the fact that
the dual space of F^(N) (direct sum) has uncountable cardinality.
However, I don't believe the issue is as straightforward as Brown
says. [He writes "A simple counting exercise will convince the
reader that ...".]
Below I'll prove a result that implies the dual space of
F^(N) (direct sum) has cardinality at least c (equal to c when
|F| is at most c) in a way that will roughly allow you to
exhibit a linearly independent subset of (F^(N))* that has
cardinality c.
A number of years ago I took a linear algebra class, and in this
class the instructor made a comment while we were in the middle
of all the usual dual space results one proves in such a class
(specifically, the last few sections of Chapter 3 from
Hoffman/Kunze's text "Linear Algebra") that V and V* are NOT
isomorphic when V is infinite dimensional. He said, however,
that the proof of this was beyond the level of our course and
so we would not prove it. At this time I had been learning some
set theory on my own and had just learned that there exists a
collection of c many almost disjoint subsets (definition is below)
of a countably infinite set. It struck me that there must be some
way of utilizing this result to prove dim(V*) > dim(V) when
dim(V) is infinite, since both the notion of a Hamal basis and the
notion of a collection of almost disjoint subsets involve the idea
of equality up to a finite set. I was able to write up a proof
during the rest of the lecture (whose notes I wound up having to
get from someone else), which of course I proudly gave to the
instructor after the class. [If only more of my guesses were
correct and I was so easily able to prove them!] He gave it back
to me the next day, after having a chance to read over it, and
said that it was neat. For a couple of years I left it alone
with all my other various writings, and then when I became more
aware of the idea of publishing math papers, it occurred to me
that this might be suitable for a short American Mathematical
Monthly article. Unfortunately, I saw essentially the same proof
somewhere shortly afterwards [I can't seem to find the specific
reference right now. (See ** below, however.)], and so there went
that idea. However, now with the internet I have an option
intermediate between "left in one's drawer" and "published"--post
it on the internet! [This is where a lot of my longer sci.math
"essay posts" have come from, by the way.]
[** Added later] I looked though some more photocopied papers
I have and found something that *might* have
been what led me to not write the result up.
The paper I found gives a short proof (the entire
paper is only half a page) that any infinite
dimensional separable Banach space has (Hamel)
dimension at least c using the fact that there
exists a collection of c many almost disjoint
subsets of the positive integers: H. Elton Lacey,
"The Hamel dimension of any infinite dimensional
separable Banach space is c", Amer. Math. Monthly
80 (1973), 298. [This fact was overlooked by
Willard when, on page 184 of his text "General
Topology", exercise 24J(7) asks the reader to
prove "If a Banach space is Aleph_0--dimensional,
it is separable."] For those interested in other
published proofs of this fact about Banach spaces,
I mention these other papers: W. R. Brauer and
R. H. Benner, "The nonexistence of a Banach space
of countably infinite Hamel dimension", Amer.
Math. Monthly 78 (1971), 895-896; Nam-Kiu Tsing,
"Infinite-dimensional Banach spaces must have
uncountable basis--an elementary proof", Amer.
Math. Monthly 91 (1984), 505-506. Also, exercise 1
on page 53 of Rudin's text "Functional Analysis"
(3'rd edition) states: "If X is an infinite-
dimensional topological vector space which is the
union of countably many finite-dimensional
subspaces, prove that X is of the first category
in itself. Prove that therefore no infinite-
dimensional F-space has a countable Hamel basis."
Finally, a proof (using the Hahn-Banach theorem)
that the dimension of any infinite dimensional
Banach space is at least c is given on page 75
of Goffman and Pedrick's text "First Course
in Functional Analysis" (1983 2'nd edition
published by Chelsea). Interestingly, I couldn't
find a proof in volume 1 of Dunford and Schwartz's
treatise, but that doesn't mean it's not there.
(I'd be quite suprised if it wasn't in DS.)
DEFINITION: Let B be a countable set and let C be a collection
of subsets of B. We say that C is almost disjoint
if the intersection of any two distinct members of
C is finite and the cardinality of each element
belonging to C is Aleph_0. [Yes, I know this notion
has a more general formulation.]
FACT: Let B be a countably infinite set. Then there exists a
collection of c many almost disjoint subsets of B.
PROOF: Two proofs are given later.
THEOREM 2: Let V be an infinite dimensional vector space over a
field F. Then the dimension of V* is at least c.
PROOF: Let B = {v1, v2, ...} be a countably infinite linearly
independent subset of V and let {B_r: r \in R} be
a collection of c many almost disjoint subsets of B.
For each real number r let f_r: V --> R be a linear
extension (which will not be unique if B isn't a basis)
of the function that equals 1 if x belongs to B_r and
equals 0 if x belongs to B - B_r. Then {f_r: r \in R}
is a subset of V*. I claim that {f_r: r \in R} is
linearly independent in V*. To this end, suppose that
(*) c1*f_r1 + c2*f_r2 + ... + cn*f_rn = 0
for some c1, c2, ..., cn in F and r1, r2, ..., rn in R.
Choose an integer N sufficiently large so that
k > N ==> (f_ri)(vk) not equal (f_rj)(vk)
whenever i not equal j. [Take N to be any integer larger
than the largest index of any basis element appearing
in the finite set UNION{B_i intersect B_j: i not equal j
and i,j are between 1 and n, inclusive}. (Note that n is
a fixed positive integer, being fixed due to equation (*)
above.)] Now fix any i between 1 and n (inclusive). Since
B_ri is infinite for each i (part of the definition of
being an almost disjoint collection), we can find k > N
such that vk \in B_ri. Evaluating both sides of (*)
(an equation in V*) above at the vector vk \in V gives
c1*(f_r1)(vk) + c2*(f_r2)(vk) + ... + cn*(f_rn)(vk) = 0
==> ci*1 = 0 [all terms except the i'th term drop out],
and so ci = 0. Since i was arbitrary (prior to being fixed),
we have c1 = c2 = ... = cn = 0. Hence, {f_r: r \in R} is
linearly independent (we've shown that every finite subset
of {f_r: r \in R} is linearly independent), which implies
that V* has dimension at least c.
REMARK: Using some elementary cardinal arithmetic it is not
difficult to use theorem 2 to show the following, where
Q = the field of rational numbers:
The vector space Q^N (direct product) over the field Q
has dimension c.
The vector space R^N (direct product) over the field R
has dimension c.
By way of contrast, note that Q^(N) and R^(N) (direct sums)
over the fields Q and R, respectively, each have
dimension Aleph_0.
TWO PROOFS THAT THERE EXISTS AN ALMOST DISJOINT COLLECTION OF c
MANY SUBSETS OF A COUNTABLY INFINITE SET.
Note that it suffices to prove the result for any specific
countably infinite set (unless you're concerned with
constructivist issues, which I'm certainly not).
1. For each irrational number between 0 and 1 we associate a
subset of the natural numbers as follows. If x = .abcdefg...,
where a, b, c, etc. are the digits of the decimal expansion of
x, let N(x) be the set {a, ab, abc, abcd, ...}. For example,
N(sqrt(1/2)) = {7, 70, 707, 7071, ...}.
Then it is not difficult to check (i) N(x) is infinite for each
irrational x in (0,1), (ii) the correspondence x <---> N(x) is
one-to-one and onto (hence, there are c many N(x)’s), and
(iii) N(x) has finite intersection with N(y) whenever x is not
equal to y.
2. We'll find an almost disjoint collection in N x N with
cardinality c. For each real number r between .9 and 1.1, let
N(r) = {(m,n) in N x N : rm - 2 < n < rm + 2}.
In other words, N(r) is the set of points in N x N that belong
to the open angular sector in the first quadrant formed by
the lines y = rx - 2 and y = rx + 2.
Then it is not difficult to check (i) N(r) is infinite for each
real number r between .9 and 1.1, (ii) the correspondence
r <---> N(r) is one-to-one and onto (hence, there are c many
N(r)’s), and (iii) N(r) has finite intersection with N(s)
whenever r is not equal to s.
Proof #2 is given in J. R. Buddenhagen, "Subsets of a countable
set", Amer. Math. Monthly 78 (1971), 536-537. I don’t know if this
proof was previously known to anyone, or even if it had been
published previously. I'd be interested in learning of an earlier
published appearance of proof #2 if anyone knows of one.
Dave L. Renfro
Panh
Si K est un corps non denombrable, il est facile (Vandermonde) de voir
que les suites (x^n) forment une famille independante non denombrable.
Dans le cas general, j'ai une solution un peu plus compliquee
Pour chaque nombre reel lambda >0, construisons une suite
dont tous les termes valent 0 ou 1, la distance entre deux 1 successifs
croissant comme n^lambda.
Je dis que cette famille (non denombrable) de suites est libre sur K:
Si une combinaison lineaire (finie, a n termes) de ces suites est
nulle, on considere le plus petit lambda qui intervient dans la
combinaison et on va suffisamment loin pour que n coefficients successifs de
cette suite ne puissent etre annules par les n-1 autres (deux ne
peuvent etre annules par le meme parce qu'ils sont trop
pres)
Dominique Bernardi
G. A. Edgar
<ull...@math.okstate.edu> wrote:
Yes... V1 (=countable product of copies of R) is naturally a
topological space. (Polish space.) The only CONTINUOUS linear
functionals are the members of V2 (=finite linear combinations of
coordinate projections). Or the only BOREL MEASURABLE
linear functionals. Or the only linear funcionals WITH THE
PROPARTY OF BAIRE. Solovay showed that it is consistent with
ZF (plus dependent choice) that every subset of a Polish space
has the property of Baire (and thus that every linear functional
on a Polish vector space has the property of Baire).
This is one precise version of the assertion that the only
linear functionals on V1 that we can actually "write down" are
the members of V2.
David C. Ullrich
>David C. Ullrich <ull...@math.okstate.edu> wrote:
Certainly. I didn't say it didn't. The question was about
what could be "written down".
> although its cardinality
>is "greater" than that of the space,and I remember that there is a
>formula for the cardinality of the dual space in Jacobson's book .
I see a proof has appeared elsewhere in the thread. But
it's really not hard to figure out the cardinality: Let's say B is a
basis for V; now we may assume that V is the space of all
functions from B to F which vanish except at finitely many
points. It's easy to see that the dual of V is exactly the space
of all functions from V to F (if L is a linear functional on
V, x is an element of V, and x is the sum over B of f_b * b,
where the f_b are scalars, then Lx = the sum of f_b * L(b).
The map taking L to the function (b -> L(b)) is an isomorphism
from the dual of V to the space of all functions from B to F.)
So the cardinality of the dual is exactly |F|^|B|.
>> We all know people who, when they make an error,
>> try to change the question to show they were right after all.
>> I hope nobody takes the following that way - that's not the
>> way it's intended. I was wrong wrong wrong, the following
>> is a _different_ question:
>
>> I wonder whether in some sense the set of all
>> linear functionals on V1 which can be "written down" is
>> V2. That's a little vague - one precise version would be to
>> ask whether it's consistent with ZF (minus AC) that the
>> dual of V1 is V2. (Or that in general if V is the space
>> of all F-valued functions on a set B then the dual is
>> the space of all functions on B vanishing except at
>> finitely many points.)
>
>--
Dave L. Renfro
[sci.math 23 Jul 00 11:08:08 -0400 (EDT)]
<http://forum.swarthmore.edu/epigone/sci.math/dwunshalclo>
wrote (in part)
> There is a short argument given in Example 6.4 on page 46 of
> William C. Brown's book "A Second Course in Linear Algebra",
> John Wiley & Sons, 1988 [QA184 .B765 1987], for the fact that
> the dual space of F^(N) (direct sum) has uncountable cardinality.
> However, I don't believe the issue is as straightforward as Brown
> says. [He writes "A simple counting exercise will convince the
> reader that ...".]
>
> Below I'll prove a result that implies the dual space of
> F^(N) (direct sum) has cardinality at least c (equal to c when
> |F| is at most c) in a way that will roughly allow you to
> exhibit a linearly independent subset of (F^(N))* that has
> cardinality c.
and David C. Ullrich <ull...@math.okstate.edu>
wrote (in part)
[Not yet at the URL given above, but his post can be
found at <http://www.deja.com/group/sci.math> as a
July 24 response to Siu Lok Shun's July 23 post.]
> I see a proof has appeared elsewhere in the thread. But it's
> really not hard to figure out the cardinality: Let's say B is
> a basis for V; now we may assume that V is the space of all
> functions from B to F which vanish except at finitely many
> points. It's easy to see that the dual of V is exactly the
> space of all functions from V to F (if L is a linear functional
> on V, x is an element of V, and x is the sum over B of f_b * b,
> where the f_b are scalars, then Lx = the sum of f_b * L(b).
> The map taking L to the function (b -> L(b)) is an isomorphism
> from the dual of V to the space of all functions from B to F.)
> So the cardinality of the dual is exactly |F|^|B|.
Ooops! Brown claimed that the DIMENSION of the dual space of F^(N)
is uncountable, not just that the dual space of F^(N)--as a set--is
uncountable. Also, what I later proved (and at that later point
stated) is that the DIMENSION of the dual space of F^(N) is at
least c. As Ullrich points out, finding the cardinality of the
dual space of F^(N) is almost immediate. Here's another
proof--essentially the same proof, actually--the proof that I
think Brown had in mind (and which leads me to suspect that in his
example Brown overlooked the distinction between the cardinality of
a space and the dimension of that space) when he wrote the stuff
I quoted from his text:
The set of all functions from a (fixed) basis B of F^(N) into F,
by the definition of cardinal exponentiation, has cardinality
|F|^|B|. Since any such function has an extension to a linear
function from F^(N) into F (and different functions from B into F
obviously give rise to different linear functions from F^(N) into
F), the cardinality of the dual space of F^(N) is at least |F|^|B|.
Moreover, since any linear function from F^(N) into F is determined
by the values it takes on B, the cardinality of the dual space of
F^(N) is at most |F|^|B|. Finally, any field has at least two
elements, and so |F|^|B| is at least 2^|B| = 2^(Aleph_0), which
is uncountable. [Regardless of the field F, |B| = dim[F^(N)]
= Aleph_0.]
Incidentally, if F is the field of real numbers, then F^(N) and
the dual space of F^(N) have the same cardinality. [F^(N) can be
identified with F union FxF union FxFxF union ..., which has
cardinality c + c^2 + c^3 + ... = c, and |F|^|B| has cardinality
c^(Aleph_0) = (2^Aleph_0)^(Aleph_0) = 2^(Aleph_0 * Aleph_0)
= 2^(Aleph_0) = c.] However, the DIMENSIONS of these two spaces
are different.
Dave L. Renfro
Mike Mccarty Sr
)
)>In article <8lanu7$sq...@hkunae.hku.hk>,
)>Siu Lok Shun <ls...@hkusua.hku.hk> wrote:
)>)Who can write down explicitly a basis for the dual space of a infinitely
)>)dimensional space, for example F^\mathbb{N}?
)>
)>Consider the analytic functions over the reals. This forms an infinitely
)>dimensional vector space. It has a basis {1,x,x^2,...}.
)
)It most certainly has *not*, on (a) the reasonable interpretation of
)your ellipsis, and (b) my guess as to what Siu Lok Shun means by "basis",
)namely, "algebraic basis" (no questions of topology, no "infinite
)sums", just pure linear algebra). Specifically, I will address this
)test question to Siu Lok Sun: "is the countable set { x^n : n is a
)non-negative integer} a basis, in your sense, of the uncountable-
)dimensional real vectorspace of real-analytic functions from R to R?"
)Please note that exp(x) is not a linear combination of elements of
)this proposed basis (again, on the definition of "linear combination"
)appropriate in linear algebra).
What do you mean by that phrase "linear combination appropriate in
linear algebra"?
Do you consider Hilbert spaces to be linear spaces? Are they vector
spaces? What is the definition of a basis on Hilbert space? What is the
definition of "linear combination" in Hilbert space? Does it involve an
integral?
I gave an example of an infinite dimensional vector space. I did *not*
attempt to answer his homework problem. I tried to get his mental juices
flowing.
)>What is the dual
)>of this space? Can you write a basis for it?
)
)However, I am happy to recycle these last two questions, in
)a new context. "Consider the real vectorspace R^\infty of all
)sequences {a_0,a_1,...,a_n,...} with only finitely many non-zero
)entries. What is the dual of this space? Can you write a basis
As you well know, finitely many non-zero entries is not necessary.
)for it?" (Notice that this space can be identified with the
)real *polynomial* functions over the reals, a subspace of
)the real-analytic functions over the reals.)
The latter of which is a (countably) infinitely dimensional vector space
with a basis {1} U {x^n: n \n Z+}, as I stated
Mike Mccarty Sr
G. A. Edgar <ed...@math.ohio-state.edu.nospam> wrote:
)<lrud...@panix.com> wrote:
)
)> However, I am happy to recycle these last two questions, in
)> a new context. "Consider the real vectorspace R^\infty of all
)> sequences {a_0,a_1,...,a_n,...} with only finitely many non-zero
)> entries. What is the dual of this space?
)
)The dual is the space of all sequences (with no restriction
)that only finitely many be non-zero).
)
)> Can you write a basis for it?"
)
)No, that requires the Axiom of Choice.
I see. So anything which requires AC is something which you cannot do.
Lee Rudolph
>Lee Rudolph <lrud...@panix.com> wrote:
...
>)>Consider the analytic functions over the reals. This forms an infinitely
>)>dimensional vector space. It has a basis {1,x,x^2,...}.
>)
>)It most certainly has *not*, on (a) the reasonable interpretation of
>)your ellipsis, and (b) my guess as to what Siu Lok Shun means by "basis",
>)namely, "algebraic basis" (no questions of topology, no "infinite
>)sums", just pure linear algebra). Specifically, I will address this
>)test question to Siu Lok Sun: "is the countable set { x^n : n is a
>)non-negative integer} a basis, in your sense, of the uncountable-
>)dimensional real vectorspace of real-analytic functions from R to R?"
>)Please note that exp(x) is not a linear combination of elements of
>)this proposed basis (again, on the definition of "linear combination"
>)appropriate in linear algebra).
>
>What do you mean by that phrase "linear combination appropriate in
>linear algebra"?
In linear algebra, as I understand the subject, no meaning is assigned
to (what are commonly notated, in calculus and other analytic subjects,
as) "infinite sums". In linear algebra, as I understand the subject,
a "linear combination" of a set B of vectors is a sum (necessarily
finite!) of scalar multiples of vectors belonging to B. Anytime
an "infinite linear combination" is being discussed (unless it's
just a finite linear combination in disguise, i.e., all but finitely
many of the scalar coefficients are 0), it's no longer happening
in "linear algebra" proper, it's happening in some context where
limits exist (and so "infinite sums" can be defined).
>Do you consider Hilbert spaces to be linear spaces? Are they vector
>spaces? What is the definition of a basis on Hilbert space?
I consider Hilbert spaces to be topological vectorspaces. The
vectorspace underlying a Hilbert space has bases. The Hilbert
space (that vectorspace, together with the inner product which
defines a *complete* metric space structrure on the underlying
set of the vectorspace) still has those "bases", of course.
It also--as you well know--has subsets which are *not* bases
in the algebraic sense, but which, thanks to the topology,
are similar and even more useful; these are (as I recall)
called "complete sets" (and the best among them are called
"complete orthonormal sets" or the like). Some people no
doubt call them "bases". I don't (except in finite-dimensional
Hilbert spaces, of course), and I don't think it's a very good
idea to do so; but I'm pedantic that way. Certainly
if my first guess was right (as it seems, from later posts,
to have been), and Siu Lok Sun was asking about "bases" in
the purely algebraic sense, then it's *very* misleading to
call those other things "bases".
>What is the
>definition of "linear combination" in Hilbert space? Does it involve an
>integral?
>
>I gave an example of an infinite dimensional vector space.
...And an example of a subset thereof which is not a basis of
that space, until and unless you (1) give the space some extra
structure (like a topology) and (2) redefine "basis" in the
context of that structure.
>I did *not*
>attempt to answer his homework problem. I tried to get his mental juices
>flowing.
Good for you. I tried to keep him from foundering in a swamp of
ill-defined terminology. Evidently one man's juicy brainpan is
another man's swamp.
>)>What is the dual
>)>of this space? Can you write a basis for it?
>)
>)However, I am happy to recycle these last two questions, in
>)a new context. "Consider the real vectorspace R^\infty of all
>)sequences {a_0,a_1,...,a_n,...} with only finitely many non-zero
>)entries. What is the dual of this space? Can you write a basis
>
>As you well know, finitely many non-zero entries is not necessary.
"Not necessary" for what, for goodness sake? The vectorspace
customarily denoted R^\infty (in, for example, Milnor and
Stasheff's book on characteristic classes) is precisely what
I said it was. It is of countably infinite dimension. It's
what I wanted to talk about.
>)for it?" (Notice that this space can be identified with the
>)real *polynomial* functions over the reals, a subspace of
>)the real-analytic functions over the reals.)
>
>The latter of which is a (countably) infinitely dimensional vector space
>with a basis {1} U {x^n: n \n Z+}, as I stated
That is *not* a "basis" in the sense of linear algebra, as I stated;
nor is the dimension of that vectorspace, as defined in linear algebra,
countably infinite. If you want to claim that {1} U {x^n: n \n Z+}
is a "basis" in the sense of topological vectorspaces, you'd better
give the "real-analytic functions over the reals" a topology. There's
plenty of candidates for the topology; maybe the most natural one, to
your taste, makes that vectorspace into a Hilbert space in which
{1} U {x^n: n \n Z+} is a complete orthonormal set, or (if we must
call it that) a "basis", but there are surely others (even other
Hilbert space structures, I'm pretty sure) in which it's not a
complete set.
Lee Rudolph
david_...@my-deja.com
It's an unfortunate fact that the word "basis" is indeed
used to mean several different things. A "basis" for a Banach
space (in particular a Hilbert space) is usually taken to be
a sequence (x_j) such that every element of the space is equal
to the sum of a unique series sum(b_j * x_j). In a context
like that the thing that you would call a basis is often called
a "Hamel basis". (Which is not to disagree with anything you
say here, it all sounds right to me - just commenting on your
"no doubt" - could be replaced by "indeed".)
> Certainly
> if my first guess was right (as it seems, from later posts,
> to have been), and Siu Lok Sun was asking about "bases" in
> the purely algebraic sense, then it's *very* misleading to
> call those other things "bases".
>
> >What is the
> >definition of "linear combination" in Hilbert space? Does it involve
an
> >integral?
> >
> >I gave an example of an infinite dimensional vector space.
>
> ...And an example of a subset thereof which is not a basis of
> that space, until and unless you (1) give the space some extra
> structure (like a topology) and (2) redefine "basis" in the
> context of that structure.
It's a little worse than that: Presumably "real analytic
functions on the line" are supposed to be functions which are
real-analytic on all of R. The monomials do not form a basis
for this space in any sense that I can see - this is because
the Maclaurin series for a function real-analytic on R need
not converge on all of R.
A person could regard the monomials as a basis for the
complex-analytic ("holomorphic") functions in the unit disc
or in the complex plane - there is a natural topology in
those spaces (given by a metric but not by a norm - convergence
in the natural metric is equivalent to uniform convergence
on every compact set), and the power series for a function
_does_ converge to the function in this topology, in these
spaces.
There are plenty of interesting/natural/useful Hilbert
spaces of holomorphic functions (in the disc or in the plane)
where the monomials are an orthonormal basis. The ones that
spring to mind do note contain _all_ the analytic functions
on the domain. I don't think it's possible to give the
space of all analytic functions in the disk a "natural"
Hilbert space structure. ("Natural" would mean rotation-invariant,
implying that the monomials are orthogonal, and would also
mean that the norm should have something to to do with the
honestly natural metric: convergence in norm implies uniform
convergence on compact sets.)
> Lee Rudolph