Halmos Chapter one Sections 15, 16 and 17: Dual Bases, Reflexivity, and Annihilators (Part II)
Point of post: In this post I complete the second half of the problems in the 15th, 16th, and 17th sections of Halmos’s book. There are some pretty neat theorems in here, in particular:
– is always embeddable into ,
-For any vector space with basis it’s true that
– That if is infinite dimensional that and thus the canonical identification between and is an isomorphism if and only if
Problem: This exercise is concerned with vector spaces that need not be finite dimensional.
a) Suppose that is a field and a set, then if and write for the function for all . The set of all such functions is a vector space with respect to this definition of linear operations, and the same is true of the set of all finitely non-zero functions (A function is finitely non-zero if is finite)
b) Every vector space is isomorphic to the set of all finitely non-zero functions on some set.
c) If is a vector space with basis , and if then there exists a unique such that for all
d) Use a), b), and c) to conclude that every vector space is isomorphic to a subspace of .
e) Which vector spaces are isomorphic to their own duals?
f) If is a l.i. subset of a vector space , then there exists a basis of containing
g) If is a set and if , write by
Let be the set of all functions together with the function . Prove that if is infinite, then is l.i. subset of the vector space of all scalar valued functions on
h) The canonical identification between and is defined for all vector spaces. Prove that is reflexive iff
a) Clearly the result is true for so we must only prove it’s true for finitely non-zero functions. But, if are finitely non-zero then and and so if then clearly and cannot be simultaneously zero. Thus, and so ‘
and is thus finitely non-zero. Also, notice that so that and so is finitely non-zero. It follows that the set of all finitely non-zero functions is a subspace of .
b) Let be a vector space with basis and let
We claim that . To see this we first find a basis for . To do this consider the set
we claim that is a basis for . To see this, first suppose that
Then, in particular we see that
from where it follows that any finite subset of is l.i. and so by definition so is . Furthermore, let . If the fact that is trivial, so assume not. Then for some . So then, we claim that
To see this, let be arbitrary. If then
since and . Now, if for some then so that
from where the conclusion follows. Thus, is a basis for . So, then define
we claim that is an isomorphism. To show it’s linear we note that if then and for some so then
from where ‘s linearity follows. Thus, it remains to show that is bijective. To see this, suppose that and may be written as the linear sum of elements of with the same notation as above. Suppose then that we see that if that and and since we (implicitly) assumed that this is a contradiction. Conversely, if then and which is a also a contradiction. It follows that . We may assume though, (with the possibility of relabeling) that (we may assume that since the sets are equal, and so in particular have the same cardinality). Thus,
from where the fact that follows from the “uniqueness of representation by basis” theorem. From there the fact that clearly follows, and so is injective.
For surjectivity, we merely notice that for some . It clearly follows then that . Thus, is an isomorphism and so as was to be proven.
c) So, let be the function described above and define for all . We may then extend to a linear functional by the following observation. Let us index by . We then know that for each it’s true that
where it’s understood that for all but finitely many . So, then define by
clearly then, for each we have that , and thus we have found one linear functional. To prove that it’s unique we merely note that if were such that for all then
for all , thus proving uniqueness.
d) Let have the basis and let be defined as in b) Then, define
where is as in c) (we see by c) that is well-defined). Noticing then that
we may conclude that is linear. To see that is injective suppose that are distinct. Then, there exists some such that , then so that . It follows that is isomorphic with it’s image , and thus
e) It was proven in the book that if then , and we will show that this is precisely when this isomorphism is true. To do this, we first show that regardless of the cardinality of the basis for that . To do this define
where is the function described in c). Using the exact same argument as in d) one may show that is both injective and linear, and so to prove that it’s a linear isomorphism it suffices to show that is surjective. So, we note as was proven many times before, that if that is the unique such that . So, in particular and as can easily be seen that . Thus, surjectivity, and consequently ‘s status as an isomorphism follow.
So, considering the above paragraph’s result that it suffices to show that if is not finite. To do this, we use a cardinality argument. We first note that if is the prime subfield of that where . This is clear though since is a basis for both. We next claim that . To see this it suffices to prove that any l.i. subset of over is l.i. as a subset of over . To do this suppose that is l.i. and . Then, as is a subfield of we may consider the vector space over , and by an earlier theorem we must have that this admits a basis . Thus, for each we may express it in the form . Thus, by earlier assumption
and since the inner sum is an element of and is l.i. over we see that the inner sum is zero, and since this was true for all we may deduce that and so is l.i. over .
Lastly, we claim that , but we can do this equally well by showing that . So, let . We clearly have that
where is the set of all finite subsets of , since we may associate with function it’s values on . Next, we appeal to a common cardinal arithmetic fact that . Thus,
where we’ve used the fact that .
Combining these inequalities gives
and so in particular
and so as desired.
Remark: The operative part of the proof was transferring from to the prime subfield . For, imagine if we’d not done that and we’d just considered . Then, if we’d get into a whole bunch of problems (whether trying to prove that the dimensionality or actual cardinality of the first space is less than the second) like trying to assert that which clearly isn’t true in general (take and
f) This was already done here
g) This was already done in general in part b)
h) You can use the other exercises as hints and literally construct an element of which isn’t in the image of the canonical identification. Or, more quickly we can notice that by the proof in e) we have that so that and since the canonical identification is a linear injection, it can’t be a surjection.
1. Halmos, Paul R. “Dual Bases, Reflexivity, Annihilators.” Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. 23-28. Print.
2.Jech, Thomas J. Set Theory. Berlin: Springer, 2003. Print.