## Halmos Chapter one Sections 15, 16 and 17: Dual Bases, Reflexivity, and Annihilators (Part I)

**Note: **For a vector space over a field I will freely switch between the notations and for the dual space of , depending which fits better.

1.

**Problem:** Define a non-zero functional such that if and that .

**Proof:** We note that and where (where is the Kronecker delta symbol, i.e. this is the standard basis for over ). Then, it suffices to define

since there is precisely one linear functional taking on the above values. So if is that functional we see that

and

as desired.

2.

**Problem:** The vectors , , and form a basis for over . If is the associated dual basis and if find , , and .

**Proof:** Recall that for a basis we define the dual basis to be the unique functionals in the dual space such that . From this we can easily see that if that . Thus, noticing that

that

3.

**Problem: **Prove that if where is an -dimensional vector space over , then the set of all vectors for which is a subspace of ; what is the dimension of that subspace?

**Proof:** For the sake of notation convenience, define . Then, if and we see that

and thus from where the fact that is a subspace follows.

Now, for the second part we claim something stronger. Namely, either or for any we have that

where the symbol merely means that

So, to prove this we first prove . This follows since

But, so that the above implies and thus . To prove we merely note that for any

but evidently is of the form and a quick computation shows that . From where our proposition follows. But, from a previous problem in a past post we know that if are any two subsets of that

But,

and

and

And so, putting this into equation one gives

Thus, to wrap this all up we remember that all of the above was based on the fact that , but if then . So, we may finally conclude that

4.

**Problem:** If whenever , then is a linear functional on ; find a basis for the subspace for

**Proof:** Since and that (this wasn’t necessary to know beforehand, but is useful to make sure we know how many things we’re looking for). So,

So the general form of an element of is

So, if we let then we get

and if we let we get

We claim that is a basis for . Clearly is a l.i. set and thus it remains to prove that . Clearly, and so we must only prove the reverse inclusion. So, let , then as said before we must have that

and so as required.

5.

**Problem:** Prove that if and if where is an -dimensional vector space over ; then there exists a non-zero vector in such that . What does this say about solutions of linear equations?

**Proof:** Note that is a subspace of of at most dimension. It follows that

where is the annihilator. Namely, since we have that is non-trivial and so there is some such that . But, as was proven in an earlier post we know that the map

is an isomorphism. Namely, there exists some (we may assume that since and is injective) such that . Thus,

and so

as desired.

We note that in particular that if

is interpreted as trying to find a common zero for the functionals associated linear functionals, the above proof shows there exists a non-zero solution.

6.

**Problem: **Suppose that $latx m<n$ and that where . Under what conditions on the scalars is it true that there exists a vector such that for . What does this say about the solutions to linear equations?

7.

**Problem:**If is an -dimensional vector space over the finite field and if then the number of -dimensional subspaces of is the same as the number of dimensional subspaces.

**Proof:** I want to prove something much stronger, namely, I’ll prove how many subspaces there are of a given dimension. But first, a technical lemma:

**Lemma:** Let be an -dimensional vector space over the finite field . Then, the number of distinct bases for is:

**Proof: **We can construct every basis for in the following fashion. We fix to be the first element 0f 0ur basis. We know though that , and so , so there are precisely choices for our initial vector . So, then choose . But, since we have that and so and thus . Thus, having chosen in our basis we may choose , and using the exact same argument as before we see there are precisely choices. Thus, since the process terminates at the choice of the th vector in our basis we see that the total number of distinct bases is the product of all the choices up to , from where the conclusion follows.

Now, let and be as before. We will prove that for each the number of subspaces of dimension is

To do this we note that any -dimensional subspace of is spanned by l.i. vectors. It’s clear using the previous argument that the numerator of is the number of ways of picking such vectors from an -dimensional space. Also, two sets of independent vectors span the same space iff they both form a basis for the resulting space. Thus, to counteract redundant counting we must divide by the number of bases for a -dimensional space. But, this is the denominator of as the lemma’s technique clearly shows. It follows that the number of -dimensional subspaces of is indeed .

Then, a simple (and I use this word lightly) calculation shows that . Think about writing the product out up to a certain point, cancelling and regrouping.

8.

**Problem:**

**a) **Prove that if is any subset of a finite-dimensional vector space, then coincides with the subspace spanned by

**b)** If and are subsets of a vectors space then

**c)** Prove that if are subspaces of a finite-dimensional vector space , then

and

**d) **Is the conclusion of **c) **true if is not finite dimensional?

**Proof:**

We prove **b) **first so that we may do **a) **in a different sort of way.

**b) **Let then and so . Thus,

**a) **We first remember that since the canonical identification

is an isomorphim, that one currently thinks of and as being “the same”. Thus, the question is really asking us to prove that

To see this we’ll prove that

is the smallest subspace of containing from where the equation will follow (since is the unique such subspace). Thus, to see that

is a subspace we must merely note that is a subspace (since the annihilator of any set is a subspace of ) and similarly is a subspace of . But, is a linear isomorphism, and thus so is and thus it clearly follows (since being a subspace is an invariant property under linear isomorphisms) that

is a subspace of . Furthermore, let , then for any we have that and thus for any we have that , but this says that

and so

Lastly, let be any subspace of such that

Then, by **b)**

and so by **b) **again

and so

Thus, is the smallest subspace of containing , namelyu

**c) **The fact that is easy.

but this clearly implies that

Conversely, if then for every we have that

But of course, this is equivalent to saying that . Thus,

To prove the second part we note trivially that . To see this, let . Then, we note that if that and so and thus . Conversely, let . Decompose the ambient space into the direct sum

and define to be the unique linear functional on or which , , and . Next, define to be the unique linear functional such that and . Evidently , , and from where it follows that and so the problem follows.

**c)** Yes. The conclusion is still true. We have not used any facts which relate to finite-dimensionality.

Hello! Can you post problem 6 because you skipped it? Thank you very much. Problem 6 from this post is actually 7, problem 7 is 8. Thanks.

Comment by Raluca E. Toscano | November 22, 2010 |

Truth be told, I’m not entirely sure of the answer right now. I’ll have to think about it. Be sure to let me know if anything strikes you!

On an upside, I added proofs for those problems I deferred to a “later post” but in all reality forgot about.

Comment by drexel28 | November 22, 2010 |

Can you actually post problem 6? Thank you.

Comment by Kira | November 22, 2010 |