## Halmos Section 21 and Section 22:Quotient Spaces and Dimension of a Quotient Space

**Point of post:** In this post I will do the problems from sections 21 and 22 in Halmos. There are some quite interesting ones, with some definite algebraic slants.

1.

**Problem:** Consider the quotient spaces obtained by reducing the space modulo various subspaces. If , the set of all th degree polynomials, is finite-dimensional? What if is the set of all even polynomials. What if consists of all polynomials divisible by ?

**Proof:** When then is not finite dimensional. To see this fix and consider suppose then that

but, by definition this implies that

But, this is the same as saying that

which is clearly impossible unless . Thus, are linearly independent vectors in and so it can’t be -dimensional. Since was arbitrary it follows that for any and is thus not finite dimensional.

Now, if is the set of all even polynomials, then since where is the set of all odd polynomials we know that and since isn’t finite dimensional (consider ) we may conclude that is not finite.

Lastly, we claim that if is the set of all polynomials divisible by that is a basis for . To see this first suppose that

then, by definition

which is evidently impossible unless . Next, let be arbitrary. We know that for some . If the result’s clear, and if not we see that

from where it follows that . Thus,

2.

**Problem: **IF and are arbitrary subsets of a vector space (not necessarily cosets of a subspace), there is nothing to stop us from defining just as addition was defined for cosets, and similarly, we may define . IF the class of all subsets of a vector space is endowed with these “linear operations” which of the axioms of a vector space are satisfies?

**Proof: **The following uses the definition of a field as is given on page three of Halmos’s book:

**A)**

1. This axiom (commutativity of the addition) is definitely true since

2. This is also true since

3. This is true since

4. This is not true. Let be a vector space over itself and consider the subset . Note then that since we must have that for all . So, in particular if then it must be true that , but this contradicts that .

**B) **

1. This is true since

2. This is also true since

3. This is true since

4.This is true since

**Problem: **

**a) **Suppose that is a subspace of a vector space . Two vectors and of are *congruent *modulo , denoted , if . Prove that is an equivalence relation on

**b) **If and if are such that and prove that .

**c) ** Prove that the equivalence classes of are precisely the cosets of .

**Proof:**

**a) **Clearly since . Also, if then and thus and so . Lastly, if and then and and thus so that

**b) **Clearly this is true since and implies that so that .

**c) **Let , we claim that . To see this let then so that for some . Thus, and thus . Conversely, if then for some and so so that and thus .

3.

**Problem:**

**a) **Suppose that is a subspace of a vector space . Prove that without appealing to dimensionality.

**b) **Show that without appealing to dimensionality.

**Proof:**

**a) **Let be a basis for and extend it to a basis . Then, we know that lifts to the dual basis for , with, in particular, be a basis for .

But, with equal verity we may start with and produce the basis for . And from there we may lift to the basis for .

From here the course of action is clear, namely let

and extend it to the map

by linearity. Clearly, by construction, we see that is linear. To see it’s injective suppose that

then, since is linearly independent we can see that . Thus, . Also, it’s clear that is surjective since if is in we have that and

**b)** Consider the mapping

Now, linearity is clear since

and

Now, to prove injectivity suppose that . Then, there exists some such that . Otherwise, for all we’d have that and so , but this (by looking at problem 2.) says that which we assumed otherwise. Thus, there exists the aforementioned and so

Lastly, to prove surjectivity let be such that . Then, we may extend any to a $\varphi^*:\mathscr{V}\to F$ by defining where is the unique representation of as the sum of an element of and . Clearly then and . Thus, is an isomorphism.

I would like to thank you for posting these nice proofs.

Do you have also a proof for problem 2 section 22? The number 2 from this post is actually problem 3 from the same section 22.

Thank you in advance! Any help will be appreciated.

Comment by Kira | November 19, 2010 |

There you go. This is the second time I’ve accidentally skipped writing up a solution. If you find any more let me know!

Comment by drexel28 | November 19, 2010 |

Please help me with problem 2 from section 22. Thank you very much!

Comment by Kira | November 19, 2010 |

1.I would like to thank you for posting these nice proofs.

Do you have also a proof for problem 2 section 22? The number 2 from this post is actually problem 3 from the same section 22.

Thank you in advance! Any help will be appreciated.

Comment by Kira | November 19, 2010 |

I’m not sure what you mean. I just did the problem I missed, above.

Comment by drexel28 | November 21, 2010 |

Thank you! I really appreciate your great work!

Comment by Kira | November 21, 2010

You’re quite welcome!

Comment by drexel28 | November 22, 2010