## Halmos Section 36: Inverses (Pt. II)

**Point of post: **This is a continuation of this post.

3.

**Problem: **Let be any (not necessarily finite dimensional) -space and . Then, if and only if

**Proof: **We first prove a set-theoretic lemma:

**Lemma: ***Let then if is injective then is injective. Similarly, if is surjective then is surjective.*

**Proof: **Suppose first that is injective but isn’t. Then, there exists such that then evidently which contradicts that is injective.

Similarly, suppose that is surjective but isn’t. We then see that which contradicts that is surjective.

The conclusion follows.

**Corollary: ***If and and are bijective then and are bijective.*

**Proof: **Since is injective we know that is injective and since is surjective we know that is surjective. Thus, is bijective.

Similarly, since is injective we know that is injective and since is surjective we know that is surjective. Thus, is bijective.

The conclusion follows.

From this it readily follows that implies that . The other direction was covered in our discussion of invertible transform.

4.

**Problem: **If is a finite dimensional -space and prove that if then .

**Proof: **You’ll recall from our post on invertible transformations that implies that has a right inverse and a left. Thus, is an epimorphism (since it’s surjective) and is a monomorphism (since it’s injective). But, from the same post we know that this implies, by the finite dimensionality of , that .

5.

**Problem: **If is an -space and and then there exists such that

and

**Proof: **Assuming that we see note that if the above two equations hold then upon addition we obtain

so that

and upon subtraction we obtain

so

solving this for gives

and solving for gives

from where the conclusion follows.

6.

**Problem:**

**a) **Prove that if is an -dimensional -space and , then if and only if is linearly independent implies is linearly independent.

**b) **Is this assumption of finite dimensionality needed here?

**Proof:**

**a) **This was proven, in some detail (recalling that an endomorphism on a finite dimensional space is an isomorphism if and only if it’s a monomorphism) here

**b) **Yes. Consider the unique endomorphism

such that for all we claim that this mapping is linear independence preserving. To see this suppose

is linearly independent. Suppose then that there existed scalars such that

this implies by definition that

but since is an integral domain this implies that

and thus from where it follows that is linearly independent. Thus, since was arbitrary it follows that is linear independence preserving. But, note that and so isn’t an epimorphism and thus not invertible.

7.

**Problem: **Show that if is a finite-dimensional -space and is such that

prove that .

**Proof: **Evidently is injective, since if then

so that . But, by a previous theorem we know that .

**References:**

1. Halmos, Paul R. *Finite-dimensional Vector Spaces,*. New York: Springer-Verlag, 1974. Print

[…] Point of post: This is a continuation of this post. […]

Pingback by Halmos Section 36: Inverses (Pt. III) « Abstract Nonsense | December 4, 2010 |