## Halmos Sections 34 and 35:Products and Polynomials (Pt. II)

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

6.

**Problem: **Let be an -dimensional -space and .

**a) **Prove that the set of all for which is a subspace of

**b) **Show that by a suitable choice of the dimension of the subspace described in** a) **can be or .

**c) **Can every subspace of be gotten this way?

**Proof:**

**a) **We must merely note that if both annihilate , then for any we have that . Thus, it follows that the set of all such is a subspace of .

**b)** Let . It’s fairly easy to show that is a subspace of . Thus, we may find a basis for it and extend this to a basis for . We claim that

is a basis for . To see this we note that evidently is linearly independent and , so it remains to show the reverse inclusion. To do this, suppose that but . We note that since that

where there is some such that . Note then that

and thus by assumption that we see that

and thus

but by construction this is only possible if

and since is a linearly independent set this implies that and so in particular which is contradictory to our assumption. Thus, we see that if then so that

which is equivalent to

from where it follows that . Therefore

From this it’s fairly easy to find examples of transformations for which and furthermore it’s clear that the only possible dimensions of are .

**c) **No. Consider . Then, there does not exist a such that has dimension since is not an integral multiple of .

7.

**Problem: **Let where is an -dimensional -space and consider the map

Prove that is a linear transformation. Can every linear transformation on that space be obtained in this manner?

**Proof: **Almost by definition we see that

To prove the second part we let

This is evidently linear since

And since was arbitrary it follows that . Note then that is also injective since implies that

and thus if were surjective we’d have that

but note that and . Thus, if it follows that and so is not surjective, if then the above reasoning leads to a dimension argument that *must *be surjective. Thus, it follows that

if and only if .

**References:**

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

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