## Halmos Sections 34 and 35:Products and Polynomials

**Point of post: **In this post we complete the problems at the end of sections 34 and 35 of Halmos’s book.

1.

**Problem: **Let

and

compute and for an arbitrary .

**Proof: **Note that

and

and so in particular in general

where the sum is the zero polynomial if and

2**.**

**Problem: **If and are linear transformations such that commutes with then

for every .

**Proof:**We first note that by induction

which, upon expansion gives

and for our particular purposes

Thus, with this in hand we may proceed by induction. Note that holds trivially for and if we assume that it holds for we see that

from where the conclusion follows.

3.

**Problem: **Suppose that is given by . Prove that

**Proof: **Recall that has basis and since a linear transformation is determined entirely on a basis it suffices to check that this equality is true on this particular basis. But,

where we’ve evidently used the Binomial Theorem in the last step.

From previous comment the problem follows.

4.

**Problem:**

**a) **If is an endomorphism on an -dimensional -space , then there exists a non-zero polynomial of degree less than or equal to which annihilates .

**Proof:**** **This is a fairly crude estimate (since for all of us know that there is a -degree polynomial which annihilates , the characteristic polynomial). Regardless, we merely note that if there did not exist a non-zero polynomial with which annihilates then for any we’d have that

which implies that has linearly independent vectors, which is clearly impossible.

5.

**Problem: **The product of linear transformations is defined only if they “match” in the following sense. Suppose that , , and are all -spaces and suppose that and . The product is defined to be the linear transformation from to given by . Interpret and prove as many of the five axioms for an associative unital algebra as you can.

**Proof: **The first axiom can be thought of as follows. If , , and , then it’s true that since function composition is associative. If and then we can say that since . Similarly, the third axiom holds. The fourth one holds trivially. The fifth one doesn’t hold well at all. Namely, there is a left and right identity elements, namely and respectively but there is no ‘single’ identity element.

**Reference:**

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

Can you post part b) of ex. 4? Thank you!

Comment by Kira | November 29, 2010 |

Jeez, you’re quick.I’ll post it ASAP. Respond to this so I remember.

Comment by drexel28 | November 29, 2010 |

Please let me know part b) of ex. 4. Thank you!

Comment by Kira | November 29, 2010