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.
compute and for an arbitrary .
Proof: Note that
and so in particular in general
where the sum is the zero polynomial if and
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.
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.
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.
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.
1. Halmos, Paul R. Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print