403/503 – Homework 6

Due Friday, April 23, unless you convince me to move it to a later date. From the textbook, Chapter 8, exercises 3, 14, 15, 18, 20, 22, 25, 26, 27, 28.

Advertisements

4 Responses to 403/503 – Homework 6

  1. Amy Griffin says:

    Dr Caicedo,

    On problem 20, where we need to show that a polynomial exists, p(T)=T^inverse for an invertible T. I’m having issues with seeing the path I need to take to solve this. I believe that I want to find:

    T(p(T))=I. So far, we have done a lot of work to find polynomials that make p(T) = 0, but I’m not sure how to get the identity. Also, I know that T has a square root and there are no eigenvalues equal to zero.

    Do you have any hints on this one?

    • Hi Amy,

      Let’s see… You have a polynomial p such that p(T)=0, say, the characteristic polynomial. Actually, and this will be useful to us, the constant term of the characteristic polynomial is not zero, right? [Show this using what you pointed out, that since T is invertible, then none of its eigenvalues is zero.]

      Let’s see an example. Suppose that the char. poly. is p(x)=x^5+3x^2+x-7.

      Then p(T)=T^5+3T^2+T-7I=0.

      Then T^5+3T^2+T = 7I.

      Then T(T^4+3T+I)=7I.

      Dividing by 7, gives us that q(x)=(x^4+3x+1)/7 has the property that q(T)=T^{-1}.

      I think the same idea should work in general. (And you see why it is important that the constant coefficient is not zero.)

    • By the way, the same idea is used in other situations. For example, in algebra, to show that if \alpha is an algebraic number, then {\mathbb Q}(\alpha)={\mathbb Q}[\alpha].

      That \alpha is algebraic means that there is a polynomial with rational coefficients that vanishes at \alpha.

      {\mathbb Q}(\alpha) is the smallest subfield of {\mathbb C} that has \alpha as an element.

      {\mathbb Q}[\alpha]=\{p(\alpha)\mid p is a polynomial with coefficients in {\mathbb Q}\}.

      For example, \sqrt2 is algebraic because it is a root of x^2-2. Then 1/\sqrt2=(\sqrt2)^{-1}=p(\sqrt2), where p(x)=x/2. Of course, p comes from the fact that if x^2-2=0, then x^2=2, or \displaystyle x\left(\frac x2\right)=1.

      (In general, if the minimal polynomial of \alpha has degree n, then every element in {\mathbb Q}(\alpha) has the form p(\alpha) where p is a polynomial with rational coefficients, and degree at most n-1.)

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: