## 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.

### 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?

• andrescaicedo says:

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.)

• andrescaicedo says:

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 $n-1.$)

• Andrew Misseldine says:

Thanks for pointing out this last observation. I didn’t see the connection between #20 and field extensions.