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.
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.
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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
Then
Then
Then
Dividing by 7, gives us that has the property that
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 is an algebraic number, then
That is algebraic means that there is a polynomial with rational coefficients that vanishes at
is the smallest subfield of that has as an element.
is a polynomial with coefficients in
For example, is algebraic because it is a root of Then where Of course, comes from the fact that if then or
(In general, if the minimal polynomial of has degree then every element in has the form where is a polynomial with rational coefficients, and degree at most )
Thanks for pointing out this last observation. I didn’t see the connection between #20 and field extensions.