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.
A kind of library
Andrés E. Caicedo
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A kind of library 
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 ![{\mathbb Q}(\alpha)={\mathbb Q}[\alpha].](https://s0.wp.com/latex.php?latex=%7B%5Cmathbb+Q%7D%28%5Calpha%29%3D%7B%5Cmathbb+Q%7D%5B%5Calpha%5D.&bg=ffffff&fg=333333&s=0&c=20201002)
That
is algebraic means that there is a polynomial with rational coefficients that vanishes at 
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.