I'm trying to understand the determinant from Axler Sheldon's paper and there is one point in the very beginning that I don't understand :S (Link below to the paper)
http://www.cs.berkeley.edu/~wkahan/MathH110/DownDets.pdf
I have attached a picture of the point I don't understand:
I have higlighted the area I don't understand with a red rectangle. Could someone clarify why can we do the factorization and the last remark (about the injectivity) made by the author?
Thank you for any help :)
On
Factoring the characteristic polynomial of a complex operator is just as doable as factoring a (more general) complex polynomial.
We know $$p(T)=0$$ Hence, $$ \forall a \in \mathbb F : p(T)(av)=0$$ This means one of the factored terms (T-bI) of p(T) maps v to 0. (Technically, the constant in the factored form could also equal 0.) This factor of p(T) also maps v=0 to 0. Thus, this factor of p(T) is not injective.
If we can factor a polynomial $p(z)$, we can also factor when $z$ is replaced by a matrix argument $T$. Proving this fact is a simple but slightly messy induction argument (we induct on the degree of the polynomial).
The reason that at least one $T-r_jI$ is not injective is that the product of injective linear operators is always injective. Thus if every $T-r_jI$ were injective, $c(T-r_1I)\cdots(T-r_mI)$ would also be injective - but this contradicts the construction of $c(T-r_1I)\cdots(T-r_mI)$, since we have $c(T-r_1I)\cdots(T-r_mI)v=0$ for $v\neq 0$.