In the textbook that I am reading, the following is how a polynomial is applied to a linear map.
Definition: For a polynomial $p(z)=a_0+a_1z+\dots+a_nz^n$ and a linear map $T\in L(V, V)$, we define the polynomial $p(T)$ to be the polynomial $p(T)=a_0I+a_1T+\dots+a_nT^n$.
My question is: Can results regarding a polynomial $p(z)$ be interpreted to be the same as for the polynomial $p(T)$? For example, suppose that a polynomial $p(z)=a_0+a_1z+\dots+a_nz^n$ over $\Bbb{C}$ has a factorisation $$p(z)=c(z-\lambda_1)\cdots(z-\lambda_m)$$
Does the polynomial $p(T)$ also have a factorisation that mimics this factorisation? Something like $$p(T)=c(T-\lambda_1I)\cdots(T-\lambda_mI)$$
I should also add that polynomials were defined to be functions $p : F\to F$ where $F$ denotes the field of real or complex numbers.
Best Answer
If the operators $T\in L(V,V)$ are linear, then yes, the factorization is valid. That's because you can just multiply out $c(T-\lambda_1 I)\cdots(T-\lambda_m I)$ and you get the same polynomial $p(T)$ as before based on how $c$ and the $\lambda_i$ combine to form the coefficients $a_k$.
As for the definition: You say that $p:F\rightarrow F$ is how the polynomial is defined. What we are doing when we are writing $p(T)$ is that we are saying "Well, we know what that polynomial is based on $F$. Let us extend it to other kinds of objects by defining $p(T)$ as a polynomial on operators". It is, strictly speaking, not the same function as the original polynomial because it has a different domain. Rather, given a polynomial $p=p_F$, we have defined a different function $p_{L(V,V)}$ by using the same polynomial coefficients as $p_F$ has. We just find it convenient to use the same symbol, even though it really is a different function.