4 - Polynomials on operators

let $F$ be a field. A function $p:F\to F$ is a polynomial (of degree $n$) if there exists $a_i\in F$, ($a_{n-1}\ne 0$) for each $i=0$ to $n-1$ such that $p(z)=\sum _{i=0}^{n-1}{a}_i{z}^i$ for each $z\in F$. The set of all polynomials of $F$ is written $\mathcal{P}(F)$. This set is a vector space, where addition of two polynomial functions $p_1+p_2$ is defined as the function that maps $z\mapsto p_1(z)+p_2(z)$ for each $z\in F$. Similarly, the product of two polynomials $p_1\cdot p_2$ is defined as the map $z\mapsto p_1(z)\cdot p_2(z)$. Finally, the scalar multiplication of a polynomial $\lambda p$, $\lambda \in F$ is defined as the map $z\mapsto \lambda p(z)$.
It is left to the reader to verify that $\mathcal{P}(F)$ with the addition and scalar multiplication defined above, is a vector space.

Let $V$ be a vector space over $F$ and $f$ an operator on $V$. Then $f^n$ is defined as $f$ composed on itself $n$ times. It's pretty clear the the usual power laws hold good. And $f^0$ is the identity operator.

Now, if we want to apply the polynomial $p\in \mathcal{P}(F)$ to the operator $f$, if $p(z)=\sum _{i=0}^{n-1}{a}_i{z}^i$, then we define $p(f):=\sum _{i=0}^{n-1}{a}_i{f}^i$.
This is well defined, because the set of operators on $V$, usually denoted $\mathcal{L}(V)$ is itself a vector space, over $F$, where (as usual) addition of linear maps is defined by pointwise addition, and scalar multiplication as pointwise scalar multiplication. We can also define multiplication on $\mathcal{L}(V)$ as function composition.
We observe that $pq(f)=p(f)q(f)=q(f)p(f)$ (informally, replacing the symbol $z$ with another symbol $f$) doesn't change anything.
It is clear that for any field polynomial $p$, the operator polynomial $p(f)$ is also an operator on $V$, that is $p(f)\in \mathcal{L}(V)$ for any $p\in \mathcal{P}(F)$ and any $f\in \mathcal{L}(V)$. Notice that $p(f)({\mathbf{u}}_{\mathbf{1}}+{\mathbf{u}}_{\mathbf{2}})=\sum _{i=1}^{n-1}{a}_i{f}^i({\mathbf{u}}_{\mathbf{1}}\mathbf{+}{\mathbf{u}}_{\mathbf{2}})=\sum _{i=1}^n{a}_i(f^i({\mathbf{u}}_{\mathbf{1}}))+\sum _{i=1}^n{a}_i(f^i({\mathbf{u}}_{\mathbf{2}}))=p(f)({\mathbf{u}}_{\mathbf{1}})+p(f)({\mathbf{u}}_{\mathbf{2}}))$
(we are using the linearity of $f^i$)
Similarly we can show that $p(f)(\lambda \mathbf{v})=\lambda p(f)(\mathbf{v})$.
Notice that using the identity polynomial $\mathbb{1}(z)=z$, we can write any operator $f$ as the polynomial operator $\mathbb{1}(f)$.

let $f$ be an operator on a vector space $V$, and $p(f)$ an operator polynomial. Then $\text{Kern}(p(f))$ and $p(f)(V)$ are both invariant under $f$.