Prove that a is an eigenvalue of $p(T)$ if $a=p(\lambda)$ for some eigenvalue $\lambda$ of $T$ Suppose $\mathbb{F}=\mathbb{C}$, $T \in L(V)$, $p \in \mathbb{P}(\mathbb{C})$ and $a \in \mathbb{C}$.
Prove that a is an eigenvalue of $p(T)$ if $a=p(\lambda)$ for some eigenvalue $\lambda$ of $T$
The answer:
Suppose that a is an eigenvalue of $p(T)$. 
Using the Fundamental Theorem of Algebra, we can factor $p(x) − a$
into linear factors over $\mathbb{C}$, i.e.
$p(x) − a = a_n(x − r_1)(x − r_2)...(x − r_n)$
and by the definition of eigenvalue, there is some $v ≠ 0$ such that
$(p(T) − aI)(v) = a_n(T − r_1)(T − r_2)...(T − r_n)(v) = 0$
Let $i$ be the first factor that kills off the image of $v$. Then $(T − r_i)(w) = 0$ for $w ≠ 0$, so $r_i$
is an eigenvalue of T. Recall that $r_i$
is a root of $p(x) − a$, so
$p(r_i) − a = 0$, $p(r_i) = a$.
Thus, $\lambda = r_i$
is the desired eigenvalue.
What I dont get is this:
1) Why are we allowed to just plug in an OPERATOR $T$ into the equation?
2) Also, is the equation with the operator $T$,  $(T-r_1)$ MULTIPLIED by $(T-r_2)$ or COMPOSITION?
3) If it is composition, how can I prove that it is in fact composition?
4) Also, when I find $r_i$, why is it that it is a root of the original equation with z as the variable? I thought that if it comes from an equation under an operator T, then you cant use it in the equation with z?
thanks!!!!!!!!!!!!
 A: 1) This is more or less just the definition of the polynomial of an operator.
If we have a polynomial $p(x) = a_nx^n + \ldots + a_1x + a_0$, we define the polynomial $p(T)$ by $a_nT^n + \ldots + a_1T + a_0*I$, where $T^k$ is composition of $T$ $k$ times. 
2) It is composition, but if you want to you can represent this by matrices and then it turns into matrix multiplication. But I believe that when the answer writes $(T - r_i)$ they really mean $(T - r_i*I)$, as an operator minus a scalar otherwise would be undefined.
3) I'm not sure what you mean by "prove that it is in fact composition". It is by definition composing these operators.
4) If they had written $T-r_iI$ it is easier to see that $r_i$ is just a complex number, so you can use it in your polynomial without any problems.
A: Given a vector space $V$ (over any field, but let's stick to the case of the field $\mathbb{C}$), we can give a ring structure in $L(V)$ by setting, for every $T,S\in L(V)$ and $x\in V$,
$$(T+S)(x)=T(x)+S(x)\quad and\quad (TS)(x)=T(S(x)),$$
so adition of linear function is given by pointwise adition and product is simply composition.
So this answers question 2): By multiplication we understand composition, so they are the same. I don't quite understand questiokn 3).
Now, given a polynomial $p(z)=\sum_{i=0}^n a_i z^i$, with $a_i\in\mathbb{C}$, we can define $p(T)=\sum_{i=0}^n a_iT^i$, where by $T^0$ we understand the identidy in $V$, which I'll denote $I_V$. This defines a ring-homomorphism $\mathbb{P}(\mathbb{C})\rightarrow L(V)$, $p\mapsto p(T)$. Also, it is usual, when working in $L(V)$, to write simply $\lambda$ instead of $\lambda I_V$, for $\lambda\in\mathbb{C}$. This answers question 1).
About question 4), the $r_i$ are found by the fundamental theorem of algebra, so they depend onlky on the polynomial $p$.
