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!!!!!!!!!!!!

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.
