Understanding the basic theorem of cyclic subspace I have stucked on understanding the following theorem. It is from the text, Linear algebra(Hoffman) p.228.


Theorem Let $\alpha$ be any non-zero vector in $V$ and let $p_\alpha$ be the $T$-annihilator of $\alpha$.
(i) The degree of $p_\alpha$ is equal to the dimension of the cyclic subspace $Z(\alpha ; T)$.
(ii) If the degree of $p_\alpha$ is $k$, then the vectors $\alpha, T\alpha, T^2 \alpha, \ldots, T^{k-1} \alpha$ form a basis for $Z(\alpha ; T)$.
(iii) If $U$ is the linear operator on $Z(\alpha ; T)$ induced by $T$, then the minimal polynomial for $U$ is $p_\alpha$.


The following explains the above notation.


Definition If $\alpha$ is any vector in $V$, the $T$-cyclic subspace generated by $\alpha$ is the subspace $Z(\alpha ; T)$ of all vectors of the form $g(T)\alpha$, $g$ in $F[x]$. If $Z(\alpha ; T) = V$, then $\alpha$ is called a cyclic vector for $T$.
Definition If $\alpha$ is any vector in $V$, the $T$-annihilator of $\alpha$ is the ideal $M(\alpha ; T)$ in $F[x]$ consisting of all polynomials $g$ over $F$ such that $g(T)\alpha = 0$. The unique monic polynomial $p_{\alpha}$ which generates this ideal will also be called the $T$-annihilator of $\alpha$.


What I don't understand is the theorem (iii). What is 'linear opertor on $Z(\alpha ; T)$ induced by $T$'? This expression is absurd for me to understand. Here I wrote the proof of (iii).


Proof of (iii) Let $U$ be the linear operator on $Z(\alpha ; T)$ obtained by restricting $T$ to that subspace. If $g$ is any polynomial over $F$, then
    \begin{align}
p_{\alpha}(U)g(T)\alpha &= p_{\alpha}(T)g(T)\alpha \\
&= g(T)p_{\alpha}(T)\alpha \\
&= g(T)0 \\
&= 0.
\end{align}
    Thus the operator $p_{\alpha}(U)$ sends every vector in $Z(\alpha ; T)$ into 0 and is the zero operator on $Z(\alpha ; T)$. bla bla...


I don't know why $p_{\alpha}(U)$ changes into $p_{\alpha}(T)$. Since $U$ is linear operator on $Z(\alpha ; T)$, the notation $p_{\alpha}(U)g(T)\alpha$ actually means $p_{\alpha}(U)(g(T)\alpha)$, doesn't it?(I mean that it is composition of the functions)
If we select $p_{\alpha} = x^2 + x + 1$ satisfying the above condition,
$$p_{\alpha}(U)g(T)\alpha = (U^2 + U +1)(g(T)\alpha)$$
There is no chance for $p_{\alpha}(U)$ to be $p_{\alpha}(T)$.
What does $U$ mean here the theorem?
 A: Hard to understand the details of the notations of the book, not having it. But as this subject is classic, I can say this: 
If $T$ is an endomorphism of some vector space $V$ over a field $F$. We can turn $V$ into an $F[x]$-module with the following definition of scalar multiplication: for any $\alpha\in V$,
$$ x\cdot \alpha= T\cdot\alpha $$
whence, for any polynomial $g\in F[x]$,
$$g(x)\cdot \alpha \stackrel{\text{def}}{=}g(T)\cdot \alpha.$$
Now $Z(\alpha;T)$ is the submodule of $V$ generated by $\alpha$:
$$\bigl\{ g(x)\cdot\alpha \mid g(x)\in F[x] \bigr\}$$
It is easy to check that $\;T\bigl(Z(\alpha;T)\bigr)\subseteq Z(\alpha;T)$, since it means that $\;x\cdot\bigl(g(x)\cdot\alpha\bigr)=h(x)\cdot\alpha$ for $\;h(x)=xg(x)$!
Thus the restriction of the endomorphism $T$ of $V$ to $Z(\alpha;T)$ is an endomorphism of the latter, and when applied to  $Z(\alpha;T)$, we may write $g(T)$ as well as $g(U)$.
A: Saying an expression is absurd for you to understand seems like an exaggerated way of saying simply that you don't understand the expression. What is meant is simply the linear operator on the subspace $Z(\alpha;T)$ obtained by restricting the given linear operator $T$ from the whole space to just $Z(\alpha;T)$.
Since one wants a linear operator (mapping some space to itself rather than to some other vector space), this kind of restriction to a subspace $W$ only makes sense if $W$ is $T$-stable, that is $T(W)\subseteq W$. Then one can define $U:w\mapsto T(w)$ for every $w\in W$; this means that $U$ is basically the same operation as $T$, but its domain and codomain have been clipped down to the subspace $W$. The reason to call this the linear operator of $W$ induced by $T$ rather than just a restriction of $T$ is that in the ordinary sense a restriction of a map just changes the domain, not the codomain (arrival set), so the restriction of $T$ to $W$ would be a map $W\to V$, which is not what is meant here. (But in the proof they use the word "restriction" anyway.)
In the proof they replace the linear operator $p_\alpha(U)$ on $W=Z(\alpha;T)$ by the linear operator $p_\alpha(T)$ on$~V$; this is justified because in the formula that operator is applied to a single vector $g(T)\alpha\in W$, and this vector implicitly changes from being an element of $W$ to being an element of the larger space$~V$ when $p_\alpha(T)$ is applied to it. (It is quite common to apply to values an operation that is defined on a larger domain than the set that value is known to live in.) The effect of applying $p_\alpha(U)$ or $p_\alpha(T)$ to some $w\in W$ is exactly the same (and the result lies in$~W$), since that holds for $U$ and $T$, and the property is easily extended across the formation of polynomials of a linear operator.
So to conclude, while $p_\alpha(U)$ and $p_\alpha(T)$ are different operators (they have different domains and different codomains), the results of applying each of them to a vector $w\in W$, for instance $w=g(T)\alpha$, will be the same.
