Prove that composition of any two elements from $L(f)$ is in $L(f)$. Let $L(V)=\{f:V\to V|f \text{ is linear map}\}$
$h=a_0+a_1x+..+a_nx^n$
the value of polynomial $h(x)$ on $f\in L(V)$ is defined like this
$h(f)=a_01_V+a_1f+a_2f^2+...+a_nf^n$
where $f^n=f\circ f^{n-1}$
$\beta_f:R[X]\to L(V)$ which is linear.Let $Im(\beta_f)=L(f)$.$L(f)$ is subspace of $L(V)$
Now the part that I have hard time understanding is this.Composition of any two elements from $L(f)$  is in $L(f)$
Book says that this follows from $\beta_f(h_1h_2)=\beta_f(h_1)\circ\beta_f(h_2)$
or which is same as $(h_1h_2)(f)=h_1(f)\circ h_2(f)$
can you explain how author got this?I know that $f_1\circ(f_2+f_3)=(f_1\circ f_2)+(f_1\circ f_3)$.But I don't know how to use this.
 A: Let, $f_1, f_2\in L(f) $ and $\lambda \in \Bbb{R}$
To show, $f_1+\lambda f_2\in L(f) $
As, $f_1\in L(f) $ , $\exists p(x) \in\Bbb{R}[X]$ such that $\beta_f(p(x)) =f_1$
and, $\exists q(x) \in\Bbb{R}[X]$ auch that $\beta_f(q(x)) =f_2$
Now, $f_1+\lambda f_2=\beta_f(p(x)) +\lambda \beta_f(q(x)) =\beta_f(p(x)+\lambda q(x)) $
Hence, $\exists p(x)+\lambda q(x)\in \Bbb{R}[X]$ such that
$\beta_f(p(x)+\lambda q(x)) =f_1 +\lambda f_2$
Hence, $f_1+\lambda f_2\in L(f) $
Hence, $L(f) $ is a subspace of $L(V) $
$f_1 , f_2\in L(f) $
To show, $f_1 \circ f_2\in L(f) $
$\beta_f(p(x)) =f_1$
$\beta_f(q(x)) =f_2$
Then, $f_1\circ f_2= \beta_f(p(x)) \circ \beta_f(p(x))= p(f) \circ q(f) =(pq)(f)= \beta_f(p(x)q(x))$
Hence, $\exists p(x)\cdot q(x)\in \Bbb{R}[X]$ such that
$\beta_f(p(x)\cdot q(x)) =f_1 \circ f_2$
Edit:
$p(x) =\sum_{i=0}^{m} a_i x^i$
$q(x) =\sum_{j=0}^{n} b_j x^j$
$(pq)(x) =\sum_{i=0}^{m} \sum_{j=0}^{n} a_i b_j x^{i+j}$
$(pq) (f) =\sum_{i=0}^{m} \sum_{j=0}^{n} a_i b_j f^{i+j}= \sum_{i=0}^{m} a_i f^i \sum_{j=0}^{n}  b_j f^{j}=p(f)q(f)$
