How to prove $(Tp)(x) = x^2p(x)$ is a linear map? In Linear Algebra Done Right, it gives an example of linear maps as follow:
multiplication by $x^2$
Define $T\in L(P(R),P(R))$ by $$(Tp)(x) = x^2p(x)$$
for $x \in \mathbb{R}$
My attempt:
$$(Tp)(x) = x^2p(x)$$
$$(Tp)(x+y) = x^2p(x+y) = x^2p(x) + x^2p(y) = (Tp)(x) + (Tp)(y)$$
I am not sure whether I can do $p(x+y) = p(x) + p(y)$. It seems I cannot. Right?
$$(Tp)(\lambda x) = x^2p(\lambda x) = \lambda x^2p(x) = \lambda(Tp)(x)$$
I am also not sure whether I can do $p(\lambda x) = \lambda p(x)$.
 A: Remember, you are adding polynomials, not independent variables like $x$ and $y$. If you want to check additivity, you need to verify that, given polynomials $p$ and $q$, that
$$T(p + q) = Tp + Tq.$$
Now, $T(p + q), T(p)$, and $T(q)$ are all polynomials themselves, so if you want to show that they are equal, you can show that they are equal at some arbitrary point $x$. That is, show
$$(T(p + q))(x) = (Tp)(x) + (Tq)(x).$$
Note that, by definition of $T$, we have that $(Tp)(x) = x^2 p(x)$ for all $x$, and similar for $q$. So, equivalently,
$$x^2(p + q)(x) = x^2 p(x) + x^2 q(x).$$
All I've done is use the definition of $T$. Then, we need to use the definition of $+$ for polynomials, which is $(p + q)(x) = p(x) + q(x)$. Note the $+$ on the left is the $+$ being defined, and adds functions. The $+$ on the right is the usual addition between real scalars, as $p(x)$ and $q(x)$ are actually just real numbers (although, unknown real numbers, since we don't know $x$, or $p$ and $q$). Therefore, we have
$$x^2 (p + q)(x) = x^2(p(x) + q(x)) = x^2p(x) + x^2q(x),$$
using the definition of polynomial addition, and the distributive law of real numbers.
Have another go at scalar multiplication. Remember the definition of polynomial scalar multiplication:
$$(\lambda \cdot p)(x) = \lambda \cdot p(x)$$
Again, on the left is the operation being defined, between a real number and a real valued polynomial. The operation on the right is ordinary multiplication between real numbers.
A: You are mistaken on the definition of $T$. Since $T$ is an application from $P(R)$ into $P(R)$ (I assume that $P(R)$ stands for the vector space of polynomials with real coefficients), the input is a polynomial $p$ and the output is also a polynomial function $T(p)$, usually written as $Tp$. 
What you have to show in order to prove that $T$  is linear is: 
1. for any $p$ and $q$ in $P(R)$, $T(p+q)=T(p)+T(q)$.
2. for any $p$in $P(R)$ and any real number $c$, $T(cp)=cT(p)$.
Then... what is that $x$ in the definition of $T$? Since $f=T(p)$ is itself a function, it is defined by expliciting its evaluation $f(x)$ at every real number $x$.
To sum up, $T$ is an application (= a function) that transforms a polynomial function $p$ into a new ploynomial application $f=T(p)$.

Let's prove the point $1$ above (point $2$ is proved similarly). Let $p$ and $q$ be two polynomials. Then, we need to check the equality of functions $T(p+q)=T(p)+T(q)$. For, consider a real number $x$. We have 
$$T(p+q)(x)=x^2(p+q)(x)=x^2(p(x)+q(x))=x^2p(x)+x^2q(x) = T(p)(x) + T(q)(x)$$
that is $T(p+q)(x) = (T(p)+T(q))(x)$. The claim $T(p+q)=T(p)+T(q)$ is proved! Make sure you understand each step of the computation above, this kind of game with definitions/formulas is important for abstract algebra.
Note: the notation $T(p)(x)$ means that the function $T(p)$ is evaluated at $x$. If you let $f=T(p)$, then it is just $f(x)$.

Note 2: in general, polynomials are not linear applications, so we don't have $p(x+y)=p(x)+p(y)$ and $p(cx)=cp(x)$.
Note 3: for simplicity, I identified polynomials and polynomial functions, though they are not exactly the same. It is fine because we are considering real coefficients here.
