How to verify that $T:P_2 \to P_3, T(p(x)) = xp(x)$ is a linear transformation? 
Show that  $T:P_2 \to P_3, T(p(x)) = xp(x)$ is a linear
  transformation.

I don't really know how to go about verifying this, here is my work so far:
In order for $T$ to be a linear transformation, the following two conditions must be met:


*

*$T(x+y) = T(x) + T(y)$

*Take $c$ as any scalar, $cT(x) = T(cx)$
Take any $(x_1,y_1), (x_2,y_2)$ in $P_2$. 
Then $T((x_1+x_2),(y_1+y_2)) xp(x) = (x_1+x_2) * ((x_1+x_2),(y_1+y_2))$
I know that is wrong but I don't know how to use the two steps mentioned above for this problem to verify that $T$ is a linear transformation.
 A: Depending on author, $P_2$ might refer to the set of degree less than or equal to $2$ polynomials in which case an arbitrary element of $P_2$ might look like $ax^2+bx+c$ and is of dimension $3$ as it has $\{x^2,x,1\}$ as a basis (2 here in reference to the highest possible power on $x$), or it might be instead that $P_2$ is in reference to the two-dimensional space of polynomials of degree strictly less than $2$ in which case an arbitrary element of $P_2$ might look like $ax+b$ (2 here in reference to the dimension of the space).
Similarly, $P_3$ would be the same but include the next higher degree power of $x$.
I will phrase this as though we were talking about the first interpretation... that $P_2 = \{ax^2+bx+c~:~a,b,c\in\Bbb R\}$ and $P_3 = \{ax^3+bx^2+cx+d~:~a,b,c,d\in\Bbb R\}$
You were tasked with showing that $T(f)=x\cdot f$, or better written $T(ax^2+bx+c)=x(ax^2+bx+c)=ax^3+bx^2+cx+0$, is a linear transformation.
Now that you have the correct objects, hopefully you can continue.  Just recall how you can algebraically manipulate polynomials.  Otherwise, this is going to be an identical problem in every way except flavor to the one of the mapping from $\Bbb R^3$ to $\Bbb R^4$  given by $[a,b,c]\mapsto [a,b,c,0]$
A: You shall show that $T(p_1(x)+p_2(x)) = T(p_1(x)) + T(p_2(x))$ and $T(cp(x)) = c T(p(x)).$
