How to show T $P_2 \, \to \, P_2$ is a linear operator. Find a basis for the kernel T? 
Let P$_2$ denote the vector space of all polynomials with real coefficients and of degree at most 2. Define a function T : $P_2$ → $P_2$ by
  $$ T(P(x)) = x^2 \frac{d^2}{dx^2}(p(x-1))+ x\frac{d}{dx}(p(x-1)) $$

How to show T $P_2 \, \to \, P_2$ is a linear operator. Find a basis for the kernel T?
I was going to take a matrice solve it to show that it has only trivial solutions where the leading 1s equal 0 but i think  it's wrong and there's no matrices to begin with.
 A: Let $P_0$, $P_1$, and $P_2$ be the spaces of polynomials of degree at most $0$, $1$, and $2$, respectively.
You have the shift operator $S:P_i\to P_i$, $S[p](x):=p(x-1)$. It is a linear operator for all $i$.
You have the derivation operators $D_1$ ($D_1[p](x):=p'(x)$) and $D_2$ ($D_2[p](x):=p''(x)$) defined on spaces $D_2: P_2\to P_0$ and $D_1: P_2\to P_1$. These operators are linear.
Also, you have multiplication operators $M_1:P_1\to P_2$ ($M_1[p](x)=xp(x)$) and $M_2:P_0\to P_2$ ($M_2[p](x)=x^2p(x)$). These operators are also linear.
Now you can rewrite your operator $T$ as
$$T[p] = M_2[D_2[S[p]]]+M_1[D_1[S[p]]],$$
i.e. a sum of two compositions of linear operators. As we know, such an object will also be a linear operator. By looking at domains and ranges of the above operators, you deduce that $T:P_2\to P_2$.
Another approach would be to show that $\forall p\in P_2$ you have $T[p]\in P_2$ and show that the operator $T$ is linear by the definition of a linear operator.
In order to find the null-space of $T$ (or kernel), you need to solve the equation $T[p]=0$. In this case, you need to solve the equation
$$x^2 p''(x-1)+xp'(x-1)=0.$$
Define a function $y(x)=p'(x-1)$ (therefore $y$ is also a polynomial), then the above equation writes as
$$xy'(x)+y=0=(xy(x))',$$
hence 
$$xy(x)=c$$and $y(x)=c/x$ with $c\in \Bbb R$. Such a function is a polynomial only when $c=0$, so $y(x)=0$ and $p'(x-1)=0$, which leads to $p$ being a constant polynomial.
