linear Transformation of polynomial with degrees less than or equal to 2 I would like to determine if the following map $T$ is a linear transformation:
\begin{align*}
T: P_{2} &\to P_{2}\\
A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2}
\end{align*}
My attempt at solving:
\begin{align}
T(p + q) &= p(x+1) + q(x+1)\\
&= \left[A_{0} + A_{1}(x+1) + A_{2}(x+1)^2\right] + \left[b_{0} + b_{1}(x+1) + b_{2}(x+1)^2\right]\\
&= \left(A_{0} + b_{0}\right) + \left(A_{1} + b_{1}\right)(x+1) + \left(A_{2} + b_{2}\right)(x+1)^2\\
&= T(p) + T(q)
\end{align}
Is this right so far? If not, what am I doing wrong?
 A: Note that you're hiding the core of the proof, which is simply that if $p,q$ are polynomials, we define their sum, which is also a polynomial, and $(p+q)(x)=\sum_{i=1}^n (a_i+b_i)x^i=p(x)+q(x)$.
Thus, if $T$ is your transformation, $$T((p+q)(x))=(p+q)(x+1)=p(x+1)+q(x+1)=T(p(x))+T(q(x))$$ and $$T(\lambda p(x))=(\lambda p)(x+1)=\lambda p(x+1)=\lambda T(p(x))$$

Maybe it is easier to see why the transformation is linear by seeing how it acts on the coefficients of $P$. If $P$ is $(a_0,a_1,a_2)=a_0+a_1x+a_2x^2$ then $$T(P)=(a_0+a_1+a_2,a_1+2a_2,a_2)$$ because $$\begin{align}a_0+a_1(x+1)+a_2(x+1)^2&=a_0+a_1x+a_1+a_2x^2+2a_2x+a_2\\&=(a_0+a_1+a_2)+(a_1+2a_2)+a_2x^2\end{align}$$
A: Why is $T(p+q)=p(x+1)+q(x+1)$? That's basically what you're trying to show.
You could try to just expand $T(p)$ and see that $T$ can be easily expressed as a matrix in base $1,x,x^2$.
A: The idea is right but you are starting from  what you have to proof
Hint : $T(p+q)=T(A_0+B_0+(A_1+B_1)x+(A_2+B_2)x^2)=...=T(p)+T(q)$.
I see that you had the right idea maybe this could make you start
