Observe what happens to the basis components under the transformation [tex]T[/tex]:
[tex]T(1)=1+1=2=2\cdot1+0\cdot x+0\cdot x^2[/tex]
[tex]T(x)=(x-1)+(x+1)=2x=0\cdot1+2\cdot x+0\cdot x^2[/tex]
[tex]T(x^2)=(x-1)^2+(x+1)^2=2x^2+2=2\cdot1+0\cdot x+2\cdot x^2[/tex]
So we can represent [tex]T[/tex] with the matrix
[tex]\mathbf T=\begin{bmatrix}2&0&2\\0&2&0\\0&0&2\end{bmatrix}[/tex]
To check this answer: If we take an arbitrary polynomial [tex]\mathbf p\in\mathbb P_2[/tex], i.e. the vector
[tex]\mathbf p=\begin{bmatrix}c\\b\\a\end{bmatrix}[/tex]
corresponding to the polynomial [tex]p(x)=c+bx+ax^2=c\cdot1+b\cdot x+a\cdot x^2[/tex], we get
[tex]\mathbf{Tp}=\begin{bmatrix}2a+2c\\2b\\2a\end{bmatrix}=(2a+2c)\cdot1+2b\cdot x+2a\cdot x^2[/tex]
Substituting [tex]\mathbf p[/tex] into the rule above, we get
[tex]p(x-1)+p(x+1)=a(x-1)^2+b(x-1)+c+a(x+1)^2+b(x+1)+c[/tex]
[tex]p(x-1)+p(x+1)=(2a+2c)+2bx+2ax^2[/tex]
as expected.
