Let T be the linear mapping on $P_2(R)$ (it is a set of polynomial to the $2^{nd}$ power) defined by $T(f(x))=f(x)+(x+1) f'(x)$, let $\beta$ be the standard ordered basis for $P_2(R)$, and let $A=[T]_{\beta}$. Find matrix A
$A=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix}$.
I know the standard basis $\beta$ is {$1,x,x^2$} and I know A is the matrix made of coordinate vectors of $\beta$. I am not quite sure how they get to the matrix A.
My thoughts are we can let $f(x)=x^2+x+1$, then T(f(x))=$3x^2+4x+2$ And I know the formula $[T]_{\beta}$=$A [T]_\beta'$. But I still don't see how the operation goes.
It's not a 'change of base' matrix: note that we are given only one basis (the standard one: $\beta=(1,x,x^2)$).
Instead, it's simply the matrix $A$ of the transformation $T$ in the basis $\beta$, that satisfies for all $f\in P_2(\Bbb R)$ $$[T(f)]_\beta \ =\ A\cdot [f]_\beta$$ If you apply this to the $i$th basis vector $e_i$, on the right hand side you simply get the $i$th column of $A$, which thus should consist of the $\beta$-coordinates of $T(e_i)$.
In the example we have $$T(1)=1+(x+1)\cdot 0=1\\ T(x)=x+(x+1)\cdot 1=1+2x\\ T(x^2) =x^2 +(x+1)\cdot(2x)=2x+3x^2$$ and these coefficients are put into the columns of $A$.