What does the notation $[T]_{B^\prime \to B}$ mean? 
Let $T:P_2 \to P_1$ be defined by $T(p(x))=p'(x) + p''(x)$ and let $B = \{1,x,x^2\} \text{ and } B'=\{1,x\}$.
Find $[T]_{B\prime \to B}$

I do not understand the notation used when saying $[T]_{\textbf{B' $\to$ B}}$.
Can anybody please explain to me how to do this? Or show me another example? I simply do not have any references to understand the notation and hence cannot go about finding the answer.
 A: I am a little bit confused by the notation. Out of context I would think that it would mean the matrix representation of the linear transformation $T$ which you get by giving the domain the basis $B'$ and the codomain the basis $B$. But in context the role of $B$ and $B'$ seems to be reversed, since $T$ maps from $P_2$ which has basis $B$ to $P_1$ which has basis $B'$.
That said, let's talk about how to compute it. Call the elements of the basis of the domain $a_j$ and the elements of the basis of the codomain $b_i$. Then the $(i,j)$ entry of this representation is the coefficient of $b_i$ in the representation of $T(a_j)$. (In other words, the $j$th column provides the representation of the image of $a_j$.) For instance in your problem $T(x)=1=1\cdot 1 + 0 \cdot x$, so the $(1,2)$ entry is $1$ and the $(2,2)$ entry is $0$.
A: Find the matrix of $T$ wrt the bases $B$ and $B'$
A: what the notation $T_{B'\to B}$ means is that you $T$ of the basis elements are expressed as a linear combination of the basis elements of $B.$ the weights serve as the column components.
here is an example. we will find what $T$ does to every basis element:
$$\begin{align}T(1) &= 0=0 \cdot 1 + 0 \cdot x \\
T(x) &= x' + x'' = 1= 1\cdot 1 + 0\cdot x\\
T(x^2) &= (x^2)' + (x^2)'' = 2 \cdot 1+ 2 \cdot x\end{align}$$  these can be written compactly as $$T\pmatrix{1&x&x^2}=\pmatrix{1&x}\pmatrix{0&1&2\\0&0&2} $$
the matrix is $$T_{B'\to B}=\pmatrix{0&1&2\\0&0&2}. $$
