Find an upper triangular matrix 
Let $T:V\rightarrow V$ such that: $T(a+bx+cx^2) = (a+b) + (b-c)x + (a+b+c)x^2$
  $B=\{1,x+1, x^2+1\}$. Find a basic $C$ such that ${[T]_{B \to C}}^{}$ is an upper triangular matrix.  

Use $T$ for each term in the basis, $B$:  
$$\begin{array}{l}
 T(1) = 1 + {x^2} \\ 
 T(x + 1) = 2(1 + {x^2}) + x \\ 
 T({x^2} + 1) = 2(1 + {x^2}) - x - 1 \\ 
 \end{array}$$
Apparently,
We get that $C = \{ 1 + {x^2},x,1\}$ is basis which satisfying out needs.  
$$\begin{array}{l}
 {[T(1)]_c} = (1,0,0) \\ 
 {[T(x + 1)]_c} = (2,1,0) \\ 
 {[T({x^2} + 1)]_c} = (2, - 1, - 1) \\ 
 \end{array}$$
I have two questions: 


*

*How could I possibly figure it out?  

*Is there another way solving it? I thought about calculating ${[T]_{B \to E}},{[T]_{E \to C}}$ (here, $E$ is the standard basis). Would that be helpful?


The transformation $T$ is invertible (just compute the determinant of its matrix in the standard basis, e.g.) so, given any basis $B = \{b_1,b_2,b_3\}$ for the space, $T(B) = \{ Tb_1, Tb_1, Tb_3\}$ is also a basis for the space (do you see why?).
The matrix representation $[T]_{B \to T(B)}$ has a very simple form (what is it?) which is certainly upper-triangular.
