How to find coordinate vector for an ordered none standart basis of polynomials? I know that it's trivial to know the coordinate vector with respect to the standart basis of polynomials.
The standart basis of $P_2[R]$ is:
$$\{1,t,t^2\}$$
A general vector spanned by the above set is of the form:
$$a+bt+ct^2$$
such that $a,b,c \in R.$
the coordinate vector with respect to the standart basis of $R^3$ is:
$$\left( \begin{array}{l}a\\b\\c\end{array} \right) = a\left( \begin{array}{l}1\\0\\0\end{array} \right) + b\left( \begin{array}{l}0\\1\\0\end{array} \right) + c\left( \begin{array}{l}0\\0\\1\end{array} \right)$$
But what if I have for example this ordered basis of $P_2[x]$ :
$$B=\{1+t,t^2,t\}$$
And I have a linear transorfmation 
$$T:{P_2}[R] \to {M_{2x2}}[R]$$
and I need to find 
$$[T]_B^E$$
This is the matrix that represent $T$ with respect to the basis $B$ and $E$ the standart basis for ${M_{2x2}}[R]$.
How can I find the general form of the coordinate vector with respect to the  basis $B$ ?
I can only work with the coordinate verctor since I'm looking for the representing matrix for $T$ with respect to this basis.
so I need to find:
$$\begin{array}{l}T({b_1})\\T({b_2})\\T({b_3})\end{array}$$
And I dont know How to find them because Ineed the coordinate vector with respend to the basis $B$.
If it was for example to find  $T[b_i] $ with respect to the standart basis of $P_2[R]$ ,
I would perform :
$$\begin{array}{l}T\left( \begin{array}{l}1\\0\\0\end{array} \right)\\T\left( \begin{array}{l}0\\1\\0\end{array} \right)\\T\left( \begin{array}{l}0\\0\\1\end{array} \right)\end{array}$$
But B is not the standart basis for $P_2[R]$ so I cant do $T$ on the three vectors above.

method $1$:
Find the change of basis matrix from $E$ to $B$:
$1+t=1(1)+1(t)+1(t^2)$ 
$t^2\;\;\;\;\,=0(1)+0(t)+1(t^2)$ 
$t\;\;\;\;\;\;= 0(1)+1(t)+0(t^2)$ 
tranpose the coefficients and put as columns in a $3x3$ matrix, you get:
$$C = \left( {\begin{array}{*{20}{c}}1&0&0\\1&0&1\\0&1&0\end{array}} \right)$$
take $C$ and multiply it by a coordinate vector corosponding to each coefficient in the polynomial. for example:
$$a+bt+ct^2$$ has the coordinate vector 
$$\left( \begin{array}{l}a\\b\\c\end{array} \right)$$
which is :
$$a\left( \begin{array}{l}1\\0\\0\end{array} \right) + b\left( \begin{array}{l}0\\1\\0\end{array} \right) + c\left( \begin{array}{l}0\\0\\1\end{array} \right)$$
so multiply $C$ times each vector and you will get:
$$\left( \begin{array}{l}1\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)$$
now as was said above me, find general solution and you will get:
$$\,x=a\\y=c\\\;\;\;\;\,\,z=b-a$$
so any coordinate vector with respect to the basis $B$
his representation as a vercor in $R^3$ is :
$$x\left( \begin{array}{l}1\\1\\0\end{array} \right) + y\left( \begin{array}{l}0\\0\\1\end{array} \right) + z\left( \begin{array}{l}0\\1\\0\end{array} \right) = a\left( \begin{array}{l}1\\1\\0\end{array} \right) + c\left( \begin{array}{l}0\\0\\1\end{array} \right) + b - a\left( \begin{array}{l}0\\1\\0\end{array} \right)$$
example :
$$\left( \begin{array}{l}2\\1\\3\end{array} \right) = 2\left( \begin{array}{l}1\\1\\0\end{array} \right) + 3\left( \begin{array}{l}0\\0\\1\end{array} \right) + (1 - 2)\left( \begin{array}{l}0\\1\\0\end{array} \right)$$
Method 2
find the coefficient as mentioned above.
and noticed that  according to the basis $B = \{1+t,t^2,t\}$
the cooridnate vectors representantiom as a basis of $R^3$ is :
$$\left\{ {\left( \begin{array}{l}1\\0\\0\end{array} \right) + \left( \begin{array}{l}0\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)} \right\} = \left\{ {\left( \begin{array}{l}1\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)} \right\}$$
