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.
 A: Suppose you have a polynomial $p + qt +rt^2$, and you want to find its coordinates with respect to the basis $\{1+t,t^2,t\}$ that you mentioned. In other words, you want to find numbers $a$, $b$, $c$ such that
$$
p + qt +rt^2 = a(1+t) + b(t^2) + c(t)
$$
Rearranging the right hand side , we get
$$
p + qt +rt^2 = a + (a+c)t + bt^2
$$
For two polynomials to be equal, the coefficients of each power of $t$ must be equal, so
\begin{align}
p &= a  \\
q &= a+c  \\
r &= b
\end{align}
Now solve for $a$, $b$, $c$.
A: We are given the basis $B=\{b_1,b_2,b_3\}$ for $P_2(\Bbb R)$ where
\begin{align*}
b_1(t) &= 1+t & b_2(t) &= t^2 & b_3(t) &= t
\end{align*}
We are also given the basis $E=\{e_1,e_2,e_3,e_4\}$ of $M_{2\times 2}(\Bbb R)$ where
\begin{align*}
e_1 &=
\left[\begin{array}{rr}
1 & 0 \\
0 & 0
\end{array}\right] &
e_2 &=
\left[\begin{array}{rr}
0 & 1 \\
0 & 0
\end{array}\right] &
e_3 &= 
\left[\begin{array}{rr}
0 & 0 \\
1 & 0
\end{array}\right] &
e_4 &=
\left[\begin{array}{rr}
0 & 0 \\
0 & 1
\end{array}\right]
\end{align*}
Finally, we are told that $T:P_2(\Bbb R)\to M_{2\times 2}(\Bbb R)$ is a linear map and we are asked to compute $[T]_B^E$. 
To do so, note that $[T]_B^E$ is
$$
[T]_B^E =
\begin{bmatrix}
\color{red}{a_{11}} & \color{blue}{a_{12}} & \color{green}{a_{13}} \\
\color{red}{a_{21}} & \color{blue}{a_{22}} & \color{green}{a_{23}} \\
\color{red}{a_{31}} & \color{blue}{a_{32}} & \color{green}{a_{33}} \\
\color{red}{a_{41}} & \color{blue}{a_{42}} & \color{green}{a_{43}} 
\end{bmatrix}
$$
where the entries are defined by the equations
\begin{array}{rcrcrcrcrc}
T(b_1) & = & \color{red}{a_{11}}\,e_1 &+&\color{red}{a_{21}}\,e_2 &+&\color{red}{a_{31}}\,e_3 &+&\color{red}{a_{41}}\,e_4 \\
T(b_2) & = & \color{blue}{a_{12}}\,e_1 &+&\color{blue}{a_{22}}\,e_2 &+&\color{blue}{a_{32}}\,e_3 &+&\color{blue}{a_{42}}\,e_4 \\
T(b_3) & = & \color{green}{a_{13}}\,e_1 &+&\color{green}{a_{23}}\,e_2 &+&\color{green}{a_{33}}\,e_3 &+&\color{green}{a_{43}}\,e_4 
\end{array}
If I understand your question correctly, then the columns of this matrix is what you're asking for.
A: 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 :
$$\left\{ {\left( \begin{array}{l}1\\0\\0\end{array} \right) + \left( \begin{array}{l}0\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)} \right\} = \left\{ {\left( \begin{array}{l}1\\1\\0\end{array} \right),\left( \begin{array}{l}0\\0\\1\end{array} \right),\left( \begin{array}{l}0\\1\\0\end{array} \right)} \right\}$$
