Determine the vectors of components For the polynomial vector space $\mathbb{R}[x]$ of degree $\leq 3$ we have the following three bases: 
$$B_1 = \{1 - X^2 + X^3, X - X^2, 1 - X + X^2, 1 - X\} , \\ 
B_2 = \{1 - X^3, 1 - X^2, 1 - X, 1 + X^2 - X^3\}, \\ 
B_3 = \{1, X, X^2, X^3\}$$ 
How can we determine the following vectors of components $\mathbb{R}^4$ ? 
$\Theta_{B_1}(b)$ for all $b \in B_1$ 
and 
$\Theta_{B_3}(b)$ for all $b \in B_1$
Could you give me  hint? 
Do we use the transformation matrix? If yes, how? 
$$$$ 
EDIT: 
I have seen the following notes : 
$\Theta_{B_1}(b\in B_1)=i$-th comlumn of idenity, since it shown always at itself, and $\Theta_{B_3}(b\in B_1)=i$-th column of $B_1$. 
Why does this hold?

Hint:
$$
B_1=
\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}
\begin{bmatrix}
1&0&1&1\\
0&1&-1&-1\\
-1&-1&1&0\\
1&0&0&0
\end{bmatrix}
$$
$$
B_2=
\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}
\begin{bmatrix}
1&1&1&1\\
0&0&-1&0\\
0&-1&0&1\\
-1&0&0&-1
\end{bmatrix}
$$
$$
B_3=
\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}
\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}
$$
Note that the columns of the matrices above are the matrices for the various bases with respect to $B_3$.

For example, consider the first vector of $B_1$, $1-X^2+X^3$
$$
\Theta_{B_3}\!\left(1-X^2+X^3\right)=\begin{bmatrix}1\\0\\-1\\1\end{bmatrix}
$$
since
$$
1-X^2+X^3=
\overbrace{\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}}^{B_3}
\begin{bmatrix}
1\\0\\-1\\1
\end{bmatrix}
$$
This is how the matrix for $B_1$ was created.
