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?
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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.