Here are my full instructions:
Find the transition matrix $P_{B'\leftarrow B}$ from $B=\left\{\left(1,1,\right),\left(2,3\right)\right\}$ to $B'=\left\{\left(1,2\right),\left(0,1\right)\right\}$, and use it to find the $B'$ coordinate matrix for the vector with $B$ coordinate matrix $\begin{bmatrix} 7 \\ -4 \\ \end{bmatrix}_B$.
So I did the following:
$\begin{bmatrix} B' & B \\ \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 & 1 & 2 \\ 2 & 1 & 1 & 3 \\ \end{bmatrix}$, $rref = \begin{bmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & -1 & -1 \\ \end{bmatrix}$
So I said $P=\begin{bmatrix} 1 & 2 \\ -1 & -1 \\ \end{bmatrix}^T$ and then $\begin{bmatrix} X \end{bmatrix}_{B'} = P\begin{bmatrix} X \end{bmatrix}_{B} = \begin{bmatrix} 1 & 2 \\ -1 & -1 \\ \end{bmatrix}\begin{bmatrix} 7 \\ -4 \\ \end{bmatrix}=\begin{bmatrix} -1 \\ -3 \\ \end{bmatrix}$
Am I correct up to this point and do I then take my coordinates and apply them to $B$?
$\endgroup$1 Answer
$\begingroup$We have, letting $P_{\mathscr{B}' \leftarrow \mathscr{B}}$ denote the transition matrix from the basis $\mathscr{B}$ to the basis $\mathscr{B}'$, $$P_{\mathscr{B}' \leftarrow \mathscr{B}} = P_{\mathscr{B}' \leftarrow \mathscr{E}}P_{\mathscr{E} \leftarrow \mathscr{B}}\\=P_{\mathscr{E} \leftarrow \mathscr{B}'}^{-1}P_{\mathscr{E} \leftarrow \mathscr{B}}\,\,,$$ where $\mathscr{E}$ is the standard basis for $\mathbb{R}^{2}$. You know each of these matrices in the final equality I have given: you need only take the inverse of the matrix with columns the basis vectors of $\mathscr{B}'$. The other matrix is just that formed by the basis vectors of $\mathscr{B}$ as columns. After you compute the desired transition matrix by taking the product of these two known matrices, you need to multiply it by the $\mathscr{B}$-coordinate vector on the right to obtain the $\mathscr{B}'$-coordinate vector. In other words, $$P_{\mathscr{B}' \leftarrow \mathscr{B}} \cdot \left[\vec{x}\right]_{\mathscr{B}} = \left[\vec{x}\right]_{\mathscr{B}'}\,\,.$$
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