Find the $\mathfrak B$-matrix for the transformation $x\mapsto Ax$, where $\mathfrak B=\left\{\begin{bmatrix}2\\3\end{bmatrix},\begin{bmatrix}3\\5\end{bmatrix}\right\}$ and $A=\begin{bmatrix}1&3\\2&4\end{bmatrix}.$
Not looking for an answer as one is already provided $\begin{bmatrix}7&12\\-1&-2\end{bmatrix}$, but I don't even know what is being asked here. I have not been able to find a similar example problem in the text or online. I may just be severely misunderstanding the content of the section.
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$\begingroup$You are given linear transformation $x\mapsto Ax$, where $A$ is a $2\times 2$ matrix with real entries, and a set of $2$ vectors $\mathfrak B$. The transformation is w.r.t. to the standard basis $B=\left\{\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\right\}$ and as you can check, the vectors of $\mathfrak B$ form a basis of $\Bbb R^2$, clearly different from the canonical basis.
Your task is to find how $A$ transforms $\Bbb R^2$ w.r.t. to the new basis $\mathfrak B$ (the $\mathfrak B$-matrix). This process is known as change of basis, and I find the following diagram quite illuminating$$\require{AMScd} \begin{CD} \Bbb R^2_B @>{A}>> \Bbb R^2_B\\ @V{M_B^{\mathfrak B}}VV @VV{M_B^{\mathfrak B}}V\\ \Bbb R^2_{\mathfrak B} @>>{\mathfrak A}> \Bbb R^2_{\mathfrak B} \end{CD} $$Here $\Bbb R^2_A$ and $\Bbb R^2_{\mathfrak B}$ refer to $\Bbb R^2$ w.r.t. to the standard basis $B$ and to the given basis $\mathfrak B$, $A$ is the given linear transformation (or rather its transformation matrix) and $\mathfrak A$ is the matrix you are asked to find. The interesting thing is the matrix $M_B^{\mathfrak B}$ which transforms the basis $B$ into the basis $\mathfrak B$, by sending $[1,0]^\top\mapsto[2,3]^\top$ and $[0,1]^\top\mapsto[3,5]^\top$. From the diagram it is clear that $\mathfrak A=M_B^{\mathfrak B}A(M_B^{\mathfrak B})^{-1}$ as we expect that it does not matter in which way we go from $\Bbb R^2_B$ to $\Bbb R^2_{\mathfrak B}$ (the diagram commutes).
Now, take $A=\begin{bmatrix}1&3\\2&4\end{bmatrix}$ and $M_B^{\mathfrak B}=\begin{bmatrix}2&3\\3&5\end{bmatrix}$. Compute $(M_B^{\mathfrak B})^{-1}$ (it should be $\begin{bmatrix}5&-3\\-3&2\end{bmatrix}$), and then you obtain $\mathfrak A$ as above. The key is the commuting diagram which, by the way, is a great tool to remember what really is going on.
$\endgroup$ 2 $\begingroup$You have$$A.\begin{bmatrix}2\\3\end{bmatrix}=\begin{bmatrix}11\\16\end{bmatrix}\text{ and }A.\begin{bmatrix}3\\5\end{bmatrix}=\begin{bmatrix}18\\26\end{bmatrix}.$$But$$\begin{bmatrix}11\\16\end{bmatrix}=7\begin{bmatrix}2\\3\end{bmatrix}-\begin{bmatrix}3\\5\end{bmatrix}\text{ and }\begin{bmatrix}18\\26\end{bmatrix}=12\begin{bmatrix}2\\3\end{bmatrix}-2\begin{bmatrix}3\\5\end{bmatrix},$$which means that the matrix of your linear transformation with respect to $\mathfrak B$ is$$\begin{bmatrix}7&12\\-1&-2\end{bmatrix}.$$
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