I'm given the bases in $\mathbb{R}^2$: $B=\{\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \end{pmatrix}\}$ and $C=\{\begin{pmatrix} -4 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 5 \end{pmatrix}\}$.
I'm trying to find both $P_{B\leftarrow C}$ and $P_{C\leftarrow B}$. I'm asked to do this with a geometric method. I already understand how to solve this computationally/algebraically.
So, to get from the basis $C$ to $B$ I could rotate $C$ some amount clockwise but I'm not convinced this would work because the angle between the vectors of $B$ is not the same as the angle between the vectors of $C$. I've solved problems like this algebraically but I'm just confused on how to approach it geometrically
HINT.- A way is as follows:
1) rotate $\begin{pmatrix}1 \\1\end{pmatrix}$ counter-clockwise of $\alpha_1=23.19859^{\circ}$ so it becomes aligned with $\begin{pmatrix}2 \\5\end{pmatrix}$.
2) similarly for $\begin{pmatrix}-1 \\2\end{pmatrix}$ of $\alpha_2=36.86989^{\circ}$ to become aligned with $\begin{pmatrix}-4 \\2\end{pmatrix}$.
You finish with two homotheties easy to determine.
(you have $\cos\alpha_1=\dfrac{\left|\begin{pmatrix}1 \\1\end{pmatrix}\cdot\begin{pmatrix}2 \\5\end{pmatrix}\right|}{\sqrt2\sqrt{29}}=\dfrac{7}{\sqrt{58}}$ and similarly...)
►For the other sens you can take rotations clockwise sens for the two angles above.