How did I obtain incorrect results for $P_{B\leftarrow C}$ and $P_{C\leftarrow B}$ geometrically? I'm given two bases in $\mathbb{R}^2$:
$B = \{\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \end{pmatrix}\}$ and $C = \{\begin{pmatrix} -4 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 5 \end{pmatrix}\}$ and I'm asked to find $P_{B\leftarrow C}$ and $P_{C\leftarrow B}$ completely geometrically.
I started out by finding them mechanically using the fact that $P_{B\leftarrow C} = (P_{S\leftarrow B})^{-1}(P_{S\leftarrow C})$, and $P_{C\leftarrow B} = (P_{S\leftarrow C})^{-1}(P_{S\leftarrow B})$ where $S$ is the standard basis in $\mathbb{R}^2$.
$P_{B\leftarrow C} = \begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix}^{-1} \begin{pmatrix} -4 & 2 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & 3 \\ 2 & 1 \end{pmatrix}$.
And: $P_{C\leftarrow B} = \begin{pmatrix} -4 & 2 \\ 2 & 5 \end{pmatrix}^{-1} \begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} -\frac{1}{8} & \frac{3}{8} \\ \frac{1}{4} & \frac{1}{4} \end{pmatrix}$.
Then, this is the work I did attempting to find these same results geometrically:
To find $P_{B\leftarrow C}$, where do the vectors $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ go? $\begin{pmatrix} -4 \\ 2 \end{pmatrix} \rightarrow \begin{pmatrix} -1 \\ 2 \end{pmatrix}$, which is a clockwise rotation by angle $\alpha_1$.
$\cos{\alpha_1} = \frac{\begin{pmatrix} -4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \end{pmatrix}}{\sqrt{20}\sqrt{5}} = \frac{8}{10} = \frac{4}{5}$. Therefore, $\sin{\alpha_1} = \frac{-3}{5}$.
Additionally, $\begin{pmatrix} 2 \\ 5 \end{pmatrix} \rightarrow \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ has a clockwise rotation by angle $\alpha_2$.
$\cos{\alpha_2} = \frac{\begin{pmatrix} 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix}}{\sqrt{2}\sqrt{29}} = \frac{7}{\sqrt{58}}$. Therefore, $\sin{\alpha_2} = \frac{-3}{\sqrt{58}}$. Now, we must multiply the rotated $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$ by a factor. $\frac{\sqrt{5}}{\sqrt{20}} = \sqrt{\frac{5}{20}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$. And we must multiply the rotated $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ by a factor $\frac{\sqrt{2}}{\sqrt{29}} = \frac{\sqrt{58}}{29}$.
Thus, $P_{B\leftarrow C} = \begin{pmatrix} \frac{2}{5} & \frac{3}{29} \\ -\frac{3}{10} & \frac{7}{29} \end{pmatrix}$.
I then followed this same methodology, obtaining: $P_{C\leftarrow B} = \begin{pmatrix} \frac{7}{2} & -\frac{6}{5} \\ \frac{3}{2} & \frac{8}{5} \end{pmatrix}$.
Obviously, these results don't match the results I obtained with mechanical calculations, so something is wrong. I feel that I may have a major misunderstanding in the geometry of change-of-basis. Perhaps I shouldn't be calculating the change-of-basis matrices by sending one set of basis vectors to the other because that may not reflect what is happening for any arbitrary vectors switching between those bases. Also, I have a feeling that my algebra may be wrong somewhere in determining the rotation matrices and then applying scalars to them. It is strange because we are trying to essentially use one transformation matrix to rotate two vectors two different angles, which I feel may be the wrong approach.
What am I missing?
 A: First a bit of review. The coordinates of a vector are the coefficients of its unique expression as a linear combination of basis vectors. That is, if we have the ordered basis $\mathcal B = \left(\mathbf b_1,\mathbf b_2\right)$, then $[\mathbf v]_{\mathcal B} = (v_1,v_2)^T$ means that $\mathbf v = v_1\mathbf b_1+v_2\mathbf b_2$. Now suppose that $\mathbf b_1 = p_{11}\mathbf c_1+p_{21}\mathbf c_2$ and $\mathbf b_2 = p_{12}\mathbf c_1+p_{22}\mathbf c_2$ for another ordered basis $\mathcal C = (\mathbf c_1,\mathbf c_2)$. Then $$\mathbf v = v_1(p_{11}\mathbf c_1+p_{21}\mathbf c_2)+v_2(p_{12}\mathbf c_1+p_{22}\mathbf c_2) = (p_{11}v_1+p_{12}v_2)\,\mathbf c_1 + (p_{21}v_1+p_{22}v_2)\,\mathbf c_2,$$ or, in matrix form, $$[\mathbf v]_{\mathcal C} = \begin{bmatrix} p_{11}v_1+p_{12}v_2 \\ p_{21}v_1+p_{22}v_2 \end{bmatrix} = \begin{bmatrix}p_{11}&p_{12}\\p_{21}&p_{22}\end{bmatrix} \begin{bmatrix}v_1\\v_2\end{bmatrix} = P_{\mathcal C\leftarrow\mathcal B}[\mathbf v]_{\mathcal B}.$$ So, the columns of the change-of-basis matrix $P_{\mathcal C\leftarrow\mathcal B}$ are the coordinates of the elements of $\mathcal B$ expressed in the basis $\mathcal C$.  
To find these coordinates geometrically, we can use the paralellogram rule for vector addition. For instance, to find $[\mathbf c_1]_{\mathcal B}$, draw a paralellogram with sides parallel to $\mathbf b_1$ and $\mathbf b_2$ that has $\mathbf c_1$ as a diagonal. The ratios of its side lengths to the lengths of the corresponding basis vectors (taking care to get the signs right) are the coordinates of $\mathbf c_1$ relative to $\mathcal B$.

The endpoints of the other diagonal, which are the vectors we’re looking for, can be computed by several methods such as intersecting pairs of lines, but in this case we can just read them off the diagram: they are $(-2,-2)^T = -2\mathbf b_1$ and $(-2,4)^T = 2\mathbf b_2$, so $[\mathbf c_1]_{\mathcal B} = (-2,2)^T$. This is the first column of $P_{\mathcal B\leftarrow\mathcal C}$, which matches what you got with a mechanical computation. I’ll leave working out the others to you.
