Finding the change of basis matricies Not quite sure what to do for this question.
Let $B_1$= {$\vec{u}_1,\vec{u}_2,\vec{u}_3$} and $C_1$= {$\vec{w}_1,\vec{w}_2,\vec{w}_3$} be two bases for $\mathbb{R}^3$ such that
$\vec{w}_1=\vec{u}_1-\vec{u}_2,$ $\vec{w}_2=\vec{u}_2-\vec{u}_3, $ $\vec{w}_3=\vec{u}_2+\vec{u}_1 $.
Find the change of coordinates matrices $P_{{C_1}\to {B_1}}$ and $P_{{B_1}\to {C_1}}$
Any hints would be greatly appreciated.
 A: (I'm not sprinking all the little \vec s through this answer.)
Interpreted in the $C_1$ basis, the vector $\begin{pmatrix}1\\0\\0\end{pmatrix}$ is $w_1$.  To indicate with respect to which basis a vector is written, we subscript it with that basis.  So we could write
$$  \begin{pmatrix}1\\0\\0\end{pmatrix}_{C_1} = w_1  \text{.}  $$
Notice that
$$  P_{C_1 \rightarrow B_1} \cdot \begin{pmatrix}1\\0\\0\end{pmatrix}_{C_1}  $$
gives $1$ times the first column of $P_{C_1 \rightarrow B_1}$ plus zero times the other columns.  We are given that (the same quantity since the $P$ matrix is being multiplied by an alias for the same vector)
$$  P_{C_1 \rightarrow B_1} \cdot w_1 = u_1 - u_2 = \begin{pmatrix}1\\-1\\0\end{pmatrix}_{B_1}  \text{.}  $$
So the first column of $P_{C_1 \rightarrow C_2}$ must be $\begin{matrix}1\\-1\\0\end{matrix}$.  Similarly, the second column must be $\begin{matrix}0\\1\\-1\end{matrix}$.  You should be able to find the last column.
To go in the other direction, solve for $u_1$, $u_2$, and $u_3$ in terms of the $w_j$s and then repeat the above analysis using these three solutions to write down the columns of $P_{B_1 \rightarrow C_1}$.
A: Given $\vec{w_1} = \vec{u_1}-\vec{u_2},\:\:\:\vec{w_2} = \vec{u_2}-\vec{u_3},\:\:\: \vec{w_3} = \vec{u_2}+\vec{u_1}$
Every vector in $\mathbb{R}^3$ can be written as a linear combination of the vectors in $C_1$, and also the vectors in $B_1$, and given a linear combination of the vectors in $C_1$, what is the corresponding vector as a linear combination of the vectors in $B_1$? There is a translation of coefficients:
$a_1\vec{w_1}+a_2\vec{w_2}+a_3\vec{w_3}= (a_1+a_3)\vec{u_1}+(-a_1+a_2+a_3)\vec{u_2}+(-a_2)\vec{u_3}$, which implies
$\begin{bmatrix} \vec{w_1}\; \big|\; \vec{w_2}\;\big|\;  \vec{w_3}\: \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} \vec{u_1}\; \big|\; \vec{u_2}\;\big|\;  \vec{u_3}\: \end{bmatrix}\begin{bmatrix} a_1+a_3 \\ -a_1+a_2+a_3 \\ -a_2 \end{bmatrix} = \begin{bmatrix} \vec{u_1}\; \big|\; \vec{u_2}\;\big|\;  \vec{u_3}\: \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & -1 & 0 \end{bmatrix}\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$
thus $P_{C_1\rightarrow B_1} = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & -1 & 0 \end{bmatrix}$, and similarly $P_{B_1\rightarrow C_1} = (P_{C_1\rightarrow B_1})^{-1}$.
If you don't believe me:
solving for $\vec{u_1}=\frac{1}{2}(\vec{w_1}+\vec{w_3}),\vec{u_2}=\frac{1}{2}(\vec{w_3}-\vec{w_1}),\vec{u_3}=\frac{1}{2}(\vec{w_3}-\vec{w_1}-2\vec{w_2})$ then
$c_1\vec{u_1}+c_2\vec{u_2}+c_3\vec{u_3} = \left(\frac{c_1 - c_2 - c_3}{2}\right)\vec{w_1} +(-c_3)\vec{w_2}+\left(\frac{c_1+c_2+c_3}{2}\right)\vec{w_3}$ and
$\begin{bmatrix} \vec{u_1}\; \big|\; \vec{u_2}\;\big|\;  \vec{u_3}\: \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} \vec{w_1}\; \big|\; \vec{w_2}\;\big|\;  \vec{w_3}\: \end{bmatrix}\begin{bmatrix} \frac{c_1 - c_2 - c_3}{2} \\ -c_3 \\ \frac{c_1+c_2+c_3}{2} \end{bmatrix} = \begin{bmatrix} \vec{w_1}\; \big|\; \vec{w_2}\;\big|\;  \vec{w_3}\: \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & -1 \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}$
where $$\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & -1 & 0 \end{bmatrix}\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & -1 \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
P.S. I may or may not of mixed up the meanings of $P_{B_1\rightarrow C_1}$ and
$P_{C_1\rightarrow B_1}$
