Find bases given that P is the change of coordinates matrix from this to this [Lay P244 Q4.7.19] Lay P289: Let $V$ be an $n$-dimensional vector space, let $W$ be an $m$-dimensional vector space, and let $T$ be any linear transformation from $V$ to $W$. To associate a matrix with $T$, choose (ordered) bases $B$ and $C$ for $V$ and $W$, respectively.
$$
[T(x)]_{c}= [[T(b_{1})]_{c}\ [T(b_{2})]_{c}\ \cdots\ [T(b_{n})]_{c}] [x]_{B}
$$
$19.$ Let $ P= \begin{bmatrix}
1 & 2 & -1\\
-3 & -5 & 0\\
4 & 6 & 1
\end{bmatrix}, $
$ v_{1}=\left\{\begin{array}{l}
-2\\
2\\
3
\end{array}\right\},\ v_{2}=\left\{\begin{array}{l}
-8\\
5\\
2
\end{array}\right\},\ v_3\ =\left\{\begin{array}{l}
-7\\
2\\
6
\end{array}\right\}
$
Textbook Question (a) : Find a basis $\{u_{1},\ u_{2},\ u_{3}\}$ for $\mathbb{R}^{3}$ such that $P$ is the change-of-coordinates matrix (I post the definition below) from $\{u_{1},\ u_{2},\ u_{3}\}$ to the basis $\{v_{1},\ v_{2},\ v_{3}\}$. 
Textbook Answer : If $C$ is the basis $\{v_{1},v_{2},v_{3}\}$, then the columns of $P$ are $[u_{1}]_{C},\ [u_{2}]_{C}$, and $[u_{3}]_{C}$. So $u_{j}=[v_{1}\ v_{2}\ v_{3}][u_{1}]_{C}$, and $\color{forestgreen}{ [u_{1}\ u_{2}\ u_{3}]=[v_{1}\ v_{2}\ v_{3}]P }$.

$1.$ I know that in part (a),  $P$ is given as $[ .. [u_{i}]_{C} ...]$, but what's the proof strategy? How do we proceed with this? I omit all calculations; I'm not asking about them in any of my questions herein.
$2.$ Whence does $ \color{forestgreen}{ [u_{1}\ u_{2}\ u_{3}]=[v_{1}\ v_{2}\ v_{3}]P }$ originate? It doesn't look like the formula at the start of my question? 

Textbook Question (b): Find a basis $\{w_{1},\ w_{2},\ w_{3}\}$ for $\mathbb{R}^{3}$ such that $P$ is the change- of-coordinates matrix from $\{v_{1},\ v_{2},\ v_{3}\}$ to $\{w_{1},\ w_{2},\ w_{3}\}$.
Textbook Answer :  Analogously to part $a.,\ [v_{1}\ v_{2}\ v_{3}]=[w_{1}\ w_{2}\ w_{3}]P$, so $[w_{1}\ w_{2}\ w_{3}]= [v_{1}\ v_{2}\ v_{3}]P^{-1}$.

$3.$ I know that in part (b), P is given as $[ .. [v_{i}]_{W} ...]$, but what's the proof strategy? How do we proceed with this? 

 A: I will only address the $2^{nd}$ part of the first question.

  
*
  
*Whence does $[u_1\;u_2\;u_3] = [v_1\;v_2\;v_3]P$ originate? It doesn't look like the formula at the start of my question? 
  

Writing something like $[u_1\;u_2\;u_3] = [v_1\;v_2\;v_3] P$ and putting $P$ on the right is "correct" and actually makes perfect sense! 
Given any two "abstract" real vector spaces $V$ and $W$ of dimension $n$ and $m$ respectively.
Let $B = ( b_1, b_2, \ldots b_n )$ be a basis for $V$ and 
$C = ( c_1, c_2, \ldots, c_m )$ be a basis for $W$.
For any $x \in V$, the $[x]_B = (x_1, x_2, \ldots, x_n)$ appear in the start of question isn't an element in $V$. Instead, it belongs to $\mathbb{R}^n$ and relates to $x \in V$ through a relation of the form:
$$x = x_1 b_1 + x_2 b_2 + \cdots + x_n b_n$$
The matrix $P = (p_{ij})_{\substack{i=1..m\\j=1..n}}$ associated with a linear transform $T : V \to W$ is defined in a similar manner. Let's say 
$$[T(x)]_c = (y_1, y_2,\ldots,y_m) \in \mathbb{R}^m \iff T(x) = y_1 c_1 + y_2 c_2 + \cdots + y_m c_m.$$
An expression like
$$[ T(x) ]_C = \Big[[T(b_1)]_C, [T(b_2)]_C, \ldots, [T(c_n)]_C\Big][x]_B$$
should be interpreted as
$$y_i = \sum_{j=1}^n p_{ij} x_j\quad\text{ for } i = 1,\ldots, m\tag{*1}$$
where $p_{ij}$ is the $i^{th}$ component of $T(b_j)$ in basis $C$.
When $V = W$ and hence $n = m$, one can associate a matrix $P$ to the identity transform
$V \ni x \xrightarrow{id} x \in V = W$ in above manner. This is the change of coordinate matrix.
Since $p_{ij}$ is the $i^{th}$ component of $b_j$ in basis $C$, we have
$$b_j = \sum_{i=1}^m p_{ij} c_i\quad\text{ for } j = 1,\ldots, n\tag{*2}$$
Notice the summation is now over the left index $i$ of the matrix $P$. 
When a matrix form of this relation is needed, some authors prefer to write this simply as $b_j = ( c P )_j$. They do this in order to stress which index of $P$ is summed over. 
One can summarize all these relations of different $j$ in a single matrix relation. Namely,
the one that you are confused about:
$$[b_1\;b_2\;\ldots\;b_n ] = [ c_1\;c_2\;\ldots\;c_n ] P\tag{*3}$$
The key to remember is


*

*In the definition of $P$ in $(*1)$, what is summed over are coordinates of a given vector and we sum over the right index $j$ of $P$.

*In an expression like $[u_1\;u_2\;u_3] = [v_1\;v_2\;v_3] P$ or that in $(*2)$ or $(*3)$, what is summed over are vectors in $V$ and the sum is over the left index $i$ of $P$.


Finally, there is one advantage in the convention used in $(*3)$. If you represent everything in a common basis and identifies the $b_i$, $c_j$ to their components/coordinates in that common basis. You can view them as ordinary column vectors.
$(*3)$ becomes an matrix equation among 3 ordinary $n \times n$ real matrices.
