I'm reading about change of basis and am confused with definitions on Wiki and less so on this site. I present the definitions which are more understandable first, then follow with Wiki. Here are the definitions which I understand:
$$
B = \{\mathbf b_1, \mathbf b_2, …, \mathbf b_n\}
$$
where $B$ is a basis of a vector space $V$. Then the coordinates $(\alpha_z)_{z \in I}$ of a vector $\mathbf x \in V$ using basis $B$ and field $F$ are:
$$
\phi_B : V \rightarrow F^n
$$
$$
\phi_B(\mathbf x) = [\mathbf x]_B
$$
$$
[\mathbf x]_B = \begin{bmatrix} \alpha_1 \cr \alpha_2 \cr … \cr \alpha_n \end{bmatrix}_B = (\alpha_1, …, \alpha_n)
$$
$$
\mathbf x = [\mathbf x]_S = \mathbf b_1 \alpha_1 + … + \mathbf b_n \alpha_n = \sum_{i=1}^n \mathbf b_i \alpha_i
$$
The basis transformation matrix or vector is denoted $P_B$ and is from basis $B$ to $S$, where $S$ is the standard basis:
$$
P_B = [\mathbf b_1 \mathbf b_2 … \mathbf b_n] = (\mathbf b_1, …, \mathbf b_n)
$$
$$
P_B = [P]_S^B = [P]^B = [[\mathbf b_1]_S [\mathbf b_2]_S … [\mathbf b_n]_S] = ( [\mathbf b_1]_S, …, [\mathbf b_n]_S)
$$
$$
\mathbf x = P_B [\mathbf x]_B
$$
$$
\mathbf x = [\mathbf x]_S = [P]_S^B [\mathbf x]_B
$$
$$
[\mathbf x]_B = P_B^{-1} \mathbf x = ([P]_S^B)^{-1} \mathbf x
$$
where $\det(P_B) \ne 0$ and the change of basis from basis $B$ to $B'$ is as follows.
Convert from standard coordinates to coordinates in $B'$:
$$
[\mathbf x]_{B'} = P_{B'}^{-1} \mathbf x
$$
Question : does above appear correct?
I'm the most confused by the Wiki definitions and explanation. The forms do not explicitly agree with above:
$$
\mathbf x_{old} = A \mathbf x_{new}
$$
is this form reversed?
Wiki then defines the change of basis formula as:
$$
B_{old} = (\mathbf v_1, …, \mathbf v_n), B_{new} = (\mathbf w_1, …, \mathbf w_j)
$$
For $j = 1, … n$ and $w_j$'s coordinates are $a_{i,j}$ over $B_{old}$
$$
\mathbf w_j = \sum_{i=1}^n a_{i,j} \mathbf v_i
$$
Question: Why are the coordinates $a_{i,j}$ two dimensional when they are one dimensional in earlier examples.
Question: Can the above form be rewritten in notation used earlier? For instance:
$$
P_{B_{old}} = [\mathbf v_0 … \mathbf v_n] = (\mathbf v_0, …, \mathbf v_n)
$$
$$
\mathbf w_j = P_{B_{old}} [\mathbf w_j]_{B_{old}}
$$
$$
[\mathbf w_j]_{B_{old}} = (a_{1,j}, …, a_{n,j})
$$
Best Answer
1) does above appear correct? Yes. Combining the equations, we have $$[\mathbf x]_B = P_B^{-1} \mathbf x= P_B^{-1} P_{B'}[\mathbf x]_{B'}=R[\mathbf x]_{B'}$$
2) No. We have $$ T(\mathbf x) = P_B \mathbf x=P_B P_B[\mathbf x]_B, $$ so the correct map is $S(\mathbf x)=P_B^{-1} \mathbf x$.
3) is this form reversed? It is not reversed, if "old" is the standard basis and "new" is $B$, then in the other notation, the wiki equation reads $$ \mathbf x= P_B[x]_B, $$ which agrees with the equation derived in the first part of your question. But i think it would be less confusing if written $\mathbf x_{new}=A^{-1}\mathbf x_{old}$.
4) Yes, the equations are equivalent, the way you wrote is just the compact matricial form of the equation. If you are not convinced of that, i recommend doing with a 2x2 or 3x3 example.
Answer to EDIT: Why are the coordinates $a_{i,j}$ two dimensional when they are one dimensional in earlier examples? Can the above form be rewritten in notation used earlier?
$a_{i,j}$ are just the coefficients of the linear combination of the vectors that produces $\mathbf w_j$, they are also the coordinates of $\mathbf w_j$ in the old basis, as you noticed. This equation is equivalent to $$ \mathbf w_j=P_{B_{old}}\mathbf [\mathbf w_j]_{B_{old}}, $$ where $[\mathbf w_j]_{B_{old}}=(a_{1,j},\dots,a_{n,j})$, note that this equation is the same as the one in display in 3). i'm not sure what you mean with "two dimensional" though.