Producing a description of an $n\times n$ matrix $A$ that implements the coordinate mapping $x\to[x]_{B}.$ Let $B=\{b_{1},\dots,b_{n}\}$ be a basis for $\mathbb{R}^n$. Produce a description of an $n\times n$ matrix $A$ that implements the coordinate mapping $x\to[x]_{B}.$

My solution:
Let $A=[I_{n}\,\, P^{-1}_{B}]=\begin{bmatrix}e_{1} & \cdots & e_{n} & b'_{1} & \cdots &b'_{n}\end{bmatrix}$, where the columns $b'_{1},\dots,b'_{n}$ corresponds to $b_{1},\dots,b_{n}$ in $[P_{B}\,\,I_{n}]=\begin{bmatrix}b_{1} & \cdots & b_{n} & e_{1} & \cdots & e_{n}\end{bmatrix}$  prior to row reducing the matrix $[P_{B}\,\,I_{n}]$ to $[I_{n}\,\, P^{-1}_{B}]$.
I'm not sure how to show how this is true, but I tried some small cases, and I think it extends to $n\times n$ matrices.
 A: I think this is along the lines of your solution.  If you are given a vector $x$ and want to write it in terms of a basis $\{b_i\}$, then you are attempting to find $c_i$ s.t.
$$x = \sum_i c_i b_i$$
As a matrix equation this can be written as
$$x = B c$$
where $B$ is a matrix whose $i$th column is $b_i$.  Then to perform the transformation you can take the inverse of $B$ on both sides (it exists because the columns of $B$ are linearly independent).  Then
$$c = B^{-1}x$$
So $B^{-1}$ is the matrix that implements the change of basis.
A: We can use the so-called matrix representation theorem. Considering
$$V = W = \mathbb{R}^n$$
$$T = I: \mathbb{R}^n \to \mathbb{R}^n$$
$$E = \{e_1, e_2, \ldots, e_n\} \subset \mathbb{R}^n$$
$$F = B \subset \mathbb{R}^n$$
Using the matrix representation theorem we have
$$[x]_B = [T(x)]_F = [T]_E^F [x]_E = Ax$$
where
\begin{equation}
A = [T]_E^F = [I]_E^B = 
\begin{pmatrix}
[e_1]_B & [e_2]_B & \ldots & [e_n]_B
\end{pmatrix}
\end{equation}
is the transition matrix from $E$ to $B$. 
