Matrices for change of basis linear transformations Find the $\mathcal{B}$ matrix of the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by reflection through the $x_1 x_2$ plane for the basis $\mathcal{B} = \{b_1, b_2, b_3\}$.
I understand the solution in general to be given by:
$$
 [~~[T(b_1)]_{\mathcal B} ~~~ [T(b_2)]_{\mathcal B} ~~~ [T(b_3)]_{\mathcal B}~~]
$$
However, since we are not changing bases from $\mathcal B$ my gut tells me that $T(b_k)$ is "already in the basis $\mathcal B $" and I shouldn't have to find it's coordinate mapping relative to $\mathcal B$ as shown in the matrix above. 
That is, since I'm not changing bases, only applying a linear transformation,  my intuition says the answer would simply be:
$$
 [~~[T(b_1)] ~~~ [T(b_2)] ~~~ [T(b_3)]~~]
$$
By the "formula" this is incorrect, but can someone explain why?
 A: Short answer:  you are changing basis ... twice!
The transformation $T$ is described geometrically in terms of the standard basis $\mathcal{E} = \{e_1, e_2, e_3\}$ for $\mathbb{R}^3$, so the matrix is
$$ \big[\,T\,\big]_{\mathcal{E}} = \left[ \begin{array}{*{3}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&-1 \end{array} \right].$$
To obtain the matrix in $\mathcal{B}$-coordinates, you must first change basis to the standard coordinates, then apply the tranformation there, and finally change basis back to the new basis.  This is called conjugation (or a similarity transformation, as you likely know):
$$ \big[\,T\,\big]_{\mathcal{B}} = \underset{\mathcal{B} \leftarrow \mathcal{E}}{\mathcal{P}} \; \big[\,T\,\big]_{\mathcal{E}} \; \underset{\mathcal{E} \leftarrow \mathcal{B}}{\mathcal{P}}$$
The trick is that when you write a vector $b_1$, you are implicitly expressing it in terms of the standard basis, so $b_1 = [\,b_1\,]_{\mathcal{E}}$, which is why the change-of-basis matrix from $\mathcal{B}$ to $\mathcal{E}$ is
$$\underset{\mathcal{E} \leftarrow \mathcal{B}}{\mathcal{P}} = \big[ \big[\,b_1\,\big]_{\mathcal{E}} \;\; \big[\,b_2\,\big]_{\mathcal{E}} \;\; \big[\,b_3\,\big]_{\mathcal{E}} \big] = \big[ b_1 \;\; b_2 \;\; b_3 \big].$$
Now, apply the matrix $T$ and give the result back in $\mathcal{B}$-coordinates, and you obtain
$$
\big[\,T\,\big]_{\mathcal{B}} = \big[ \big[\,T(b_1)\,\big]_{\mathcal{B}} \;\; \big[\,T(b_2)\,\big]_{\mathcal{B}} \;\; \big[\,T(b_3)\,\big]_{\mathcal{B}} \big]
.$$
