In proving that $f(w)f(v) = vw$ as an orthogonal transformation, why does $v^T = v$? I've a proof that says if $f: \mathbb{R}^3 \to \mathbb{R}^3, ~ v \mapsto Av$, where $A$ is an orthogonal matrix, then for $v,w \in \mathbb{R}^3$, we have $f(w)f(v) = (Av)(Aw) = (Av)^T(Aw) = v^TA^TAw = v^Tw = vw$. This is from the proof that the map is an isometry. Could someone explain the last step? It seems to be saying $v^T = v$ but I'm not sure how that's true. Thanks. 
 A: There could be some confusion in general going on. E.g. $f(w)$ and $f(v)$ are column vectors in $\mathbb{R}^3$, so that $f(w)f(v)$ does not make sense in a matrix/vector multiplication way, as we'd try to calculate the matrix product
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\begin{pmatrix}x'\\y'\\z'\end{pmatrix}$$
which is ill-defined.
There is however one way I could read the statement, namely involving the classical dot product over real vectors, where you'd have the relationship $v\cdot w=v^\top w$. Then, to write this out with the standard notation for the dot product, $\langle\cdot,\cdot\rangle$, you'd have:
$$
\langle f(w),f(v)\rangle=\langle Aw,Av\rangle=(Aw)^\top(Av)=w^\top A^\top Av=w^\top v=\langle w,v\rangle
$$
EDIT: Note, that there is a slight twist in the way you wrote it, if I'm not mistaken. This makes also sense as note that $f$ is a so called orthogonal transformation, i.e. it is a linear transformation that additionally preserves a corresponding scalar product, i.e. it is an isometry of the standard euclidean vector space $(\mathbb{R}^3,\langle\cdot,\cdot\rangle)$ and thus preserves lengths, angles, etc.
Preservation of the scalar product means exactly that $\langle w,v\rangle=\langle f(w),f(v)\rangle$ for all $w,v\in\mathbb{R}^3$ and you with this verified the statement that an orthogonal matrix induces an orthogonal map. There is actually(up to orthonormal bases), a 1-to-1 correspondence between orthogonal maps and orthogonal matrices.
