In my last question, it became clear I could use $v' = q v q^*$ for quaternion coordinate transformation.
However, I believe I've interpreted this incorrectly. Firstly, do I understand correctly that $q$ n this instance is the 4x4 matrix form of the quaternion, not the quaternion vector?
In my example, I have the vector: $$ v = \begin{bmatrix}0 \\0 \\ 20\end{bmatrix}$$
And the quaternion (in form $[w, x,y,z]$) $$ q = \begin{bmatrix}0.7102 \\-0.3828 \\-0.3319 \\0.4887 \end{bmatrix}$$
Which, following $v' = qvq^*$, would result in the equation:
$$v' = \begin{bmatrix}0.7102 & 0.3828 & 0.3319 & -0.4887 \\ -0.3828 & 0.7102 & -0.4887 & -0.3319 \\ -0.3319 & 0.4887 & 0.7102 & 0.3828 \\ 0.4887 & 0.3319 & -0.3828 & 0.7102 \end{bmatrix} \begin{bmatrix}0 \\ 0 \\ 0 \\ 20\end{bmatrix}\begin{bmatrix}0.7102 & -0.3828 & -0.3319 & 0.4887 \\ 0.3828 & 0.7102 & 0.4887 & 0.3319 \\ 0.3319 & -0.4887 & 0.7102 & -0.3828 \\ -0.4887 & -0.3319 & 0.3828 & 0.7102\end{bmatrix}$$
This results in the vector:
$$v' = \begin{bmatrix}-13.88 \\ -9.43 \\ 10.88 \\ 0.18 \end{bmatrix}$$
I would expect the last three elements of this vector to represent my transformed vector. Yet the vector has a w component, indicating the vector has been transformed into the fourth dimension, and taking only the 3d components would not work. Thus do I need to perform some kind of projection into 3D space of this 4D vector, in order to end up with $v'$?
For the sake of not leaving questions unanswered, I’ll sum up the comments here.
It looks like you’ve mistaken a quaternion product for a matrix product in the answer to the linked question. The product of two quaternions is just the result of applying the distributive law of multiplication over addition and the rules for the products of the basis quaternions $\mathbf i$, $\mathbf j$ and $\mathbf k$ to the algebraic product of quaternions of the form $a+b\mathbf i+c\mathbf j+d\mathbf k$. I find it easier to remember what the final result looks like if I separate the quaternions into their scalar and vectors parts, $r=a$ and $\vec v=(b,c,d)$, respectively. The quaternion product then looks like $$(r_1,\vec v_1)(r_2,\vec v_2)=(r_1r_2-\vec v_1\cdot\vec v_2,r_1\vec v_2+r_2\vec v_1+\vec v_1\times\vec v_2).$$ Historically, this product was a motivation for the definition of the dot and cross products of vectors.