Pick a quaternion

*p* = *p*_{0} + *p*_{1}*i* + *p*_{2}*j* + *p*_{3}*k*

and consider the function that acts on quaternions by multiplying them on the left by *p*.

If we think of *q* as a vector in *R*^{4} then this is a linear function of *q*, and so it can be represented by multiplication by a 4 × 4 matrix *M*_{p}.

It turns out

How might you remember or derive this matrix? Consider the matrix on the left below. It’s easier to see the pattern here than in *M*_{p}.

You can derive *M*_{p} from this matrix.

Let’s look at the second row, for example. The second row of *M*_{p}, when multiplied by *q* as a column vector, produces the *i* component of the product.

How do you get an *i* term in the product? By multiplying the *i* component of *p* by the real component of *q*, or by multiplying the real component of *p* times the *i* component of *p*, or by multiplying the *i*/ *j* component of *p* by the *j* component of *q*, or by multiplying the *i*/*k* component of *p* by the *k* component of *q*.

The other rows follow the same pattern. To get the *x* component of the product, you add up the products of the *x*/*y* term of *p* and the *y* term of *q*. Here *x* and *y* range over

{1, *i*, *j*, *k*}.

To get *M*_{p} from the matrix on the right, replace 1 with the real component of *p*, replace *i* with the *i* component of *p*, etc.

As a final note, notice that the off-diagonal elements of *M*_{p} are anti-symmetric:

*m*_{ij} = –*m*_{ji}

unless *i* = *j*.

Coming late to the game here, but this paper

https://ieeexplore.ieee.org/ielx7/8860/4359257/09415699.pdf?tp=&arnumber=9415699&isnumber=4359257&ref=aHR0cHM6Ly9zY2hvbGFyLmdvb2dsZS5jb20v

detailing the math and functionality of a cross-spherical 3D gear might have some practical application questions addressable by the working mathematician.

(for what it’s worth it’s absolutely on fire on LinkedIn)