TITLE: Need help determining the pairs of quaternions that anticommute
QUESTION [6 upvotes]: I tried to solve another exercise and I would be grateful if someone could tell me if my answer is right. This is the exercise:

Characterize the pairs $p,q \in \mathbb H$ such that $pq = -qp$.

I can see that the unit quaternion paris $(i,k), (i,j), (j,k)$ anticommute.
Since the real part of two purely imaginary quaternions is the dot product and the imaginary part the vector product it follows that purely imaginary quaternions anticommute.
If $a+p, b+q$ are two quaternions with real parts $a,b$ then $(a+p)(b+q) = ab + aq + bp + pq$. The only way for this to anticommute is if $ab=0$. Therefore I conclude that all pairs of purely imaginary quaternions anticommute.
Is this correct?

REPLY [2 votes]: We write the product as 
$$ pq = (r_p r_q - v_p \cdot v_q, r_p v_q + r_q v_p + v_p \times v_q)$$
Considering the imaginary part we see that since the crossproduct is antisymmetric, $\operatorname{Im}{pq}=-\operatorname{Im}{qp}$ if and only if $r_pv_q + r_q v_p = 0$.
This can happen in exactly 3 ways: if either $r_p = r_q = 0$, if $v_p = v_q = 0$ or if $v_p, v_q$ are linearly dependent. 
If $r_p = r_q = 0$, $p,q$ are purely imaginary. Since the real part of the product of two purely imaginary quaternions is given by the dot product and the dot product is symmetric the real part commutes if and only if $v_p \cdot v_q = 0$. This happens if and only if the angle between the two vectors is $90$ degrees. Hence $v_p, v_q$ must be orthogonal if $r_p=r_q =0$. 
If $v_p = v_q = 0$ then we are multiplying real numbers. These anticommute if and only if they are zero. Hence in this case, $p=q=0$.
If $v_p,v_q$ are linearly dependent with
$$ r_p v_q + r_q v_p = 0$$
Then also
$$ r_p v_q \cdot v_p + r_q \|v_p\|^2 = 0$$
The real part $r_p r_q - v_p \cdot v_q $ vanishes if and only if $r_p r_q = v_p \cdot v_q$ which can happen only if
$$ r_p^2 r_q + r_q \|v_p\|^2 = r_q (r_p^2 + \|v_p\|^2)= 0$$
hence $r_q$ must be zero. Doing the same argument with $v_q$ yields $r_p=0$. Hence if $v_p$ and $v_q$ are linearly dependent the product $pq$ anticommutes if both real parts are zero. But in this case we have shown that $v_p$ and $v_q$ must be orthogonal, a contradiction to the linear dependence of $v_p,v_q$. Hence $v_p, v_q$ cannot be linearly dependent.