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Under what conditions for the real-valued coefficients of v is the product qv equal to vq where both q and v are quaternions?
One way to answer this question is from first principles. Let
q = a + bi + cj + dk
and
v = e + fi + gj + hk
Then,
qv
= ae - bf - cg - dh
+ i(af + be + ch - dg)
+ j(ag - bh + ce + df)
+ k(ah + bg - cf + de)
= ae - bf - cg - dh + afi + bei + agj +cej + ahk + dek
+ i(ch - dg)
+ j(df - bh)
+ k(bg - cf)
and (keeping in mind that reals commute under multiplication)
vq
= ae - bf - cg - dh
+ i(af + be - ch + dg)
+ j(ag + bh + ce - df)
+ k(ah - bg + cf + de)
= ae - bf - cg - dh + afi + bei + agj + cej + ahk + dek
- i(ch-dg)
- j(df-bh)
- k(bg-cf)
Let m = ae - bf - cg - dh + afi + bei + agj +cej + ahk + dek and n = i(ch - dg) + j(df - bh) + k(bg - cf). Then,
qv = m + n
and
vq = m - n.
For qv to equal vq, we must have n = 0. For n to equal zero, each of its parts must equal zero also, that is,
ch = dg
df = bh
bg = cf
so that either
c/g = d/h
d/h = b/f
b/f = c/g
i.e. (bi + cj + dk) is a multiple (not necessarily integer) of (fi + gj + hk)
or
b = 0, c = 0, d = 0,
or
f = 0, g = 0, h = 0.
Therefore, the nonreal part of v must be a multiple (possibly zero) of the nonreal part of q. More concisely, quaternions q and v commute under multiplication if and only if
v = pq + r
where p and r are real.
A neater proof, requiring more background, can also be given.
Quaternions are often regarded as the combination of a scalar part (the real part of the quaternion) and the vector part (the part containing i, j, and k.) That is, the quaternion a + bi + cj + dk corresponds to the combination of the scalar a and the vector
qv = ac + ad + bc - b·d + b×d.
In the product qv, the order of the factors in each of these five products is reversed. That is, the product is now
vq = ca + da + cb - d·b + d×b.
The first four products commute, whereas the last (a cross product) does not. Therefore,
vq = ac + ad + bc - b·d + d×b.
so that qv = vq if and only if b×d = d×b. It is known that the cross product is anticommutative, meaning that b×d = -d×b. Therefore, this condition reduces to b×d = θ, the zero vector, meaning that one is a scalar multiple of the other (i.e. they lie along the same line.) This gives the same result as the purely algebraic derivation above.
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