Quaternion product as a matrix product

Posted on by John

Pick a quaternion

p = p0 + p1i + p2j + p3k

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 R4 then this is a linear function of q, and so it can be represented by multiplication by a 4 × 4 matrix Mp.

It turns out

M_p = \begin{bmatrix} p_0 & {-}p_1 & {-}p_2 & {-}p_3 \\ p_1 & \phantom{-}p_0 & {-}p_3 & \phantom{-}p_2 \\ p_2 & \phantom{-}p_3 & \phantom{-}p_0 & {-}p_1 \\ p_3 & {-}p_2 & \phantom{-}p_1 & \phantom{-}p_0 \\ \end{bmatrix}

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 Mp.

\begin{pmatrix} 1/1 & 1/i & 1/j & 1/k \\ i/1 & i/i & i/j & i/k \\ j/1 & j/i & j/j & j/k \\ k/1 & k/i & k/j & k/k \\ \end{pmatrix} = \begin{bmatrix} 1 & {-}i & {-}j & {-}k \\ i & \phantom{-}1 & {-}k & \phantom{-}j \\ j & \phantom{-}k & \phantom{-}1 & {-}i \\ k & {-}j & \phantom{-}i & \phantom{-}1 \\ \end{bmatrix}

You can derive Mp from this matrix.

Let’s look at the second row, for example. The second row of Mp, 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 Mp 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 Mp are anti-symmetric:

mij = –mji

unless i = j.

One thought on “Quaternion product as a matrix product”

Comments are closed.