Matrix representation of the Quaternions?

Question

Can anyone explain how why the matrix representation of the quaternions using real matrices is constructed as such?

Answer 1

I am not sure exactly what you mean by asking "... why the ...". "Why" questions can be hard to answer satisfactory in math.

The claim is that the Quaternions $\mathbb{H}$ are isomorphic (as $\mathbb{R}$-algebras) to the given set of matrices. The isomorphism looks like this:

$$ \phi: a + bi + cj + dk \longmapsto \begin{pmatrix}a & b & c & d \\ -b & a & -d & c \\ -c &d &a& - b\\ -d& -c & b& a\end{pmatrix}. $$ To "understand" why this is true, you "simply" check that this is an isomorphism.

You check for example that $\phi$ is bijective, which is clear from the construction.

Then you check that $\phi$ is an algebra homomorphism, so you need for $x,y\in \mathbb{H}$ and $\lambda \in \mathbb{R}$:

1. $\phi(xy) = \phi(x)\phi(y)$ for $x,y\in\mathbb{H}$
2. $\phi(x+y) = \phi(x) + \phi(y)$
3. $\phi(\lambda x) = \lambda\phi(x)$

The last two are not difficult to check. The first one requires a bit of work.

Even though this does not answer the minus signs are where there are in the matrix, I highly recommend that you try to prove that $\phi$ is a homomorphism. This exercise will make you more familiar with the Quaternions.

But note if you check property $3$ above you would need (as a special case) $$ \phi((bi)(bi)) = \phi(ib)\phi(ib). $$ That is you would need $$ \begin{pmatrix} -b^2 & 0 & 0& 0 \\ 0 & -b^2 & 0 & 0 \\ 0 &0 &-b^2& 0\\ 0& 0 & 0& -b^2\end{pmatrix} = \begin{pmatrix} 0 & b & 0& 0 \\ -b & 0 & 0 & 0 \\ 0 &0 &0& -b\\ 0& 0 & b& 0\end{pmatrix}\begin{pmatrix} 0 & b & 0& 0 \\ -b & 0 & 0 & 0 \\ 0 &0 &0& -b\\ 0& 0 & b& 0\end{pmatrix}. $$ So here you can see that you "need" the minus on all the $b$'s. In this case it comes down to the fact that $i^2 = -1$.