Can anyone explain how the complex matrix representation of a quaternions is constructed?

Question

I am reading some properties of quaternionic matrices and I am unable to understand how can we got such matrix representation. please help in this regards.

Answer 1

Your question is essentially the aim of representation theory: representing the elements of a group as matrices.

https://en.m.wikipedia.org/wiki/Group_representation

Answer 2

The ring of quaternions $\mathbb{H}$ is isomorphic to the ring of matrices with complex entries of the form $A =\begin{pmatrix} x & y \\ - \bar{y} & \bar{x} \end{pmatrix} $

For a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ the isomorphism is given by: $$ z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} \quad \mapsto \quad \mathbf{Z}= a\mathbf{U}+b\mathbf{I}+c\mathbf{J}+d\mathbf{K} \quad a,b,c,d \in \mathbb{R} $$ with: $$ \mathbf{U}= \left( \begin{array}{ccccc} 1&0 \\ 0 &1 \end{array} \right) \qquad \mathbf{I}= \left( \begin{array}{ccccc} i&0 \\ 0 &-i \end{array} \right) \qquad \mathbf{J}= \left( \begin{array}{ccccc} 0&1 \\ -1 &0 \end{array} \right) \qquad \mathbf{K}= \left( \begin{array}{ccccc} 0&i \\ i &0 \end{array} \right) $$ We can easy see that: $$ \mathbf{I}^2=\mathbf{J}^2=\mathbf{K}^2=\mathbf{I}\mathbf{J}\mathbf{K}=-\mathbf{U} $$ and $$ z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} \quad \mapsto \quad \mathbf{Z}= \left( \begin{array}{ccccc} a+ib&c+id \\ -c+id &a-ib \end{array} \right) $$ with $$ \mbox{det}(\mathbf{Z})= \left | \left( \begin{array}{ccccc} a+ib&c+id \\ -c+id &a-ib \end{array} \right) \right |= a^2+b^2+c^2+d^2=|z|^2 $$

Answer 3

The key is as Jyrki states in the comments: view $\mathbb{H}$ as a right vector space over $\mathbb{C}$. That means we apply scalars from the right, instead of the left. This way, if $q\in\mathbb{H}$ is a scalar and $L_q(x)=qx$ is the left-multiplication-by-$q$ map, it is $\mathbb{C}$-linear in the sense that $L_q(x\lambda)=L_q(x)\lambda$ for all $x\in\mathbb{C}$ and complex scalars $\lambda\in\mathbb{C}$. As a complex vector space, $\mathbb{H}$ is $2$-dimensional with basis $\{1,\mathbf{j}\}$.

With this idea, every $L_q$ corresponds to a $2\times 2$ complex matrix:

$$ \begin{cases} L_1(1)=1+\mathbf{j}0 \\ L_1(\mathbf{j})=0+\mathbf{j}1 \end{cases} \implies \quad L_1\leftrightarrow \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} $$

$$ \begin{cases} L_{\mathbf{i}}(1)=\mathbf{i}+0\mathbf{j} \\ L_{\mathbf{i}}(\mathbf{j})= 0-\mathbf{j}\mathbf{i} \end{cases} \implies \quad L_{\mathbf{i}}\leftrightarrow \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} $$

$$ \begin{cases} L_{\mathbf{j}}(1)=0+\mathbf{j}1 \\ L_{\mathbf{j}}(\mathbf{j})=-1+\mathbf{j}0 \end{cases} \implies \quad L_{\mathbf{j}}\leftrightarrow \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$

$$ \begin{cases} L_{\mathbf{k}}(1)=0-\mathbf{ji} \\ L_{\mathbf{k}}(\mathbf{j})=-\mathbf{i}+\mathbf{j}0 \end{cases} \implies \quad L_{\mathbf{k}}\leftrightarrow \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} $$

The correspondence $\mathbf{q}\leftrightarrow L_{\mathbf{q}}$ preserves addition and multiplication, hence is an $\mathbb{R}$-algebra homomorphism from $\mathbb{H}$ into $M_2(\mathbb{C})$. The image is then an isomorphic copy of $\mathbb{H}$.

Other conventions are possible. For instance, we could associate $\mathbf{q}\mapsto R_{\overline{\mathbf{q}}}$ (notice that conjugation).

Answer 4

I know this is an old question, but I'd like to share a (hopefully) more basic method of deriving the matrices.

Using the notation $U,I,J,K$ for the quarternions, we know $I$ is a root of $X^2+1=0$, so $I$ has an eigenvector $v$ with eigenvalue $i$. Now let $w=Jv$ (we want the quarternions to live in $GL_2(\mathbb{C})$ so we hope that span$\{v,w\}$ is fixed under $Q_8$).

$Iw=IJv=-JIv=-Jiv=-iw$

so $w$ is an eigenvector of $I$ with eigenvalue $-i$. In particular, it must be linearly independent from $v$. A quick check gives us:

$Jw=J^2v=-v$

$Kv=IJv=Iw=-iw$

$Kw=IJw=-Iv=-iv$

So the matrices of $U,I,J,K$ with basis $\{v,w\}$ are:

$$U=\begin{pmatrix}1&0\\0&1\end{pmatrix}\ I=\begin{pmatrix}i&0\\0&-i\end{pmatrix}\ J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\ K=\begin{pmatrix}0&-i\\-i&0\end{pmatrix}$$