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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.

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    $\begingroup$ What do you mean by quaternionic matrices? Do you mean the subject of this question: math.stackexchange.com/q/1850231/11323 $\endgroup$ – Kimball Feb 17 '17 at 17:53
  • $\begingroup$ yes I have no Idea how this matrix is formulated from quaternion and what is the proof of this result. $\endgroup$ – baam Feb 18 '17 at 5:59
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    $\begingroup$ Multiplication from the left by a fixed quaternion $q$ is a mapping from $\Bbb{H}$ to itself. Such a mapping is linear over $\Bbb{C}$, if you view $\Bbb{H}$ as a vector space over $\Bbb{C}$ with $\Bbb{C}$ acting from the right. It is important to go from left to right as otherwise the two actions won't commute. If $h$ is an arbitrary quaternion and $z$ an arbitrary complex number, the linearity of left multiplication by $q$ amounts to the rule $$q(hz)=(qh)z,$$ which holds, because the multiplication of quaternions is associative. $\endgroup$ – Jyrki Lahtonen Feb 18 '17 at 8:55
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    $\begingroup$ @Kimball: You seem to understand what is going on. Do you mind fleshing that out to an answer. IMHO the current answers miss the point. Emilio Novati has the right matrices, but they seem to fall from heaven. $\endgroup$ – Jyrki Lahtonen Feb 18 '17 at 9:00
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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 $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{i}$ the isomorphism is given by: $$ z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{i} \quad \mapsto \quad \mathbf{Z}= a\mathbf{U}+b\mathbf{I}+c\mathbf{J}+d\mathbf{K} \quad a,b,c,d \in \mathbb{R} $$ 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 $$ z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{i} \quad \mapsto \quad \mathbf{Z}= \left( \begin{array}{ccccc} a+ib&c+id \\ -c+id &a-ib \end{array} \right) $$ with $$ \mbox{det}(\mathbf{Z})= \left | \left( \begin{array}{ccccc} a+ib&c+id \\ -c+id &a-ib \end{array} \right) \right |= a^2+b^2+c^2+d^2=|z|^2 $$

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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

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    $\begingroup$ It's really a more basic linear algebra question. $\endgroup$ – Kimball Feb 17 '17 at 17:51
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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 conjugation).

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