Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Since quaternions $\mathbb{H}$ have a matrix representation as elements of $\text{SU}(2,\mathbb{C})$ as the following

$$ 1 \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad \mathrm i \mapsto \begin{pmatrix} \mathrm i_{\mathbb C} & 0 \\ 0 & -\mathrm i_{\mathbb C} \end{pmatrix},\quad \mathrm j \mapsto \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},\quad \mathrm k \mapsto \begin{pmatrix} 0 & \mathrm i_{\mathbb C} \\ \mathrm i_{\mathbb C} & 0 \end{pmatrix}, $$ I always wondered if there is also matrix representation of the octonions?

How is the non-associativity realised with matrices?

share|cite|improve this question
You shpuld probably be explicit about the sense in which quaternions have a matrix representation as SU(2,C)... H is an algebra and SU(2,C) is a group! – Mariano Suárez-Alvarez Jan 4 '12 at 18:47
Matrix multiplication is associative, so what do you hope for? – Grigory M Jan 4 '12 at 18:48
@GrigoryM: I don't hope, I wonder. So the answer is NO? – draks ... Jan 4 '12 at 19:01
What you mean, probably, is that there is a representation of $\mathbb H$ as a real subalgebra of the algebra $M_2(\mathbb C)$ of $2\times 2$ matrices. $\mathrm{SU}(2,\mathbb C)$ has little to do with it. – Mariano Suárez-Alvarez Jan 4 '12 at 19:08
@MarianoSuárez-Alvarez: Would $\text{su}(2,\mathbb{C})$ suit better? – draks ... Jan 4 '12 at 22:09
up vote 7 down vote accepted

A $\mathbb R$-linear function $\phi:\mathbb O\to A$ to any real associative algebra which is multiplicative and unitary (so that $\phi(xy)=\phi(x)\phi(y)$ for all $x$, $y\in \mathbb O$, and $\phi(1)=1$) has to vanish on the bilateral ideal $I$ generated by the elements of the form $$(x\cdot y)\cdot z-x\cdot(y\cdot z)$$ with $x$, $y$, $z\in\mathbb O$. Now, this ideal is not zero because $\mathbb O$ is not associative, and therefore $I=\mathbb O$, because $\mathbb O$ has no non-trivial bilateral ideals (it is a division thing).

It follows that the map $\phi$ is in fact zero.

share|cite|improve this answer

M. Zorn (Abh.Mat.Sem.Hamburg,9,395, 1933) modified matrix multiplication for 2x2 "vector-matrices" with two real-numbers as diagonal entries and two 3d-vectors off the diagonal.

See also Matrix Representation of Octonions and Generalizations by Jamil Daboul and Robert Delbourgo, (9 June 1999)

share|cite|improve this answer

The non-associativity is realized by representing 4 of the matrices as ket matrices. To multiply on one side versus the other the ket matrix needs to be converted to its bra representation. This is where the non-associativity comes in.

share|cite|improve this answer
With ket matrix you mean the matrix in a vectorized form, like putting coloumns on top of each other? – draks ... May 14 '14 at 12:24
Check out octonions on Wikipedia. You can see under the matrix section. There are 4 ket matrices. All other elements are regular matrices. – Scott May 14 '14 at 13:19
thanks, I missed that on the wiki page... – draks ... May 14 '14 at 13:21

We have two Octonions algebra, one is a division algebra (with no zero divisors) and a split algebra (with zero divisors). Both are constructed by Cayley-dicson process over a field $F$. If you pick the split algebra of octonions you can show that it is isomorphic to a non-assossiative matrix algebra called Zorn matrix-vector algebra, $M(F)$ where the elements are: $\left[ \begin{array}{cc} a &\textbf{x}\\ \textbf{y} & b \end{array}\right]$, where $\textbf{x,y}\in F^3$.

The addition is defined term by term and the multiplication defined as $\left[ \begin{array}{cc} a_1 &\textbf{x}_1\\ \textbf{y}_1 & b_1 \end{array}\right] \left[ \begin{array}{cc} a_2 &\textbf{x}_2\\ \textbf{y}_2 & b_2 \end{array}\right] = \left[ \begin{array}{cc} a_1a_2+\textbf{x}_1.\textbf{y}_2 & a_1\textbf{x}_2+b_2\textbf{x}_1-\textbf{y}_1\wedge\textbf{y}_2\\ a_2\textbf{y}_1+b_1\textbf{y}_2+\textbf{x}_1\wedge\textbf{x}_2 & b_1b_2+\textbf{y}_1.\textbf{x}_2 \end{array}\right]$. Where $\wedge$ denote the exterior product over $F^3$.

It is easy to show that it is non-assossiative. And, as we have the general linear group, the special linear group and the projective linear group over the assossiative matrix algebra, here we have the concept of general linear loop, special linear loop and projective linear loop over M(K).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.