Matrix Representation of Octonions 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?
 A: 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)
A: 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).  
A: 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.
A: 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.
A: I tried to convert different dimensional base units to matrices. The reals are trivial, the complexes are easy, the quaternions are interesting, but solvable and the octonions are so tricky, that it does not work. My matrix looks like this:
$ M_{n m} = i_x^{o-1} o (i_n i_m)$
So e. g.
M = $ \begin{pmatrix} 0 & -1 & -2 & -3 \\ 1 & 0 & -3 & 2 \\ 2 & 3 & 0 & -1 \\ 3 & -2 & 1 & 0\end{pmatrix}$
for the quaternions, it can be for the octonions, too.
But to make sense of all these matrices, we should convert it to another (in this case 4) matrices with full of 0, 1 and -1 numbers. And if we multiplicate them, we get each other, right? I tested it, and it works at quaternions, but so does not at octonions. I am really sorry.
Maybe the multiplication table of octonions is wrong or the previous person was right talking about its problematics of the associativity.
