Matrix representation of the automorphisms of the octonion's algebra without Lie's theory It is known that if a function $g$ is an automorphism of the algebra of octonions then there is an orthogonal basis of a form: $1,e_1, e_2, e_3=e_1e_2, e_4, e_5=e_1e_4, e_6=e_2e_4, e_7=e_3e_4$, where all $e_1,...,e_7$ are in $Im \mathbb O$, $e_2 \bot e_1$ and $e_4 \bot e_1,e_2, e_1e_2$ and there are $\phi_1, \phi_2, \phi_3$ with sum $\pi$ such that 
$$
g(e_1)=e_1,\\
g(e_2)=\cos \phi_1  e_2+\sin \phi_1 e_3, \\
g(e_3)=-\sin \phi_1  e_2+\cos \phi_1 e_3,\\
g(e_4)=\cos \phi_2 e_4+\sin \phi_2 e_5, \\
g(e_5)=-\sin \phi_2 e_4+\cos \phi_2 e_5,\\
g(e_6)=\cos \phi_3 e_6+\sin \phi_3 e_7, \\
g(e_7)=-\sin \phi_3 e_6+\cos \phi_3 e_7.
$$
How to prove this fact using the theory of composition algebras and linear algebra, but not using Lie groups nor Lie algebras? (It is clear that $g$ have to be isometry, $g(x)=x$ for $x\in \mathbb R$, $g(Im \mathbb O)=Im \mathbb O$).
 A: The answer for this question can be concluded from my answer on this question:
 About some property of automorphism of octonions
Does it use Lie theory ? We have used the fact that orthogonal mapping on $\mathbb R^7$ has fixed point. This fact can be proved using linear algebra, i believe.
It remains to prove that sum of the angles is $\pi$. I thought that sum of the angles should be zero.
To prove this we need to consider basis $\langle 1, i, u, iu \rangle$ and products of these by $v$ (notation from my answer to the other question; $i,u,v$ are perpendicular imaginary unit octonions and $v$ is perpendicular also to $iu$). 
EDIT 2017-08-20
One idea is to use following formula for octonion multiplication. Let's define octonions as pairs $(a,\mathbf v)$ where $a$ is complex number and $\mathbf v$ vector in $\Bbb C^3$. Then octonion multiplication can be defined as $$(a,\mathbf v)(b,\mathbf w)=(ab-\mathbf {v\cdot w},a\mathbf w+\bar b \mathbf v + \mathbf {v \times w})$$
It can be proved by applying twice Cayley-Dickson formula and complex conjugation to last coefficient. BTW, I have placed this formula already in one question on MO, but it was closed by administrators. I would like to have this formula on MO, so I don't forget it.
We can see that complex cross product is preserved by $SU_3$.
Second idea is to observe that subgroup of $G_2$ which fix $i$ is generated by set $M=\{(L_iR_x)^2:x \text{ imaginary perpendicular to i}\}$.
By $L_i$ I denote matrix of left multiplication by octonion $i$, $R_x$ denote right multiplication by octonion $x$. Element $(L_iR_x)^2$ is identity on quaternion subalgebra $\langle i, x\rangle$ and minus identity on perpendicular 4-space. Set $M$ is topologically $\mathbb CP^2$ and generates $SU_3$. 
