Let$\DeclareMathOperator\Ker{Ker}$Let $\mathbb O'$ be the split octonion algebra over $\mathbb R$. For each nonzero divisor of zero $x\in \mathbb O'$ $\mbox{($x \neq 0, N(x)=0$)}$($x \neq 0$, $N(x)=0$) the kernel of the left multiplication by $x$, $Ker L_x=\{y\in \mathbb O': xy=0\}$$\Ker L_x=\{y\in \mathbb O': xy=0\}$ is a maximal totally isotropic subspace in $\mathbb O'$ and is equal to $\overline{x} \mathbb O'$ (since both $Ker L_x$$\Ker L_x$, $\overline{x} \mathbb O'$ are totally isotropic (hence with dimension $\leq 4$) and $dim Ker L_x+Im L_x=8$$\dim \Ker L_x+\operatorname{Im} L_x=8$). Similarly, $Ker R_x=\{y\in \mathbb O': yx=0\}=\mathbb O' \overline{x}$$\Ker R_x=\{y\in \mathbb O': yx=0\}=\mathbb O' \overline{x}$ is maximal totally isotropcisotropic, where $R_x(y)=yx$ for $y\in \mathbb O'$.
My question: is it maybe true that for each maximally totally isotropic $V\subset \mathbb O'$ there is a nonzero zero's divisor $x$ in $\mathbb O'$ such that $V=x\mathbb O'$ or $V=\mathbb O' x$? Is $x$ unique up to nonzero real constant?
