I have an issue in understanding page 202 of Arnold's ODE book (3rd edition).
As shown in the image above, he proves the existence of a quadratic form $r^2$on $^\mathbf R \mathbb C^n$ such that $L_{^\mathbf R Az}r^2 \gt0$ (provided $z\neq 0$).
Here $L$ designates directional derivative and $^\mathbf R\mathbb C^n$ is the realification(decomplexification) of $\mathbb C^n$.
How is $L_{^\mathbf R Az}r^2$ defined? I guess that $^\mathbf R\mathbb C^n$ is identified with $\mathbb R^{2n}$ and $L_{^\mathbf R Az}r^2$ is defined to be the dot product of $^\mathbf RAz$ and $grad(r^2)$. Am I correct?
Why is (3) true? That is, why does$$L_{^\mathbf R Az}(z,\bar z)=(Az,\bar z)+(z,\bar {Az})$$hold?
For 1. Yes, in general the directional derivative can be defined by $\langle \mathrm{grad}(r^2),^\mathbb{R}Az \rangle$. For 2, see that he defines $r^2:=\sum^n_{i=1}z_k\bar{z}_k$. Then the "realified" directional derivative of $r^2$, with $A:\mathbb{C}^n\to \mathbb{C}^n$, becomes \begin{equation} \Big\langle \frac{\partial r^2}{\partial z},Az\Big\rangle+\Big\langle \frac{\partial r^2}{\partial \bar{z}},\bar{Az}\Big\rangle = \langle\bar{z},Az\rangle+\langle z,\bar{Az}\rangle. \end{equation} Note that this works since his proposed $(z,\bar{z})$ as a basis for $^\mathbb{R}\mathbb{C}^n$. To see that this makes sense, compare $(z+\bar{z})/2$ with $(z-\bar{z})/2$ for some $z\in \mathbb{C}$.