I am reading the Morse theoretic proof of the Lefschetz Hyperplane theorem in Griffiths-Harris and I am missing a transition. They claim that since the matrix $$\dfrac{1}{4}\left(\left(\dfrac{\partial^2}{\partial x_i\partial x_j}+\dfrac{\partial^2}{\partial y_i \partial y_j}\right) + i\left(\dfrac{\partial^2}{\partial x_i \partial y_j}-\dfrac{\partial^2}{\partial x_j\partial y_i}\right)\right)\log||s||^2$$
has $n$ negative eigenvalues, so does the Hessian $$H(\varphi)= \begin{pmatrix} \dfrac{\partial^2}{\partial x_i\partial x_j} & \dfrac{\partial^2}{\partial x_i \partial y_j}\\ \dfrac{\partial^2}{\partial x_j\partial y_i} & \dfrac{\partial^2}{\partial y_i \partial y_j} \end{pmatrix}\log ||s||^2$$ (there can be a permutation between the top right and bottom left corners).
I dont really understand why this is true. Any help would be appreciated.
Denote by $L=\left(\dfrac{\partial f}{\partial z_i\partial \bar{z_j}}\right)$ the Levi matrix of a smooth function $f$, by $H(f)$ its Hessian and by $X\in M_{2n,n}(\mathbb{C})$ the matrix \begin{equation*} X=\begin{pmatrix} Id\\ iId \end{pmatrix}. \end{equation*} We then have \begin{equation*} L=\overline{X}^tH(f)X \end{equation*} It follows that if $v\in\mathbb{C}^n$ verifies $\bar{v}^tLv = \lambda |v|^2$ for $\lambda\in\mathbb{R}\setminus \{0\}$ then \begin{eqnarray*} \overline{(Xv)}^tH(f)(Xv) &=& \bar{v}^tLv \\ &=& \lambda |v|^2\\ &=& \dfrac{\lambda}{2}|Xv|^2 \end{eqnarray*} We deduce that if $L$ is negative (positive) then $H(f)$ has $n$ negative (positive) eigenvalues.