i.e. $$\label{eq:EVP}
  \lambda\phi = {\mathbf{C}}\phi = (i{\mathbf{C}}_1 +{\mathbf{C}}_2) \phi \ .$$ Take the complex inner product of this equation with $\phi$, to obtain $$\overline{\lambda} \langle\phi,\phi\rangle = \langle {\mathbf{C}}\phi, \phi\rangle \ ,$$ using $\langle\phi,\psi\rangle = \overline{\phi}^\top \psi$ for all $\phi,\psi\in{\mathbb{C}}^n$. Its real part satisfies $$\label{eq:Re:lambda}
  \Re (\lambda) \langle\phi,\phi\rangle = \langle{\mathbf{C}}_2 \phi,\phi\rangle \ ,$$ due to the assumptions on the matrices ${\mathbf{C}}_1$ and ${\mathbf{C}}_2$. Moreover, there exists a skew-Hermitian matrix ${\mathbf{K}}$ such that ${\mathbf{C}}_2 +[{\mathbf{K}},{\mathbf{C}}_1]$ is positive definite by Proposition \[prop:equivalence\]. We multiply equation  with $i{\mathbf{K}}$ and take the inner product with $\phi$ such that $$\overline{\lambda} \langle i{\mathbf{K}}\phi,\phi\rangle = \langle i{\mathbf{K}}{\mathbf{C}}\phi,\phi\rangle \ .$$ Its real part satisfies $$\label{eq:Re:lambda:2}
  2\Re (\lambda) \langle i{\mathbf{K}}\phi,\phi\rangle = \langle({\mathbf{C}}_1 {\mathbf{K}}-{\mathbf{K}}{\mathbf{C}}_1)\phi,\phi\rangle - i\langle({\mathbf{C}}_2 {\mathbf{K}}+{\mathbf{K}}{\mathbf{C}}_2)\phi,\phi\rangle \ ,$$ since ${\mathbf{C}}_1$, ${\mathbf{C}}_2$ and $i{\mathbf{K}}$ are Hermitian matrices. Moreover, $$\begin{gathered}
 \label{est:1}
  2\Re (\langle i{\mathbf{K}}{\mathbf{C}}_2 \phi,\phi\rangle)
   = \langle({\mathbf{C}}_2 \ i{\mathbf{K}}+ i{\mathbf{K}}{\mathbf{C}}_2)\phi,\phi\rangle
   = \langle \sqrt{{\mathbf{C}}_2}\ i{\mathbf{K}}\phi,\sqrt{{\mathbf{C}}_2}\phi\rangle +\langle \sqrt{{\mathbf{C}}_2}\phi,\sqrt{{\mathbf{C}}_2}\ i{\mathbf{K}}\phi\rangle \\
     \leq 2{\| {\sqrt{{\mathbf{C}}_2}\phi} \|} {\| {\sqrt{{\mathbf{C}}_2}\ i{\mathbf{K}}\phi} \|}
   \leq 2M {\| {\sqrt{{\mathbf{C}}_2}\phi} \|} {\| {\phi} \|}
     \leq \epsilon {\| {\phi} \|}^2 +\frac{M^2}{\epsilon} \langle{\mathbf{C}}_2 \phi,\phi\rangle
 \end{gathered}$$ for any positive $\epsilon$. Here we used $M:=\|\sqrt{{\mathbf{C}}_2}\ i{\mathbf{K}}\|$ and ${\| {\sqrt{{\mathbf{C}}_2}\phi} \|}^2 = \langle{\mathbf{C}}_2 \phi,\phi\rangle$ since ${\mathbf{C}}_2\geq 0$. Combining equations  and  as $2\cdot$$-\alpha\cdot$ for some constant $\alpha>0$ to 