A Hermitian-definite pencil has a real spectrum and a B-orthonormal eigenbasis

By [F2], $L$ is lower triangular with positive diagonal. [F3] gives $\det L\mathrm{\ne }0$, and [F4] therefore makes $L$ invertible. Hence $C=L^{-1}A{L}^{-\ast }$ is defined. Taking adjoints and using $A=A^{\ast }$ gives $C^{\ast }=L^{-1}A{L}^{-\ast }=C$.

Regard $C$ as a real or complex endomorphism with the Euclidean inner product. In the real case [F5] applies; in the complex case $C=C^{\ast }$ implies $C$ is normal, so [F6] applies. Thus there are orthonormal $y_i$ and real ${\lambda }_i$ with $C{y}_i={\lambda }_i{y}_i$.

Set $x_i=L^{-\ast }{y}_i$. Then $A{x}_i=A{L}^{-\ast }{y}_i=LC{y}_i={\lambda }_{i}L{y}_i={\lambda }_{i}B{x}_i,$ and $x_i^{\ast }B{x}_j=y_i^{\ast }{L}^{-1}L{L}^{\ast }{L}^{-\ast }{y}_j=y_i^{\ast }{y}_j={\delta }_{ij}\mathrm{.}$ The invertible change of variables $L^{-\ast }$ sends the basis $(y_i)$ to a basis $(x_i)$.

For every $\lambda \in \mathbb{C}$, $A-\lambda B=L(C-\lambda I)L^{\ast }$, so $\det (A-\lambda B)=\det (L)\det (C-\lambda I)\det (L^{\ast })\mathrm{.}$ The diagonalisation from step 2.1 expresses $C=UD{U}^{\ast }$ with $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ and $U$ unitary. By [F7], $\det (C-\lambda I)=\det (U)\det (D-\lambda I)\det (U^{\ast })$ and $\det (U)\det (U^{\ast })=\det (U{U}^{\ast })=\det I=1$. Applying [F3] to the diagonal matrix gives $\det (C-\lambda I)={\prod }_i({\lambda }_i-\lambda )$. Also $\det (L)$ and $\det (L^{\ast })$ are nonzero by step 1.1, so the finite roots of $\det (A-\lambda B)$ are exactly the real ${\lambda }_i$. Finally $\det (B)=\det (L)\det (L^{\ast })\mathrm{\ne }0$, so $[1:0]$ is not a homogeneous root.