Considering a similar question than in Generalized eigenvalue problem with nonsymmetric, positive definite matrices: Let $A,B \in \mathbb{R}^{n\times n}$ be matrices, which
- have real entries,
- are positive definite (i.e. $\forall x \in \mathbb{R}\setminus\{0\}: x^\top A x > 0, x^\top B x>0$),
- are not symmetric.
Is the matrix $$ B^{-1}A $$ diagonalizable, i.e. does a regular matrix $T \in \mathbb C^{n\times n}$ exist, such that $$ B^{-1}A = T \Lambda_{B^{-1} A} T^{-1}$$ where $\Lambda_{B^{-1} A}$ is a diagonal matrix with the complex eigenvalues of $B^{-1} A?$
EDIT: In general, this statement is wrong for nonsymmetric $A$ and symmetric $B$, see comments. What assumptions on $A,B$ are needed?
Would it be true for $B$ being symmetric? In this case, one can use that $B$ is diagonalizable, i.e. $B= U \Lambda_{B} U^\top$ with orthogonal $U \in \mathbb R^{n \times n}.$ Moreover, the matrix $B^{-1} A$ is positive definite. How could we proceed?
Do you have any suggestions?
Many thanks in advance!