Significance of Rayleigh quotient of a non-hermitian matrix? I've got a non-hermitian, but diagonalizable matrix $A$ (with respect to a matrix $K$); all its eigenvalues are real, but they need not be simple.
At the same time, the quadratic form associated with $A$ is positive-valued and it dominates the quadratic form of another, Hermitian matrix $B$. Can one conclude that the eigenvalues of $A$ are greater/equal than the eigenvalues of $B$? (Actually, I am only interested in the lowest eigenvalue.) Can I enforce this behavior by a suitable assumption on the diagonalizing matrix $K$?

By saying the quadratic form associated with $A$ dominates the quadratic form of $B$, I presume you mean $v^* A v \ge v^* B v$ for all vectors $v$.
If $v$ is an eigenvector for $A$ with eigenvalue $\lambda$, normalized so $\|v\|=1$, then
$v^* B v \le v^* A v = \lambda$. That implies the lowest eigenvalue of $B$ is at most $\lambda$.
