Consequences of the properties of B on the results of a Generalized Eigenvalue $\lambda$ B v=Av I'm trying to find a good source for the consequences of the properties of the matrix $B$ for the generalized eigenvalue problem:
$\lambda B v = A v \Leftrightarrow \textrm{eig}(B^{-1}A)$
For example, for a general matrix $A$, if $B$ can be expressed as $B=\alpha I$ where $\alpha$ is a real value and $I$ the identity matrix, then
$\textrm{eig}(B^{-1}A) = \textrm{eig}((\alpha I)^{-1}A) = \frac{1}{\alpha}\textrm{eig}(A)$
(obviously, taking $\alpha =0$ would be silly)
Questions:


*

*What if $B$ is diagonal and positive definite?

*What if $B$ is positive definite?
I am particularly interested in knowing how the stability of $A$ (i.e. $\textrm{Re}[\textrm{eig}(A)]$) will be affected by premultiplying with $B^{-1}$.
Thank you so much.
 A: If $A$ is stable (positive/negative), that is, the real parts of its eigenvalues are positive/negative, then premultiplying it with a symmetric positive definite $B$ does not necessarily preserve the positiveness/negativeness of the eigenvalues. 
Consider matrices
$$
A = \begin{bmatrix}
1 & -5 \\ 0 & 1
\end{bmatrix},
\quad
B = 
\begin{bmatrix}
2 & -1 \\ -1 & 2
\end{bmatrix}.
$$
The matrix $A$ is positive stable (all its eigenvalues are equal to 1) and $B$ is positive definite. However, the eigenvalues of $B^{-1}A$ are
$$
-\frac{1}{6}\pm\frac{\sqrt{11}}{6}\mathrm{i}.
$$
For the diagonal $B$, you can construct a counter-example from the example above. The eigenvalue/eigenvector factorization of $B$ is
$$
B = VDV^T, \quad
V = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & -1\\1 & 1\end{bmatrix},
\quad
D=\begin{bmatrix}1 & 0 \\0 & 3\end{bmatrix}.
$$
So $B^{-1}A=UD^{-1}U^TA$ which has the same eigenvalues as $D^{-1}(U^TAU)$. The matrix 
$$
\tilde{A}=U^TAU=\frac{1}{2}\begin{bmatrix}-3&-5\\5&7\end{bmatrix}
$$
has again the eigenvalues equal to 1, $D$ is SPD diagonal, but the eigenvalues of $D^{-1}\tilde{A}$ are again
$$
-\frac{1}{6}\pm\frac{\sqrt{11}}{6}\mathrm{i}.
$$
NOTE: What is in a sense preserved is the positive real property (and negative equivalently). Sometimes the matrix $A$ is called positive real if $x^TAx>0$ for all nonzero $x$ ($A$ here need not to be symmetric). Then the matrix $B^{-1/2}AB^{-1/2}$, which is similar to $B^{-1}A$, is also positive real. Consequently, $B^{-1}A$ is positive stable.
If the matrix is positive real, then it is positively stable. Not vice versa though. The matrix $A$ above is not positive real, because taking $x=[1,1]^T$ gives $x^TAx=-3$.
