I have a block matrix over $\mathbb{R}$
$$T=\begin{bmatrix} A & PA \\\ -E & 0 \end{bmatrix}$$
Here $A$ is a $n\times n$ matrix, $P$ is a projector (i.e. $P^2=P$), $E$ is $n\times n$ identity matrix and $0$ is $n\times n$ zero block. $A$ is non-invertible, diagonalizable (with real eigenvalues) matrix but it is not diagonalizable simultaneously with $P$, and they do not commute.
Is it possible to say something about eigenvalues of $T$ with eigenvalues of $A$ being given? I don't think so, because the equation $\det(\lambda^2E-\lambda A -PA)=0$ for eigenvalues does not look promising and nothing more is coming to mind.
But suppose all eigenvalues of $A$ are nonnegative. Is it possible to conclude that real parts of eigenvalues of $T$ also all nonnegative?
For extreme cases $P=E$ or $P=0$ it is true, the first one gives $\frac{\lambda\pm\sqrt{\lambda^2-4\lambda}}{2}$ as eigenvalues of $T$, the second one just gives $\lambda$. Is it also true for general $P$?