I’ve the next cubic eigenvalue drawback, the place eigenvalue $lambda$ is taken into account within the nonlinear sense:

$$ { lambda^three start{bmatrix} I & 0 0 & I finish{bmatrix} + lambda^2 start{bmatrix} alpha I+beta D & -beta I beta I & alpha I+beta D finish{bmatrix} + lambda alpha beta start{bmatrix} D & -I-M^{-1}C 0 & D finish{bmatrix} + kalphabeta start{bmatrix} 0 & 0 M^{-1}C & Zero finish{bmatrix} } start{pmatrix} pmb{p} pmb{q} finish{pmatrix} = 0 , $$

the place $alpha, beta, okay$ are constructive scalars, $D,M$ are $ntimes n$ diagonal matrices with constructive components on its diagonals, $C$ is $ntimes n$ constructive semi-definite matrix (non-diagonal), $lambda neq 0$ is the advanced non-trivial eigenvalue and $(pmb{p}^T, pmb{q}^T)$ is the corresponding eigenvector.

For which matrix $D$ part angles of $pmb{p}$ are reverse to $pmb{q}$ (specifically, $p_k = |p_k|e^{iphi_k}$ and $q_k = |q_k|e^{-iphi_k}, okay=1,cdots,n$)?

I think it’s legitimate provided that $D = dI$, i.e. has the identical components on its diagonal. For instance, if $D=dI$, then $pmb{p}$ would be the eigenvector $pmb{psi}$ of $M^{-1}C$ that might be choosen to be actual and $pmb{q}= – frac{betalambda(lambda + alpha) + alphabeta mu}{lambda(lambda+alpha)(lambda+beta d)} pmb{psi}$, the place $mu$ is the correspoding eigenvalue of $M^{-1}C$ such that every components of $pmb{q}$ can have the identical part angle $phi_0 = angle frac{betalambda(lambda + alpha) + alphabeta mu}{lambda(lambda+alpha)(lambda+beta d)}$, and the pair $(pmb{p} = e^{-ifrac{phi_0}{2}}pmb{psi}, pmb{q}= – e^{-ifrac{phi_0}{2}}frac{betalambda(lambda + alpha) + alphabeta mu}{lambda(lambda+alpha)(lambda+beta d)} pmb{psi})$ can have the alternative part angles. I have no idea how one can present $D=dI$ is important or discover conterexample with $Dneq d I$.