Building off of this post, What are the relations between eigenvectors of $A$ and its adjoint $A^*$?
How can I derive the bi-orthogonality condition for a generalized eigenvalue problem?
Generalized eigenvalue problem:
$Ax = \lambda Bx$
Its adjoint:
$A^H y = \mu B^H y$.
An inner product is: $<a|b> \equiv a^H b$
$A, B$ are complex square matrices. $H$ is the complex conjugate transpose operation.
My attempt:
- The adjoint problem must satisfy: $<y|(A-\lambda B)x>=<(A^H-\lambda^* B^H)y|x>$ which equals zero because $(A-\lambda B)x = 0$
- Another way to see the adjoint problem is: $<(A^H -\mu B^H)y|x>=<y|(A-\mu^* B) x> = 0$
- By inspection, we can equate $<(A^H-\lambda^* B^H)y|x>=<(A^H-\mu B^H)y|x>$
- We can see that $y^H A x$ cancels out from above
- The remaining term is: $y^H (\mu^* - \lambda) Bx=0$
- This is equivalent to $<y|(\mu^* -\lambda)Bx>=<(\mu-\lambda^*)y|Bx>$
- Finally, the bi-orthogonality condition is $(\mu-\lambda^*)y^H Bx = (\mu-\lambda^*)<y|Bx>=0$
- This demonstrates that if the typical and adjoint eigenproblems are connected by a shared eigenvalue, the eigenvectors are not orthogonal. However, if the eigenvectors are not connected by a shared eigenvalue, then the eigenvectors are orthogonal. This also shows that the typical and adjoint eigenvalues are different by a complex conjugate.