I am reading up on the generalized Schur decomposition as a means to solve the generalized eigenvalue problem

$A\nu = \lambda B \nu $,

With $A$ and $B$ matrices, $\lambda$ the eigenvalues and $\nu$ the eigenvectors. I have understood so far that the decomposition occurs as the following

$A = LRZ^\text{T}$,

$B = LSZ^\text{T}$,

where ${L}$, ${Z}$ are unitary and ${R}$, ${S}$ are upper triangular and represent the Schur forms of ${A}$ and ${B}$ respectively. I understand where the eigenvalues come from, but something I am unclear on is how to obtain the corresponding generalized eigenvectors, $\nu$ using this decomposed form?

  • $\begingroup$ The generalized eigenvectors make the kernel of $A-\lambda B$. $\endgroup$ – Berci Mar 5 '18 at 15:52
  • $\begingroup$ I think I see it now. Convert $A-\lambda B$ to row-echelon form and then back substitute? $\endgroup$ – Yeti Mar 5 '18 at 16:09

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