# Generalized eigenvectors from left and right Schur vectors

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?

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