common eigenvector of a diagonal element cross-section
Let for all , so that the problem is to find a common eigenvector of whose corresponding eigenvalue for is . It is sufficient to find such a common eigenvector in the case that is the least for which for all , because if some smaller also has this property then one can solve the corresponding problem for the submatrices consisting of rows and columns through of , and then pad the common eigenvector of these submatrices with zeros to get a common eigenvector of the original .
By the existence of a characteristic matrix of a diagonal element cross-section (http://planetmath.org/CharacteristicMatrixOfDiagonalElementCrossSection) there exists a matrix in the unital algebra generated by such that if for all , and otherwise; in other words that matrix satisfies and for all . Since it is also upper triangular it follows that the matrix has rank (http://planetmath.org/RankLinearMapping) , so the kernel of this matrix is one-dimensional. Let be such that ; it is easy to see that this is always possible (indeed, the only vector in this nullspace with th is the zero vector). This is the wanted eigenvector.
To see that it is an eigenvector of , one may first observe that commutes with this , since the unital algebra of matrices to which belongs is commutative (http://planetmath.org/Commutative). This implies that since . As is one-dimensional it follows that for some . Since is upper triangular and this must furthermore satisfy , which is indeed what the eigenvalue was claimed to be. ∎
|Title||common eigenvector of a diagonal element cross-section|
|Date of creation||2013-03-22 15:30:41|
|Last modified on||2013-03-22 15:30:41|
|Last modified by||lars_h (9802)|