simultaneous triangularisation of commuting matrices over any field
Let denote the (column) vector whose th position is and where all other positions are . Denote by the set . Denote by the set of all matrices over , and by the set of all invertible elements of . Let be the function which extracts the th diagonal element of a matrix, i.e., .
Let be a field, let be pairwise commuting matrices, and let be a field extension of in which the characteristic polynomials of all split (http://planetmath.org/SplittingField). Then there exists some such that
is upper triangular for all , and
if are such that and for all , then for all as well.
The proof relies on two lemmas.
Let be a field, let be pairwise commuting matrices, and let be a field extension of in which the characteristic polynomials of all split. Then there exists some nonzero which is an eigenvector of for all .
For any sequence of upper triangular pairwise commuting matrices and every row index , there exists such that
This is by induction on . The induction hypothesis is that given pairwise commuting matrices , whose characteristic polynomials all split in , and a sequence of arbitrary scalars , there exists some such that:
is upper triangular for all .
If some are such that and for all , then .
If some is such that for all , then for all .
For this hypothesis is trivially fulfilled (all matrices are upper triangular). Assume that it holds for and consider the case .
It is easy to see that condition 1 implies that must be an eigenvector that is common to all the matrices. If there exists a nonzero vector such that for all then this is such a common eigenvector, and in that case let for all . Otherwise there by Lemma 1 exists a vector such that for some . Either way, one gets a suitable candidate for and eigenvalues that incidentally will satisfy for all .
Let be arbitrary vectors such that is a basis of . Let be the matrix whose th column is for .11By imposing extra conditions on the choice of the basis (such as for example requesting that it is orthonormal) at this point, one can often prove a stronger claim where the choice of is restricted to some smaller group of matrices (for example the group of orthogonal matrices), but this requires assuming additional things about the fields and . Then is invertible and for each the first column of is
for all and .
Now let be the matrix formed from rows and columns though of . Since by expansion (http://planetmath.org/LaplaceExpansion) along the first column, it follows that the characteristic polynomial of splits in . Furthermore all the have side and commute pairwise with each other, whence by the induction hypothesis there exists some such that every is upper triangular. Let . Then the submatrix consisting of rows and columns through of is equal to and hence contains no nonzero subdiagonal elements. Furthermore the first column of is equal to the first column of and thus the are all upper triangular, as claimed.
It also follows from the induction hypothesis that can be chosen such that for all if there is any for which for all and more generally if are such that for all then similarly for all . This has verified condition 2 of the induction hypothesis. For the remaining condition 3, one may first observe that if there is some such that for all then by Lemma 2 there exists a nonzero such that for all . This means will fulfill the condition for choice of , and hence as claimed.
The theorem now follows from the principle of induction. ∎
|Title||simultaneous triangularisation of commuting matrices over any field|
|Date of creation||2013-03-22 15:29:38|
|Last modified on||2013-03-22 15:29:38|
|Last modified by||lars_h (9802)|