simultaneous triangularisation of commuting matrices over any field


Let 𝐞i denote the (column) vector whose ith position is 1 and where all other positions are 0. Denote by [n] the set {1,,n}. Denote by Mn(𝒦) the set of all n×n matrices over 𝒦, and by GLn(𝒦) the set of all invertible elements of Mn(𝒦). Let di be the function which extracts the ith diagonal element of a matrix, i.e., di(A)=𝐞iTA𝐞i.

Theorem.

Let K be a field, let A1,,ArMn(K) be pairwise commuting matricesMathworldPlanetmath, and let L be a field extension of K in which the characteristic polynomialsMathworldPlanetmathPlanetmath of all Ak split (http://planetmath.org/SplittingField). Then there exists some PGLn(L) such that

  1. 1.

    P-1AkP is upper triangular for all k=1,,r, and

  2. 2.

    if i,j,l[n] are such that ilj and di(P-1AkP)=dj(P-1AkP) for all k=1,,r, then dl(P-1AkP)=dj(P-1AkP) for all k=1,,r as well.

The proof relies on two lemmas.

Lemma 1.

Let K be a field, let A1,,ArMn(K) be pairwise commuting matrices, and let L be a field extension of K in which the characteristic polynomials of all Ak split. Then there exists some nonzero uLn which is an eigenvectorMathworldPlanetmathPlanetmathPlanetmath of Ak for all k=1,,r.

Lemma 2.

For any sequenceMathworldPlanetmath R1,,RrMn(L) of upper triangular pairwise commuting matrices and every row index i[n], there exists vLn{0} such that

Rk𝐯=di(Rk)𝐯for all k[r].
Proof.

This is by inductionMathworldPlanetmath on n. The induction hypothesis is that given pairwise commuting matrices A1,,ArMn(), whose characteristic polynomials all split in , and a sequence of arbitrary scalars μ1,,μr, there exists some PGLn() such that:

  1. 1.

    P-1AkP is upper triangular for all k=1,,r.

  2. 2.

    If some i,j[n] are such that i<j and dj(P-1AkP)=di(P-1AkP) for all k[r], then di+1(P-1AkP)=di(P-1AkP).

  3. 3.

    If some j[n] is such that dj(P-1AkP)=μk for all k[r], then d1(P-1AkP)=μk for all k[r].

For n=1 this hypothesisMathworldPlanetmathPlanetmath is trivially fulfilled (all 1×1 matrices are upper triangular). Assume that it holds for n=m and consider the case n=m+1.

It is easy to see that condition 1 implies that P𝐞1 must be an eigenvector that is common to all the matrices. If there exists a nonzero vector 𝐮1n such that Ak𝐮1=μk𝐮1 for all k=1,,r then this is such a common eigenvector, and in that case let λk=μk for all k=1,,r. Otherwise there by Lemma 1 exists a vector 𝐮1n{𝟎} such that Ak𝐮1=λk𝐮1 for some {λk}k=1r. Either way, one gets a suitable candidate 𝐮1 for P𝐞1 and eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath λ1,,λr that incidentally will satisfy d1(P-1AkP)=λk for all k[r].

Let 𝐮2,,𝐮nn be arbitrary vectors such that {𝐮i}i=1n is a basis of n. Let U be the n×n matrix whose ith column is 𝐮i for 1in.11By imposing extra conditions on the choice of the basis {𝐮i}i=1n (such as for example requesting that it is orthonormal) at this point, one can often prove a stronger claim where the choice of P is restricted to some smaller group of matrices (for example the group of orthogonal matricesMathworldPlanetmath), but this requires assuming additional things about the fields 𝒦 and . Then U is invertiblePlanetmathPlanetmathPlanetmath and for each k the first column of Bk=U-1AkU is

U-1AkU𝐞1=U-1Ak𝐮1=λkU-1𝐮1=λk𝐞1.

Furthermore

BjBk=U-1AjUU-1AkU=U-1AjAkU==U-1AkAjU=U-1AkUU-1AjU=BkBj

for all j and k.

Now let Ak be the matrix formed from rows and columns 2 though n of Bk. Since det(Ak-xI)=det(Bk-xI)=(λk-x)det(Ak-xI) by expansion (http://planetmath.org/LaplaceExpansion) along the first column, it follows that the characteristic polynomial of Ak splits in . Furthermore all the Ak have side m=n-1 and commute pairwise with each other, whence by the induction hypothesis there exists some PGLn-1() such that every P-1AkP is upper triangular. Let P=U(100P). Then the submatrixMathworldPlanetmath consisting of rows and columns 2 through n of P-1AkP is equal to P-1AkP and hence contains no nonzero subdiagonal elements. Furthermore the first column of P-1AkP is equal to the first column of Bk and thus the P-1AkP are all upper triangular, as claimed.

It also follows from the induction hypothesis that P can be chosen such that d2(P-1AkP)=d1(P-1AkP)=λk=d1(P-1AkP) for all k[r] if there is any j2 for which dj(P-1AkP)=dj-1(P-1AkP)=λk=d1(P-1AkP) for all k[r] and more generally if 2i<j are such that dj(P-1AkP)=di(P-1AkP) for all k[r] then similarly di+1(P-1AkP)=di(P-1AkP) for all k[r]. This has verified condition 2 of the induction hypothesis. For the remaining condition 3, one may first observe that if there is some i[n] such that di(P-1AkP)=μk for all k[r] then by Lemma 2 there exists a nonzero 𝐯n such that P-1AkP𝐯=μk𝐯 for all k[r]. This means P𝐯 will fulfill the condition for choice of 𝐮1, and hence d1(P-1AkP)=λk=μk as claimed.

The theoremMathworldPlanetmath now follows from the principle of induction. ∎

Title simultaneous triangularisation of commuting matrices over any field
Canonical name SimultaneousTriangularisationOfCommutingMatricesOverAnyField
Date of creation 2013-03-22 15:29:38
Last modified on 2013-03-22 15:29:38
Owner lars_h (9802)
Last modified by lars_h (9802)
Numerical id 4
Author lars_h (9802)
Entry type Theorem
Classification msc 15A21
Related topic CommutingMatrices