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)=𝐞iT⁢A⁢𝐞i.


Let K be a field, let A1,…,Ar∈Mn⁢(K) be pairwise commuting matricesMathworldPlanetmath, and let L be a field extension of K in which the characteristic polynomialsMathworldPlanetmathPlanetmath of all Ak split ( Then there exists some P∈GLn⁢(L) such that

  1. 1.

    P-1⁢Ak⁢P is upper triangular for all k=1,…,r, and

  2. 2.

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

The proof relies on two lemmas.

Lemma 1.

Let K be a field, let A1,…,Ar∈Mn⁢(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 u∈Ln which is an eigenvectorMathworldPlanetmathPlanetmathPlanetmath of Ak for all k=1,…,r.

Lemma 2.

For any sequenceMathworldPlanetmath R1,…,Rr∈Mn⁢(L) of upper triangular pairwise commuting matrices and every row index i∈[n], there exists v∈Ln∖{0} such that

Rk⁢𝐯=di⁢(Rk)⁢𝐯 for all k∈[r].

This is by inductionMathworldPlanetmath on n. The induction hypothesis is that given pairwise commuting matrices A1,…,Ar∈Mn⁢(ℒ), whose characteristic polynomials all split in ℒ, and a sequence of arbitrary scalars μ1,…,μr∈ℒ, there exists some P∈GLn⁢(ℒ) such that:

  1. 1.

    P-1⁢Ak⁢P is upper triangular for all k=1,…,r.

  2. 2.

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

  3. 3.

    If some j∈[n] is such that dj⁢(P-1⁢Ak⁢P)=μk for all k∈[r], then d1⁢(P-1⁢Ak⁢P)=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 P⁢𝐞1 must be an eigenvector that is common to all the matrices. If there exists a nonzero vector 𝐮1∈ℒn such that Ak⁢𝐮1=μk⁢𝐮1 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 𝐮1∈ℒn∖{𝟎} such that Ak⁢𝐮1=λk⁢𝐮1 for some {λk}k=1r⊆ℒ. Either way, one gets a suitable candidate 𝐮1 for P⁢𝐞1 and eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath λ1,…,λr that incidentally will satisfy d1⁢(P-1⁢Ak⁢P)=λk for all k∈[r].

Let 𝐮2,…,𝐮n∈ℒn be arbitrary vectors such that {𝐮i}i=1n is a basis of ℒn. Let U be the n×n matrix whose ith column is 𝐮i for 1⩽i⩽n.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 ℒ. Then U is invertiblePlanetmathPlanetmathPlanetmath and for each k the first column of Bk=U-1⁢Ak⁢U is




for all j and k.

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

It also follows from the induction hypothesis that P can be chosen such that d2⁢(P-1⁢Ak⁢P)=d1⁢(P′⁣-1⁢Ak′⁢P′)=λk=d1⁢(P-1⁢Ak⁢P) for all k∈[r] if there is any j⩾2 for which dj⁢(P-1⁢Ak⁢P)=dj-1⁢(P′⁣-1⁢Ak′⁢P′)=λk=d1⁢(P-1⁢Ak⁢P) for all k∈[r] and more generally if 2⩽i<j are such that dj⁢(P-1⁢Ak⁢P)=di⁢(P-1⁢Ak⁢P) for all k∈[r] then similarly di+1⁢(P-1⁢Ak⁢P)=di⁢(P-1⁢Ak⁢P) 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-1⁢Ak⁢P)=μk for all k∈[r] then by Lemma 2 there exists a nonzero 𝐯∈ℒn such that P-1⁢Ak⁢P⁢𝐯=μk⁢𝐯 for all k∈[r]. This means P⁢𝐯 will fulfill the condition for choice of 𝐮1, and hence d1⁢(P-1⁢Ak⁢P)=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