simultaneous block-diagonalization of upper triangular commuting matrices

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 1.

Let K be a field, let n be a positive integer, and let be an equivalence relationMathworldPlanetmath on [n] such that if ij and ikj then ki. Let A1,,ArMn(K) be pairwise commuting upper triangular matricesMathworldPlanetmath. If these matrices and are related such that

ijif and only ifdi(Ak)=dj(Ak) for all k[r],

then there exists a matrix BGLn(K) such that:

  1. 1.

    If 𝐞iTB-1AkB𝐞j0 then ij and ij.

  2. 2.

    If ij then 𝐞iTB-1AkB𝐞j=𝐞iTAk𝐞j.

Condition 1 says that if an element of B-1AkB is nonzero then both its row and column indices must belong to the same equivalence classMathworldPlanetmath of , i.e., the nonzero elements of B-1AkB only occur in particular blocks ( along the diagonal, and these blocks correspond to equivalence classes of . Condition 2 says that within one of these blocks, B-1AkB is equal to Ak.

The proof of the theorem requires the following lemma.

Lemma 2.

Let a sequence A1,,ArMn(K) of upper triangular matrices be given, and denote by A the unital ( algebraMathworldPlanetmathPlanetmath ( generated by these matrices. For every sequence λ1,,λrK of scalars there exists a matrix CA such that

di(C)={1if di(Ak)=λk for all k[r],0𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

for all i[n].

The proof of that lemma can be found in this article (

Proof of theorem.

The proof is by inductionMathworldPlanetmath on the number of equivalence classes of . If there is only one equivalence class then one can take B=I.

If there is more than one equivalence class, then let S be the equivalence class that contains n. By Lemma 2 there exists a matrix C in the unital algebra generated by A1,,Ar (hence necessarily upper triangular) such that di(C)=1 for all iS and di(C)=0 for all i[n]S. Thus C has a


where C22 is a |S|×|S| matrix that has all 1s on the diagonal, and C11 is a (n-|S|)×(n-|S|) matrix that has all 0s on the diagonal.

Let k[r] be arbitrary and similarly decompose

Cn= (D11D120D22), Ak= (A11A120A22).

One can identify D11=(C11)n and D22=(C22)n, but due to the zero diagonal of C11 and the fact that the of these matrices are smaller than n, the more striking equality D11=0 also holds. As for D22, one may conclude that it is invertiblePlanetmathPlanetmathPlanetmath.

Since the algebra that Cn belongs to was generated by pairwise commuting elements, it is a commutativePlanetmathPlanetmathPlanetmath ( algebra, and in particular CnAk=AkCn. In of the individual blocks, this becomes


Now let

D=(ID120D22), so that D-1=(I-D12D22-10D22-1)

and consider the matrix D-1AkD. Clearly


so that the positions with row not in S and column in S are all zero, as requested for B-1AkB. It should be observed that the choice of D is independent of k, and that the same D thus works for all the Ak.

In order to completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof, one applies the induction hypothesis to the restrictionPlanetmathPlanetmathPlanetmathPlanetmath of to [n]S and the corresponding submatricesMathworldPlanetmath of D-1AkD, which satisfy the same conditions but have one equivalence class less. This produces a block-diagonalising matrix B for these submatrices, and thus the sought B can be constructed as D(B00I). ∎

Title simultaneous block-diagonalization of upper triangular commuting matrices
Canonical name SimultaneousBlockdiagonalizationOfUpperTriangularCommutingMatrices
Date of creation 2013-03-22 15:29:42
Last modified on 2013-03-22 15:29:42
Owner lars_h (9802)
Last modified by lars_h (9802)
Numerical id 5
Author lars_h (9802)
Entry type Theorem
Classification msc 15A21