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)=𝐞iT⁢A⁢𝐞i.

Theorem 1.

Let K be a field, let n be a positive integer, and let ∼ be an equivalence relationMathworldPlanetmath on [n] such that if i∼j and i⩽k⩽j then k∼i. Let A1,…,Ar∈Mn⁢(K) be pairwise commuting upper triangular matricesMathworldPlanetmath. If these matrices and ∼ are related such that

i∼j if and only if di⁢(Ak)=dj⁢(Ak)⁢ for all ⁢k∈[r]⁢,

then there exists a matrix B∈GLn⁢(K) such that:

  1. 1.

    If 𝐞iT⁢B-1⁢Ak⁢B⁢𝐞j≠0 then i∼j and i⩽j.

  2. 2.

    If i∼j then 𝐞iT⁢B-1⁢Ak⁢B⁢𝐞j=𝐞iT⁢Ak⁢𝐞j.

Condition 1 says that if an element of B-1⁢Ak⁢B is nonzero then both its row and column indices must belong to the same equivalence classMathworldPlanetmath of ∼, i.e., the nonzero elements of B-1⁢Ak⁢B 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-1⁢Ak⁢B is equal to Ak.

The proof of the theorem requires the following lemma.

Lemma 2.

Let a sequence A1,…,Ar∈Mn⁢(K) of upper triangular matrices be given, and denote by A the unital ( algebraMathworldPlanetmathPlanetmath ( generated by these matrices. For every sequence λ1,…,λr∈K of scalars there exists a matrix C∈A 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 i∈S 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 Cn⁢Ak=Ak⁢Cn. In of the individual blocks, this becomes


Now let

D=(ID120D22)⁢, so that ⁢D-1=(I-D12⁢D22-10D22-1)

and consider the matrix D-1⁢Ak⁢D. Clearly


so that the positions with row not in S and column in S are all zero, as requested for B-1⁢Ak⁢B. 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-1⁢Ak⁢D, 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⁢(B′00I). ∎

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