simultaneous upper triangular block-diagonalization of commuting matrices

Let ${𝐞}_i$ denote the (column) vector whose $i$th position is $1$ and where all other positions are $0$. Denote by $[n]$ the set $\{1,\mathrm{\dots },n\}$. Denote by ${\mathrm{M}}_n(𝒦)$ the set of all $n\times n$ matrices over $𝒦$, and by ${\mathrm{GL}}_n(𝒦)$ the set of all invertible elements of ${\mathrm{M}}_n(𝒦)$.

Theorem 1.

Let $\mathrm{K}$ be a field, let $A_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{A}_r\mathrm{\in }{\mathrm{M}}_n\mathit{}\mathrm{(}\mathrm{K}\mathrm{)}$ be pairwise commuting matrices, and let $\mathrm{L}$ be a field extension of $\mathrm{K}$ in which the characteristic polynomials of all $A_k$ split. Then there exists an equivalence relation $\mathrm{\sim }$ on $\mathrm{[}n\mathrm{]}$ and a matrix $R\mathrm{\in }{\mathrm{GL}}_n\mathit{}\mathrm{(}\mathrm{L}\mathrm{)}$ such that:

1. If $i\sim j$ and $i⩽k⩽j$ then $k\sim i$.

2. If $i\sim j$ then ${𝐞}_i^{\mathrm{T}}{R}^{-1}{A}_{k}R{𝐞}_i={𝐞}_j^{\mathrm{T}}{R}^{-1}{A}_{k}R{𝐞}_j$.

3. If ${𝐞}_i^{\mathrm{T}}{R}^{-1}{A}_{k}R{𝐞}_j\ne 0$ then $i⩽j$ and $i\sim j$.

In other words there exists a simultaneous upper triangular block-diagonalisation of the matrices $A_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{A}_r$ in which each block is characterised by the particular values of the diagonal elements.

The proof of this theorem is the obvious combination of the following two lemmas.

Lemma 2.

Let $\mathrm{K}$ be a field, let $A_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{A}_r\mathrm{\in }{\mathrm{M}}_n\mathit{}\mathrm{(}\mathrm{K}\mathrm{)}$ be pairwise commuting matrices, and let $\mathrm{L}$ be a field extension of $\mathrm{K}$ in which the characteristic polynomials of all $A_k$ split. Then there exists some $P\mathrm{\in }{\mathrm{GL}}_n\mathit{}\mathrm{(}\mathrm{L}\mathrm{)}$ such that

1. $P^{-1}{A}_{k}P$ is upper triangular for all $k=1,\mathrm{\dots },r$, and

2. if $i,j,l\in [n]$ are such that $i⩽l⩽j$ and ${𝐞}_i^{\mathrm{T}}{P}^{-1}{A}_{k}P{𝐞}_i={𝐞}_j^{\mathrm{T}}{P}^{-1}{A}_{k}P{𝐞}_j$ for all $k=1,\mathrm{\dots },r$, then ${𝐞}_l^{\mathrm{T}}{P}^{-1}{A}_{k}P{𝐞}_l={𝐞}_j^{\mathrm{T}}{P}^{-1}{A}_{k}P{𝐞}_j$ for all $k=1,\mathrm{\dots },r$ as well.

Let $B_k=P^{-1}{A}_{k}P$ for all $k=1,\mathrm{\dots },r$ and define

$$i\sim j\mathit{\hspace{1em}}\text{if and only if}\mathit{\hspace{1em}}{𝐞}_i^{\mathrm{T}}{P}^{-1}{A}_{k}P{𝐞}_i={𝐞}_j^{\mathrm{T}}{P}^{-1}{A}_{k}P{𝐞}_j\text{ for all }k\in [r]\text{.}$$

Lemma 3.

Let $\mathrm{L}$ be a field, let $n$ be a positive integer, and let $\mathrm{\sim }$ be an equivalence relation on $\mathrm{[}n\mathrm{]}$ such that if $i\mathrm{\sim }j$ and $i\mathrm{⩽}k\mathrm{⩽}j$ then $k\mathrm{\sim }i$. Let $B_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{B}_r\mathrm{\in }{\mathrm{M}}_n\mathit{}\mathrm{(}\mathrm{L}\mathrm{)}$ be pairwise commuting upper triangular matrices. If these matrices and $\mathrm{\sim }$ are related such that

$$i\sim j\mathit{\hspace{1em}}\mathit{\text{if and only if}}\mathit{\hspace{1em}}{𝐞}_i^{\mathrm{T}}{B}_k{𝐞}_i={𝐞}_j^{\mathrm{T}}{B}_k{𝐞}_j\mathit{\text{ for all }}k\in [r]\mathit{\text{,}}$$

then there exists a matrix $Q\mathrm{\in }{\mathrm{GL}}_n\mathit{}\mathrm{(}\mathrm{L}\mathrm{)}$ such that:

1. If ${𝐞}_i^{\mathrm{T}}{Q}^{-1}{B}_{k}Q{𝐞}_j\ne 0$ then $i\sim j$ and $i⩽j$.

2. If $i\sim j$ then ${𝐞}_i^{\mathrm{T}}{Q}^{-1}{B}_{k}Q{𝐞}_j={𝐞}_i^{\mathrm{T}}{B}_k{𝐞}_j$.

The wanted $R$ is then $PQ$.

Title	simultaneous upper triangular block-diagonalization of commuting matrices
Canonical name	SimultaneousUpperTriangularBlockdiagonalizationOfCommutingMatrices
Date of creation	2013-03-22 15:29:35
Last modified on	2013-03-22 15:29:35
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	JordanCanonicalFormTheorem
Related topic	IfABInM_nmathbbCABBAThenQHAQT_1QHBQT_2