# simultaneous upper triangular block-diagonalization of commuting matrices

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

###### Theorem 1.

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

1. 1.

If $i\sim j$ and $i\leqslant k\leqslant j$ then $k\sim i$.

2. 2.

If $i\sim j$ then $\mathbf{e}_{i}^{\mathrm{T}}\!R^{-1}A_{k}R\mathbf{e}_{i}=\mathbf{e}_{j}^{% \mathrm{T}}\!R^{-1}A_{k}R\mathbf{e}_{j}$.

3. 3.

If $\mathbf{e}_{i}^{\mathrm{T}}\!R^{-1}A_{k}R\mathbf{e}_{j}\neq 0$ then $i\leqslant j$ and $i\sim j$.

In other words there exists a simultaneous upper triangular block-diagonalisation of the matrices $A_{1},\ldots,A_{r}$ 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 $\mathcal{K}$ be a field, let $A_{1},\ldots,A_{r}\in\mathrm{M}_{n}(\mathcal{K})$ be pairwise commuting matrices, and let $\mathcal{L}$ be a field extension of $\mathcal{K}$ in which the characteristic polynomials of all $A_{k}$ split. Then there exists some $P\in\mathrm{GL}_{n}(\mathcal{L})$ such that

1. 1.

$P^{-1}A_{k}P$ is upper triangular for all $k=1,\ldots,r$, and

2. 2.

if $i,j,l\in[n]$ are such that $i\leqslant l\leqslant j$ and $\mathbf{e}_{i}^{\mathrm{T}}\!P^{-1}A_{k}P\mathbf{e}_{i}=\mathbf{e}_{j}^{% \mathrm{T}}\!P^{-1}A_{k}P\mathbf{e}_{j}$ for all $k=1,\ldots,r$, then $\mathbf{e}_{l}^{\mathrm{T}}\!P^{-1}A_{k}P\mathbf{e}_{l}=\mathbf{e}_{j}^{% \mathrm{T}}\!P^{-1}A_{k}P\mathbf{e}_{j}$ for all $k=1,\ldots,r$ as well.

Let $B_{k}=P^{-1}A_{k}P$ for all $k=1,\ldots,r$ and define

 $i\sim j\quad\text{if and only if}\quad\mathbf{e}_{i}^{\mathrm{T}}\!P^{-1}A_{k}% P\mathbf{e}_{i}=\mathbf{e}_{j}^{\mathrm{T}}\!P^{-1}A_{k}P\mathbf{e}_{j}\text{ % for all }k\in[r]\text{.}$
###### Lemma 3.

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

 $i\sim j\quad\text{if and only if}\quad\mathbf{e}_{i}^{\mathrm{T}}\!B_{k}% \mathbf{e}_{i}=\mathbf{e}_{j}^{\mathrm{T}}\!B_{k}\mathbf{e}_{j}\text{ for all % }k\in[r]\text{,}$

then there exists a matrix $Q\in\mathrm{GL}_{n}(\mathcal{L})$ such that:

1. 1.

If $\mathbf{e}_{i}^{\mathrm{T}}\!Q^{-1}B_{k}Q\mathbf{e}_{j}\neq 0$ then $i\sim j$ and $i\leqslant j$.

2. 2.

If $i\sim j$ then $\mathbf{e}_{i}^{\mathrm{T}}\!Q^{-1}B_{k}Q\mathbf{e}_{j}=\mathbf{e}_{i}^{% \mathrm{T}}\!B_{k}\mathbf{e}_{j}$.

The wanted $R$ is then $PQ$.

Title simultaneous upper triangular block-diagonalization of commuting matrices SimultaneousUpperTriangularBlockdiagonalizationOfCommutingMatrices 2013-03-22 15:29:35 2013-03-22 15:29:35 lars_h (9802) lars_h (9802) 4 lars_h (9802) Theorem msc 15A21 JordanCanonicalFormTheorem IfABInM_nmathbbCABBAThenQHAQT_1QHBQT_2