PlanetMath: simultaneous triangularisation of commuting matrices over any field

Let $\mathbf {e}_i$ denote the (column) vector whose

th position is

and where all other positions are 0. Denote by

the set $\{1,\dotsc,n\}$ . 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 $\mathrm{M}_n(\mathcal {K})$ . Let

be the function which extracts the

th diagonal element of a matrix, i.e., $d_i(A) = \mathbf {e}_i^{\mathrm{T}}\! A \mathbf {e}_i$ .

Proof. This is by induction on

. The induction hypothesis is that given pairwise commuting matrices $A_1,\dotsc,A_r \in \mathrm{M}_n(\mathcal {L})$ , whose characteristic polynomials all split in $\mathcal {L}$ , and a sequence of arbitrary scalars $\mu_1,\dotsc,\mu_r \in \mathcal {L}$ , there exists some $P \in \mathrm{GL}_n(\mathcal {L})$ such that:

$P^{-1} A_k P$ is upper triangular for all $k=1,\dotsc,r$ .
If some $i,j \in [n]$ are such that and $d_j(P^{-1} A_k P) = d_i(P^{-1} A_k P)$ for all $k \in [r]$ , then $d_{i+1}(P^{-1} A_k P) = d_i(P^{-1} A_k P)$ .
If some $j \in [n]$ is such that $d_j(P^{-1} A_k P) = \mu_k$ for all $k \in [r]$ , then $d_1(P^{-1} A_k P) = \mu_k$ for all $k \in [r]$ .

For

this hypothesis is trivially fulfilled (all $1 \times 1$ matrices are upper triangular). Assume that it holds for

and consider the case

It is easy to see that condition 1 implies that $P\mathbf {e}_1$ must be an eigenvector that is common to all the matrices. If there exists a nonzero vector $\mathbf {u}_1 \in \mathcal {L}^n$ such that $A_k \mathbf {u}_1 = \mu_k \mathbf {u}_1$ for all $k=1,\dotsc,r$ then this is such a common eigenvector, and in that case let $\lambda_k=\mu_k$ for all $k=1,\dotsc,r$ . Otherwise there by Lemma 1 exists a vector $\mathbf {u}_1 \in \mathcal {L}^n \setminus \{\mathbf {0}\}$ such that $A_k \mathbf {u}_1 = \lambda_k \mathbf {u}_1$ for some $\{\lambda_k\}_{k=1}^r \subseteq \mathcal {L}$ . Either way, one gets a suitable candidate $\mathbf {u}_1$ for $P\mathbf {e}_1$ and eigenvalues $\lambda_1,\dotsc,\lambda_r$ that incidentally will satisfy $d_1(P^{-1} A_k P) = \lambda_k$ for all $k \in [r]$ .

Let $\mathbf {u}_2,\dotsc,\mathbf {u}_n \in \mathcal {L}^n$ be arbitrary vectors such that $\{\mathbf {u}_i\}_{i=1}^n$ is a basis of $\mathcal {L}^n$ . Let be the $n \times n$ matrix whose th column is $\mathbf {u}_i$ for $1 \leqslant i \leqslant n$ .¹Then is invertible and for each the first column of $B_k = U^{-1} A_k U$ is

$\displaystyle U^{-1} A_k U \mathbf {e}_1 = U^{-1} A_k \mathbf {u}_1 = \lambda_k U^{-1} \mathbf {u}_1 = \lambda_k \mathbf {e}_1$ .

Furthermore

$\begin{multline*} B_j B_k = U^{-1} A_j U U^{-1} A_k U = U^{-1} A_j A_k U = \\ = U^{-1} A_k A_j U = U^{-1} A_k U U^{-1} A_j U = B_k B_j \end{multline*}$

for all

and

Now let be the matrix formed from rows and columns though of . Since $\det(A_k -\nobreak xI) = \det( B_k -\nobreak xI ) = (\lambda_k -\nobreak x) \det( A_k' -\nobreak xI )$ by expansion along the first column, it follows that the characteristic polynomial of splits in $\mathcal {L}$ . Furthermore all the have side and commute pairwise with each other, whence by the induction hypothesis there exists some $P' \in \mathrm{GL}_{n-1}(\mathcal {L})$ such that every $P'^{-1} A_k' P'$ is upper triangular. Let $P = U \left(\begin{smallmatrix}1& 0 \\ 0& P' \end{smallmatrix}\right)$ . Then the submatrix consisting of rows and columns through of $P^{-1} A_k P$ is equal to $P'^{-1} A_k' P'$ and hence contains no nonzero subdiagonal elements. Furthermore the first column of $P^{-1} A_k P$ is equal to the first column of and thus the $P^{-1} A_k P$ are all upper triangular, as claimed.

It also follows from the induction hypothesis that can be chosen such that $d_2(P^{-1} A_k P) = d_1(P'^{-1} A_k' P') = \lambda_k = d_1(P^{-1} A_k P)$ for all $k \in [r]$ if there is any $j \geqslant 2$ for which $d_j(P^{-1} A_k P) = d_{j-1}(P'^{-1} A_k' P') = \lambda_k = d_1(P^{-1} A_k P)$ for all $k \in [r]$ and more generally if $2 \leqslant i < j$ are such that $d_j(P^{-1} A_k P) = d_i(P^{-1} A_k P)$ for all $k \in [r]$ then similarly $d_{i+1}(P^{-1} A_k P) = d_i(P^{-1} A_k P)$ for all $k \in [r]$ . This has verified condition 2 of the induction hypothesis. For the remaining condition 3, one may first observe that if there is some $i \in [n]$ such that $d_i(P^{-1} A_k P) = \mu_k$ for all $k \in [r]$ then by Lemma 2 there exists a nonzero $\mathbf {v} \in \mathcal {L}^n$ such that $P^{-1} A_k P \mathbf {v} = \mu_k \mathbf {v}$ for all $k \in [r]$ . This means $P \mathbf {v}$ will fulfill the condition for choice of $\mathbf {u}_1$ , and hence $d_1(P^{-1} A_k P) = \lambda_k = \mu_k$ as claimed.

The theorem now follows from the principle of induction. $\qedsymbol$

Footnotes