Ref:

• “Linear Algebra and Matrices” by Helene Shapiro. Chapter on Simultaneous Triangularization.

• Math StackExchange answer here.

Thm [Commuting matrices share an eigenvector]

Let ${ A, B \in \mathbb{C} ^{n \times n} . }$ Suppose ${ AB = BA . }$
Then ${ A, B }$ share an eigenvector.

Pf: Let ${ \lambda }$ be an eigenvalue of ${ A . }$

Consider any eigenvector

\[{ v \in \ker(A - \lambda I) . }\]

Note that

\[{ AB v = B A v . }\]

Hence

\[{ A (Bv) = \lambda (Bv) . }\]

Hence

\[{ Bv \in \ker(A - \lambda I) }\]

as well.

Hence

\[{ \ker(A - \lambda I) \, \text{ is a (nontrivial) B-invariant subspace.} }\]

Hence ${ B \vert _{\ker(A - \lambda I)} }$ admits an eigenvector ${ w . }$

Note that ${ w }$ is a common eigenvector of ${ A, B , }$ as needed. ${ \blacksquare }$

We can strengthen above result.

Thm [Commuting matrices are simultaneously triangularizable]

Let ${ A, B \in \mathbb{C} ^{n \times n} . }$ Suppose ${ AB = BA . }$
Then ${ A, B }$ are simultaneously triangularizable.

Pf: We will proceed by induction on ${ n . }$

Note that ${ A, B }$ have a common eigenvector ${ v _1 . }$

Construct a basis ${ P = (v _1; v _2 , \ldots, v _n) . }$

Note that

\[{ {\begin{aligned} &\, AP = P {\begin{pmatrix} m _{11} &M _{12} \\ 0 &M _{22} \end{pmatrix}} \\ &\, BP = P {\begin{pmatrix} n _{11} &N _{12} \\ 0 &N _{22} \end{pmatrix}} . \end{aligned}} }\]

Note that since ${ A, B }$ commute, we have

\[{ {\begin{pmatrix} m _{11} &M _{12} \\ 0 &M _{22} \end{pmatrix}} {\begin{pmatrix} n _{11} &N _{12} \\ 0 &N _{22} \end{pmatrix}} = {\begin{pmatrix} n _{11} &N _{12} \\ 0 &N _{22} \end{pmatrix}}{\begin{pmatrix} m _{11} &M _{12} \\ 0 &M _{22} \end{pmatrix}} . }\]

Hence

\[{ M _{22} N _{22} = N _{22} M _{22} . }\]

Hence by induction hypothesis, there is an invertible matrix ${ Q }$ such that

\[{ Q ^{-1} M _{22} Q, \quad Q ^{-1} N _{22} Q }\]

are upper triangular.

Note that

\[{ \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} ^{-1} \begin{pmatrix} m _{11} &M _{12} \\ 0 &M _{22} \end{pmatrix} \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} , \quad \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} ^{-1} \begin{pmatrix} n _{11} &N _{12} \\ 0 &N _{22} \end{pmatrix} \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} }\]

are now upper triangular.

Hence

\[{ \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} ^{-1} P ^{-1} A P \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} , \quad \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} ^{-1} P ^{-1} B P \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} }\]

are upper triangular.

Hence

\[{ \left( P \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} \right) ^{-1} A \left( P \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} \right), \quad \left( P \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} \right) ^{-1} B \left( P \begin{pmatrix} 1 &0 \\ 0 &Q \end{pmatrix} \right) }\]

are upper triangular, as needed. ${ \blacksquare }$