proof that commuting matrices are simultaneously triangularizable

Proof by induction on $n$ , order of matrix.
For $n=1$ we can simply take $Q=1$ . We assume that there exists a common unitary matrix $S$ that triangularizes simultaneously commuting matrices , $(n-1)\times(n-1)$ .
So we have to show that the statement is valid for commuting matrices, $n\times n$ . From hypothesis $A$ and $B$ are commuting matrices $n\times n$ so these matrices have a common eigenvector.
Let $Ax=\lambda x$ , $Bx=\mu x$ where $x$ be the common eigenvector of unit length and $\lambda$ , $\mu$ are the eigenvalues of $A$ and $B$ respectively. Consider the matrix, $R=\begin{pmatrix}x&X\end{pmatrix}$ where $X$ be orthogonal complement of $x$ and $R^{H}R=I$ , then we have that

R^{H}AR=\begin{pmatrix}\lambda&x^{H}AX\\ 0&X^{H}AX\end{pmatrix}

R^{H}BR=\begin{pmatrix}\mu&x^{H}BX\\ 0&X^{H}BX\end{pmatrix}

It is obvious that the above matrices and also $X^{H}BX$ , $X^{H}AX$ , $(n-1)\times(n-1)$ matrices are commuting matrices. Let $B_{1}=X^{H}BX$ and $A_{1}=X^{H}AX$ then there exists unitary matrix $S$ such that $S^{H}B_{1}S=\bar{T}_{2},\,S^{H}A_{1}S=\bar{T}_{1}.$ Now $Q=R\begin{pmatrix}1&0\\ 0&S\end{pmatrix}$ is a unitary matrix, $Q^{H}Q=I$ and we have

Q^{H}AQ=\begin{pmatrix}1&0\\ 0&S^{H}\end{pmatrix}R^{H}AR\begin{pmatrix}1&0\\ 0&S\end{pmatrix}=\begin{pmatrix}\lambda&x^{H}AXS\\ 0&\bar{T}_{1}\end{pmatrix}=T_{1}.

Analogously we have that

Q^{H}BQ=T_{2}.

Title	proof that commuting matrices are simultaneously triangularizable
Canonical name	ProofThatCommutingMatricesAreSimultaneouslyTriangularizable
Date of creation	2013-03-22 15:27:08
Last modified on	2013-03-22 15:27:08
Owner	georgiosl (7242)
Last modified by	georgiosl (7242)
Numerical id	10
Author	georgiosl (7242)
Entry type	Proof
Classification	msc 15A23