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 ProofThatCommutingMatricesAreSimultaneouslyTriangularizable 2013-03-22 15:27:08 2013-03-22 15:27:08 georgiosl (7242) georgiosl (7242) 10 georgiosl (7242) Proof msc 15A23