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, where $X$ be orthogonal complement of $x$ and $R^{H}R=I$, then we have that

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 $A_1=X^{H}AX$ thenthere exists unitary matrix $S$ such that $S^H{B}_{1}S={\overline{T}}_2,S^H{A}_{1}S={\overline{T}}_1.$ Now is a unitary matrix,$Q^{H}Q=I$ and we have

Analogously we have that

$$Q^{H}BQ=T_2.$$