TITLE: Triangularizing a matrix with function entries
QUESTION [6 upvotes]: Hi Everybody!
Given a matrix, with smooth functions as arguments is there any result which say about its triangularization?
I know that, the question is in affirmative for diagonalizing a matrix which has distinct eigenvalues. Infact, we can show by Implicit function theorem that eigenvalues can be chosen to be $\mathcal{C}^{1}$functions. We can also calculate projection operators for the matrix using Cauchy integral formula. 
But, I have not found any result for triangularization. An answer or any references would be great help. Thank you. 
ADDED LATER: I have found in the Micheal Taylor's book 'Pseudo-differential operators' Page.72 a claim that for a matrix $K(\xi)$ there is always a measurable unitary matrix $U(\xi)$ such that $U^{-1}(\xi)K(\xi)U(\xi)$ is an upper triangular matrix. In this case $K(\xi)$ is smooth matrix in $\xi.$ He does not say exactly the name of this result though.

REPLY [9 votes]: You cannot triangularize smoothly the parametrized matrix
$$A(z)=\begin{pmatrix} 0 & 1 \\\\ z & 0 \end{pmatrix}$$
about $z=0$. The eigenvalues, square roots of $z$ aren't smooth and, above all, cannot be distinguished one from the other: when $z$ runs over a circle $C(0;\epsilon)$ in the complex plane, and an eigenvalue is selected continuously, its sign changes after $2\pi$.