# Triangularizing a matrix with function entries

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.

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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 $2\pi$.