Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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 $2\pi$.

share|improve this answer

I am pretty sure this question (actually, the Jordan canonical form question) is studied at great length in Kato's perturbation theory for linear operators book (chapter 1). Kato is also very careful about explaining the techniques, so you should be able to adapt whatever he does to the specific setting you are interested in.

share|improve this answer
    
Thanks! I actually checked it. He does not talk about triangularizbility as such. He discusses eigenvalue of the operator valued functions and their singularities. Is it a straightforward to go from this analysis to actually converting the given matrix to triangular form? –  Uday May 14 '12 at 15:46
    
He DOES talk a lot about the Jordan canonical form (he decomposes the matrix into the diagonal and nilpotent parts), which is even better than triangular (since you seem to be happy with complex numbers, over the reals obviously things are different...) –  Igor Rivin May 14 '12 at 16:05
    
Pardon me if I could not get this from the book. But, from what I have made out he discusses Jordan form only on page 42. The discussion there seemed like only for the case of complex numbers and not function entries. Feel free to correct me. Thank you, once again. –  Uday May 14 '12 at 16:16
    
I was quoting from memory -- that part does introduce the Dunford integral, which is the main technical tool, but the uses for perturbation purposes are actually in chapter 2, sections 5,6 –  Igor Rivin May 14 '12 at 17:44
    
I will check this. Thanks a million! –  Uday May 14 '12 at 18:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.