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

Given a matrix with function entries, which are smooth and homogeneous, and having smooth eigenvalues, can we find a conjugating matrix with smooth and homogeneous entries that triangularize the given matrix? For instance, given $A(x)$ is an $N\times N$ matrix with entries $a_{ij}(x)$ that are smooth and homogeneous in $x$ of order $1.$ Also, given that the eigenvalues of $A(x)$ are smooth. Find an invertible(may be in small neighborhood) matrix $E(x)$ with smooth entries such that $E^{-1}(x)A(x)E(x)$ is upper-triangular.

Sometime back I had asked a question on triangularizing a function matrix. Now, it is clear to me that it is possible to find, by Schur decomposition, a triangularizing matrices which are measurable. Also, one of the answers posted for that question was, it is not always possible to uniformly triangularize especially for certain matrices with non-differentiable eigenvalues. The question in this post is directed towards smoothness and homogenity of such matrices under the condition that they have smooth eigenvalues.

I would be grateful for any reference or insight in this direction. Thank you.

share|improve this question
2  
Did you have a look to Kato's book Perturbation Theory for Linear Operators, volume 132 of Grundlehren der mathematischen Wissenschaften ? –  Denis Serre Dec 12 '12 at 7:15
1  
@Denis Serre Yes. Kato's book discusses Jordan form. But, I find that questions about Jordan form and triangular form are a bit different. For example, the matrix $$ \left(\begin{array}{cc} 1&z\\ 0&1 \end{array}\right) $$ is trivially triangulariable with smooth entries but cannot be written in Jordan form at $z=0$. –  Uday Dec 12 '12 at 9:30
add comment

2 Answers 2

This is not a complete answer, but the paper

Kreiss, H. O., Über Matrizen die beschränkte Halbgruppen erzeugen, Math. Scand. 7(1959), 71-81.

contains a lot of results in this direction. Unfortunately, at the moment I do not have access to it to check...

share|improve this answer
    
@András Bátkai Thank you for the reference. I will try to get this paper. Is there an English translation of this paper? –  Uday Dec 12 '12 at 9:33
    
I am sorry to say that I cannot find an English reference... –  András Bátkai Dec 23 '12 at 12:19
    
I have figured out a way to translate the paper. Thank you once again. –  Uday Dec 27 '12 at 15:37
add comment

Not precisely what you are asking, but if you look at continuous functions (instead of smooth and homogeneous ones), Grove and Pedersen ["Diagonalizing Matrices over $C(X)$", Journal of Functional Analysis 59, 65--89, 1984] prove the following. $N \times N$ matrices can be diagonalized for all $N$ if and only if $X$ is a sub-Stonean topological space with $\dim X \leq 2$ and $X$ carries no nontrivial $G$-bundles over any closed subset, for $G$ a symmetric group or the circle group.

share|improve this answer
add comment

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.