Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Omega \subset \mathbb{R}^n$ open, $A(x)$($x\in \Omega$) symetrical regular real matrix continuous in x.

The question is: Are there continuous fuctions $D,S : \Omega \rightarrow \mathbb{R}^{n\times n}$. That $D(x)$ is diagonal, $S(x)$ orthogonal and $A(x) = S^T(x)D(x)S(x)$for all $x\in \Omega$.

Basically I'm looking for continuous eigenvalue decomposition.

Let's have one more restriction and that $D_{ii}(x) \leq D_{jj}(x)$ for all $i \leq j$. With this D is given uniquely i.e. $D_{ii}(x)$ is i-th eigenvalue of $A(x)$. When $A(x)$'s eigenvalues are pairwise different and we request the first non zero element in each column to be positive than $S(x)$ is given uniguely too(edited based on comment).

$\Omega_0 = \{ x:A(x)\text{ has pairwise different eigenvalues}\}$. Remark: $\Omega_0$ is open.

So $S| _{\Omega_0}$ is uniquely given on $\Omega_0$.

Can $S| _{\Omega_0}$ be continuously extended to $\overline \Omega_0 \cap \Omega$ ?(I don't know)

If so than I would like to use something like Tiezte extension theorem. To extend $S| _{\overline \Omega_0 \cap \Omega}$ from $\overline \Omega_0 \cap \Omega$ to $\Omega$ and get $S$.

share|improve this question
1  
Just a little issue: even if $A(x)$'s eigenvalues are pairwise disjoint you may find $2^n$ different $S(x)$: in fact if $S(x)$ works, so does the matrix $S'(x)$ obtained from $S(x)$ by changing the sign of one (or more) column(s). –  AndreasT Mar 2 '13 at 9:33
1  
The problem may be fixed by requesting the first non zero value of each column of $S$ to be (e.g.) positive. –  AndreasT Mar 2 '13 at 9:35

1 Answer 1

up vote 0 down vote accepted

Ok I found counterexample suppose matrix $\Omega = (-1,1)$

$$A(x)=\left(\begin{matrix} \cos \theta(x) & -\sin \theta(x) \\ \sin \theta(x) & \cos \theta(x) \end{matrix} \right) \left(\begin{matrix} 1 & 0 \\ 0 & 1+|x| \end{matrix} \right) \left(\begin{matrix} \cos \theta(x) & \sin \theta(x) \\ -\sin \theta(x) & \cos \theta(x) \end{matrix} \right)$$

where $\theta(x) = \frac{1}{|x|}$, and $\theta(0) = 0$

For this $A(x)$ one cannot find continuous $S(x)$ on $(-1,1)$

share|improve this answer

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.