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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 $S$.

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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
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
up vote 1 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 $(-1,1)$

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