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Let $A\in \mathcal{M}_n(\mathbb{R})$ be a diagonal positive matrix. We assume that $A$ is generic (in a sense to clarify). Let $\lambda \in\mathbb{R}$ and $U\in O_n(\mathbb{R})$ ($UU^T=I$) be such that $A+\lambda U$ has a multiple singular value (that is $A²+\lambda(UA+AU^T)$ has a double eigenvalue). Q1. Has $U$ necessarily an entry equal to $\pm 1$ on its diagonal ? Q2. Do there exist only a finite number of corresponding values of $\lambda$ ? Thanks in advance.

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For almost all choices of two real symmetric matrices $M=M^T$ and $N=N^T$, none of the matrices $M+xN$ (for $x \in \mathbb R$) has double eigenvalues. This is a surprising phenomenon known as eigenvalue avoidance. References: Lax's Linear Algebra and its Applications (the book, not the journal with the same name) and eprints.maths.ox.ac.uk/1175. –  Federico Poloni Jun 21 '13 at 19:17
    
Hi Frederico. Here $N$ depends on $M$ and you cannot use the "almost all" argument. Moreover, clearly there are solutions $(\lambda ,U)$ such that at least an entry $\pm 1$ is on the diagonal of $U$. –  loup blanc Jun 21 '13 at 21:34
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Yes, I agree - this result does not solve your problem, but I thought it was a remark worth mentioning about a related case. –  Federico Poloni Jun 21 '13 at 21:54
    
What do you assume exactly on $A$ ? –  Dietrich Burde Jun 26 '13 at 12:36
    
Hi Dietrich. The entries of $A$ are indeterminates. $A=diag(a_1,\cdots,a_n)$ where the $(a_i)_i$ are positive indeterminates. I hope that the result is true if $U$ is orthogonal and, then, I replace in my question $U\in SO_n(\mathbb{R})$ with $U\in O_n(\mathbb{R})$. –  loup blanc Jun 26 '13 at 17:22
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