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Let $A$ be the set of all $n \times n$ symmetric real matirix and $f \in C(\mathbb R , A)$. Then whether there is a $g \in C(\mathbb R , O(n))$ such that for all $t \in \mathbb R$, $g{(t)^{ - 1}}f(t)g(t)$ is a diagonal matrix?

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No. Consider $f(t) = \pmatrix{t & 0\cr 0 & 2t\cr}$ for $t \ge 0$ and $\pmatrix{0 & t\cr t & 0\cr}$ for $t \le 0$. For $t > 0$, $g(t)$ must be $\pmatrix{\pm 1 & 0\cr 0 & \pm 1\cr}$ or $\pmatrix{0 & \pm 1\cr \pm 1 & 0 \cr}$, but for $t < 0$ all entries of $g(t)$ must be $\pm 1/\sqrt{2}$.

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