Willow Pratt
2022-07-12
Answered

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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Melina Richard

Answered 2022-07-13
Author has **14** answers

No. Consider $f(t)=\left(\begin{array}{cc}t& 0\\ 0& 2t\end{array}\right)$ for $t\ge 0$ and $\left(\begin{array}{cc}0& t\\ t& 0\end{array}\right)$.

For $t>0$, $g(t)$ must be $\left(\begin{array}{cc}\pm 1& 0\\ 0& \pm 1\end{array}\right)$ or $\left(\begin{array}{cc}0& \pm 1\\ \pm 1& 0\end{array}\right)$, but for $t<0$ all entries of $g(t)$ must be $\pm 1/\sqrt{2}$

For $t>0$, $g(t)$ must be $\left(\begin{array}{cc}\pm 1& 0\\ 0& \pm 1\end{array}\right)$ or $\left(\begin{array}{cc}0& \pm 1\\ \pm 1& 0\end{array}\right)$, but for $t<0$ all entries of $g(t)$ must be $\pm 1/\sqrt{2}$

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