One can prove that the set $\{ Tr(OM) /$ $O$ $\in$ $ O_n ( \mathbb R ) $ $\}$ has a maximum for a certain orthogonal matrix $O$, furthermore, the application $f$ : $M_n (\mathbb R)$ $\mapsto$ $\mathbb R$
such as that $f(M) = max( \{ Tr(OM) /$ $O$ $\in$ $ O_n ( \mathbb R ) $ $\}$ is continuous.
And for all $M \in S_n ^{+}(\mathbb R)$, $f(M) = Tr(M) $
So in order to prove that for $O\in$ $ O_n ( \mathbb R ) $ : $f(M) =Tr( OM)$ iff $OM \in S_n ^{+}(\mathbb R)$, one can prove that $f(M) = Tr(M)$ iff $M \in S_n ^{+}(\mathbb R)$.
I tried to prove that if $f(M) = Tr(M)$ then at least $M$ is symmetric, which should obviously be the fist step but to no avail.
Any ideas?
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AymaneMaaitat
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Maybe von Neumann's trace inequality can be useful here? All you have to do is exhibit an orthogonal matrix that saturates its upper bound. – VHarisop May 04 '21 at 22:07
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if you don't know von Neumann trace you can do this with polar decomposition, basically mimic this https://math.stackexchange.com/questions/3518238/optimizing-traceqtz-subject-to-qtq-i/ – user8675309 May 04 '21 at 22:34