Let $\Sigma$ be a real $d \times d$-matrix and $C$ be a real, symmetric, positive definite $d \times d$-matrix. Does it then hold, that $$ \text{tr}(\Sigma^2C) \geq 0 \quad ?$$
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3$\begingroup$ Try $d=2$, $\Sigma=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ and $C=I$. $\endgroup$– user647486Apr 3, 2019 at 20:26
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$\begingroup$ The trace is then $0$, which is not a counterxample. $\endgroup$– Joker123Apr 3, 2019 at 20:29
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$\begingroup$ @Joker123 The trace is $-2$. $\endgroup$– Robert IsraelApr 3, 2019 at 20:30
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$\begingroup$ Such that you know. The example takes advantage that the algebra of matrices of the form $\begin{pmatrix}a&b\\-b&a\end{pmatrix}$ is the same as that of the complex numbers $a+ib$. The $\operatorname{tr}(\cdot)$ on them is the same as $2\operatorname{Re}(\cdot)$. So, that $\Sigma$ is like $i$, $C$ is $1$, and $\Sigma^2C$ is $i^2\cdot 1=-1$. Since $2\operatorname{Re}(i^2\cdot 1)=-2$, that is why we expect that $\operatorname{tr}(\Sigma^2C)=-2$. $\endgroup$– user647486Apr 3, 2019 at 20:36
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$\begingroup$ The trace is now -2 after editing. Thanks for the counterexample! $\endgroup$– Joker123Apr 3, 2019 at 20:50
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