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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    $\begingroup$ Try $d=2$, $\Sigma=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ and $C=I$. $\endgroup$ – user647486 Apr 3 at 20:26
  • $\begingroup$ The trace is then $0$, which is not a counterxample. $\endgroup$ – Joker123 Apr 3 at 20:29
  • $\begingroup$ @Joker123 The trace is $-2$. $\endgroup$ – Robert Israel Apr 3 at 20:30
  • $\begingroup$ 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 $\operatorname{tr}(\Sigma^2C)=-2$. $\endgroup$ – user647486 Apr 3 at 20:36
  • $\begingroup$ The trace is now -2 after editing. Thanks for the counterexample! $\endgroup$ – Joker123 Apr 3 at 20:50

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