I need to show that for a matrix $\mathcal{Q} \in \mathcal{R}^{n\times n}$ with $Q^TQ = I$ and $H \in \mathcal{R}^{n\times n}$ positive semidefinite, we have the relation $Trace(HQ) \leq Trace(H)$.
Considering the SVD decomposition of H : H = $U\Sigma V^*$
I start the proof with the following : Trace(HQ) = Trace($U\Sigma V^*Q$) = Trace$((V^*QU)\Sigma)$
I the define : $X = V^*QU$ orthogonal, it's column all have unit length. This implies that $X_{ii} \leq 1$, since otherwise the norm of column i would be too big.
Since $\sigma_{i} \geq 0\quad \forall i$, this argument shows that we can maximize tr(RC) by taking $X = I_{n\times n}$.
Finally, i found that Trace(HQ) = Trace$((V^*QU)\Sigma)$ = Trace$(X\Sigma) \leq$ Trace($I\Sigma$) $\leq Trace(\Sigma)$
But i don't know how to make the last step to prove the relation.
Thanks,
$H$ is real symmetric. You may use orthogonal diagonalisation instead of SVD.