I was reading through this paper and on page 6, there is a lemma proving $(\det A)(\det B) \leq [(\operatorname{tr} AB)/n]^n$ for two positive semideifinite matrices $A$ and $B$.
I get every single line until the part that concluded $\operatorname{tr} AB = \sum \lambda_l \mu_l $. Especially, what's bugging me is the very last equation. Before the equality sign we have $$ \sum S_{li} (S^T)_{il} (D_1)_{ii} (D_2)_{ll} $$ and after the equality sign we have $$ \sum \delta^{il} (D_1)_{ii} (D_2)_{ll} $$ Since S is a symmetric orthonormal matrix, $\ S_{li} = (S^T)_{il}$, so I believe the first term should just be $ s^2_{li} (D_1)_{ii} (D_2)_{ll} $. But it doesn't make sense that $ s^2_{li} = \delta^{il} $, since the orthonormal matrix just confirms that the 2-norm of any 'row' or 'column' is 1, not that the single $ s^2_{li} $ is $1$ for $l=i$ and $0$ otherwise.
There must be something I am missing. Can anyone help me filling the gap between those two line, please?
P.S. The paper is about elliptic equation, but the lemma is for general positive semidefinite matrices, so I put tag just for these type of questions.
$S=R^TQ$ is orthogonal, not symmetric as they claimed, but that is inconsequential to the proof.
Anyway, it is not true that $\operatorname{tr}(SD_1S^TD_2)=\operatorname{tr}(D_1D_2)$. For example, $D_1=\operatorname{diag}(1,2)=D_2$, and $S=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Then $SD_1S^T=\operatorname{diag}(2,1)$ and so we end up with $4$ on LHS and $5$ on RHS.
A much easier proof of the result they want is simply to use $\sqrt{A}B\sqrt{A}$ is symmetric positive semidefinite, has determinant $\det A\det B$ and trace $\operatorname{tr}AB$. The result follows from AM-GM on its eigenvalues.