$ \newcommand{\blkdiag}{\operatorname{blkdiag}} \newcommand{\tr}{\operatorname{tr}} $Suppose $G_1$ is a $m_1 \times n$ matrix, and $G_2$ is a $m_2 \times n$ matrix. $D$ is a $n \times n$ diagonal matrix with $\{d_1, d_2, \ldots, d_n\}$ along the diagonal such that $d_1 \neq d_2 \neq \ldots \neq d_n$. And $C$ is a $n \times m$ matrix.
Let $m = m_1 + m_2$. Then $$A = \blkdiag(G_1DG_1^T, G_2DG_2^T)$$ is a $m \times m$ matrix, and
$$B = C^TC$$ is a $m \times m$ matrix.
I'm interested in manipulating the expression for $\tr(AB)$ such that I can pull out $\sum_{i=1}^n d_i$ outside of the trace operator somehow.
I know this can be done for the following: suppose $A^* = G_1DG_1^T$ is a $m_1 \times m_1$ matrix, and $B^* = C^{*T}C^*$ is a $m_1 \times m_1$ matrix where $C^*$ is $n \times m_1$. Then we have \begin{align*} \tr(A^*B^*) &= \tr(G_1DG_1^TC^{*T}C^*)\\ &= \tr(DG_1^TC^{*T}C^*G_1)\\ & \sum_{i=1}^n d_i \tr(e_i^TG_1^TC^{*T}C^*G_1e_i) \end{align*} where $e_i$ is $i$th unit vector in $\mathbb{R}^n$.
Can something similar be done for $\tr(AB)$ where $A$ is now a block diagonal matrix? I know that $\tr(A) = \tr(G_1DG_1^T) +\ tr(G_2DG_2^T)$, but I'm not sure how to work with $\tr(AB)$ so that I might be able to pull $\sum_{i=1}^n d_i$ outside of the trace operator.
I assume that “$\operatorname{blkdiag}(\dots)$” denotes a block diagonal matrix with the respective block cells. Put $G=\operatorname{blkdiag}(G_1, G_2)$, and $D’=\operatorname{blkdiag} (D,D)$. Then $A = GD’G^T$ so $$\operatorname{tr}(AB)= \operatorname{tr}(GD’G^TB)= \operatorname{tr}(D’G^TBG).$$