$A=({B}\otimes{I_{k}})C$ where $B$ is a $N$x$r$ matrix with rank $r$, and $C$ is a $rk$x$rk$ symmetric matrix

$M=DAE$ where $D$ is a $Nk$ x $Nk$ symmetric matrix and $E$ is a $rk$x$rk$ symmetric matrix.

$F=ln(det(I_{rk}-M'M))+Tr(M'GM)-Tr(G)$ where $G$ is a symmetric $Nk$x$Nk$ matrix.

How can I end up with $$F=-\sum_{n=1}^{r}ln(\lambda_{n})+\sum_{n=1}^{r}(\lambda_{n}-1)-\sum_{n=1}^{N}(\lambda_{n})$$ with $r(r-1)/2$ restrictions involving the $rk$x$rk$ matrix $M$, given what follows below?

$rk(rk-1)/2$ restrictions are imposed involving $M$ say $m_{n}'m_{l}=0$ for all $l\ne n = 1,2,...,rk$ where $m_{n}$ is a column of $M$.

Using these, $F$ becomes $$F=\sum_{n=1}^{rk}ln(1-m_{n}'m_{l})+\sum_{n=1}^{rk} m_{n}'Gm_{n}-Tr(G)$$

Taking first derivatives with respect to $m_{n}$ gives $$G m_{n}=(\frac{1}{1-m_{n}'m_{n}}) m_{n}, n=1,2,...,rk$$ and so the eigenvalues $\lambda_{n}=\frac{1}{1-m_{n}'m_{n}}$ and $$F=-\sum_{n=1}^{rk}ln(\lambda_{n})+\sum_{n=1}^{rk}(\lambda_{n}-1)-\sum_{n=1}^{Nk}(\lambda_{n})$$