I agree with your assumption that the eigenvalues and eigenvectors don't have much to say about the $q_{i}$ and $\phi _{i}.$ I didn't follow the reference you gave, other than it seems the case of $D\leq N$ is of interest, which I assume below. As a Gram matrix, the elements of $L$ give the dot products of the columns of $B$. The diagonal entries of $L$ are just the $q_{i}^{2}$ but there is already loss of information about the absolute orientations of the $\phi _{i}$. (If all the $B_{i}$ were rotated by some angle by premultiplying $B$ by a rotation matrix, then the new matrix $L$ is unchanged.)

The eigenvalues and eigenvectors have less information. Since $k=\operatorname{% rank}(B)=\operatorname{rank}(L)\leq D,$ there are only $k$ non-zero eigenvalues, so in the eigendecomposition, there are only $k$ non-zero terms. This means the $N-k$ eigenvectors corresponding to the zero eigenvalues essentially contain no useful information - one can take arbitrary linear combinations and they will still be eigenvectors. So you only have $k$ eigenvalues and eigenvectors that are useful.

There are some relationships between the eigenvalues and the $q_{i}$. The trace of $L$ is both the sum of the eigenvalues and the sum of the $q_{i}^{2}$

$$ \sum_{i}\lambda _{i}=\sum_{i}q_{i}^{2}. $$

The diagonal entries majorize the eigenvalues, so if the eigenvalues and $% q_{i}^{2}$ are arranged in increasing order then the sum of the $k$ smallest $% q_{i}^{2}$ is less than or equal to the sum of the $k$ smallest eigenvalues, with equality for $k=N$ (the above relationship). But since most eigenvalues are zero, most of these inequalities follow anyway from the positivity of the $q_{i}^{2}$.

It is also true that

$$ \sum_{i}\lambda _{i}^{2}=\sum_{i}\left\vert L_{ij}\right\vert ^{2} $$

though that is a rather indirect relationship to the individual dot products.

I agree with your assumption that the eigenvalues and eigenvectors don't have much to say about the $q_{i}$ and $\phi _{i}.$ I didn't follow the reference you gave, other than it seems the case of $D\leq N$ is of interest, which I assume below. As a Gram matrix, the elements of $L$ give the dot products of the columns of $B$. The diagonal entries of $L$ are just the $q_{i}^{2}$ but there is already loss of information about the absolute orientations of the $\phi _{i}$. (If all the $B_{i}$ were rotated by some angle by premultiplying $B$ by a rotation matrix, then the new matrix $L$ is unchanged.)

The eigenvalues and eigenvectors have less information. Since $k=\operatorname{% rank}(B)=\operatorname{rank}(L)\leq D,$ there are only $k$ non-zero eigenvalues, so in the eigendecomposition, there are only $k$ non-zero terms. This means the $N-k$ eigenvectors corresponding to the zero eigenvalues essentially contain no useful information - one can take arbitrary linear combinations and they will still be eigenvectors. So you only have $k$ eigenvalues and eigenvectors that are useful.

There are some relationships between the eigenvalues and the $q_{i}$. The trace of $L$ is both the sum of the eigenvalues and the sum of the $q_{i}^{2}$

$$ \sum_{i}\lambda _{i}=\sum_{i}q_{i}^{2}. $$

The diagonal entries majorize the eigenvalues, so if the eigenvalues and $% q_{i}^{2}$ are arranged in increasing order then the sum of the $k$ smallest $% q_{i}^{2}$ is greater than or equal to the sum of the $k$ smallest eigenvalues, with equality for $k=N$ (the above relationship). But since most eigenvalues are zero, most of these inequalities follow anyway from the positivity of the $q_{i}^{2}$.

It is also true that

$$ \sum_{i}\lambda _{i}^{2}=\sum_{i}\left\vert L_{ij}\right\vert ^{2} $$

though that is a rather indirect relationship to the individual dot products.