Given matrix $A\in\mathbb{R}^{l\times m}$ and diagonal matrices with positive diag. elements $B=\mathrm{diag}\{b_1,\ldots,b_m\}$ and $C=\mathrm{diag}\{c_1,\ldots,c_l\}$.
Consider the following matrix: $D=BA^TCAB$. Since the matrix $D$ is symmetric positive semidefinite, its diagonal elements are always nonnegative. Some of its off-diagonal elements might be positive and some might be negative.
I want to understand on what conditions the sign of the off-diagonal elements of $D$ depends on?
My attempt:
Let $E=A^TA$. Since both $B$ and $C$ are diagonal matrices with positive diagonal elements, the sign of the off-diagonal elements of $D$ and $E$ are same. So we can concentrate on matrix $E$ instead.
Perform singular value decomposition on $A=USV^T$. Then $E=A^TA=VS^TSV^T$. Here $S^TS$ is diagonal matrix with nonnegative diagonal elements. If $S=[I_l\quad 0_{m-l}]$ or $S=[I_m\quad 0_{l-m}]^T$, then off-diagonal elements of $E$ will be equal to zero. However, in general, the sign of $e_{ij}$ depends on $\sum_{k=1}^{m}(s_kv_{ki}v_{kj})$, where $e_{ij},s_i,v_{ij}$ are entries of $E,S,V$, respectively.
So the sign of the off-diagonal elements of $D$ depends on singular values and right singular vectors of $A$.
I just wonder if we can say more? If someone gives us matrix $A$, without computing and seeing $D$, just by looking at matrix $A$ and its SVD, can we say for example if there are any negative off-diagonal elements in $D$, how many of them and where are they located?