Let $A$ be a $m\times n$ matrix, with $m>n$, such that $A^TA$ is positive definite. Let $D$ be a diagonal matrix such that $A^TDA$ is rank deficient, i.e. $\det(A^TDA)=0$. Can anything be said about the matrix $D$?
A few things that I can immediately observe are:
1) $D$ can not be of the form $a I$, for any $a\ne 0$;
2) The diagonal elements of $D$ cannot be all positive or all negative.
However, it is not clear to me if $D$ can have a only few entries equal to $0$ (but not all), or a few positive and and a few negative (but not all simultaneously), or the combination of these. Is there any systematic way to determine a necessary condition that $D$ has to satisfy in order to respect the determinant condition? Please share some ideas and possible references if any. Thanks in advance.
As soon as $D$ has at least one zero entry, we can say that $$ \DeclareMathOperator{\rk}{rank} \rk(A^TDA) \leq \rk(D) < n $$ which implies that $\det (A^TDA) = 0$.
Suppose that $$ D = \pmatrix{D_1 \\&-D_2} $$ where $D_1$ and $D_2$ have all positive entries. Conformally break $A$ up into the block matrix $$ A = \pmatrix{A_1\\A_2} $$ we then have $$ A^TDA = A_1^TD_1A_1 - A_2^TD_2A_2 $$ If the row spaces of $A_1$ and $A_2$ have a trivial intersection, then this matrix is necessarily invertible. However, this matrix need not generally be invertible. For instance, with $$ A = \pmatrix{I\\I}, \quad D = \pmatrix{I\\&-I} $$ we find that $A^TDA = 0$.
Note that if $D$ has entries $d_i$ and $A$ has rows $a_i^T$, then we can write $$ A^TDA = \sum_{i=1}^n d_i \, a_ia_i^T $$