2
$\begingroup$

For matrices $D\in C^{d×p}$ and $E\in C^{d×p}$ with $d>p$, if $D$ is a full column matrix, for what condition that $D \odot E$ is also a full column matrix where $\odot$ denotes the Hadamard product.

$\endgroup$
2
  • 1
    $\begingroup$ Are you looking for necessary conditions or sufficient conditions? $\endgroup$ – Zach Teitler Apr 22 at 2:40
  • $\begingroup$ Thanks. I am looking for a necessary condition. I mean, I am trying to find a constraint for the matrix $E$ to make sure $D \odot E$ is a full column matrix. $\endgroup$ – tianhang Apr 22 at 3:26
0
$\begingroup$

Say $E$ contains a nonzero transversal (I welcome any suggestions of better terminology) if there is a collection of entries of $E$ with the properties (1) no two of the entries are in the same row or the same column (they are in "rook position") (2) the collection includes one entry in each row of $E$ or one entry in each column of $E$, whichever is smaller (3) the entries of $E$ in those positions are nonzero.

If $D$ and $D \odot E$ have full rank, then $E$ contains a nonzero transversal. Obviously: $D \odot E$ has a maximal minor of full rank, so there is at least one nonzero transversal (nonzero term in the permutation expansion of the determinant of that minor), and if $D \odot E$ has nonzero entries in those positions then so does $E$.

That is a very weak condition but in some sense it's the strongest that can hold. If $E$ is any matrix that contains a nonzero transversal, then there exists $D$ such that both $D$ and $D \odot E$ have full rank. Namely, give $D$ entries $1$ in positions of the nonzero transversal of $E$, all other entries $0$.

So if the question was "Given a matrix $E$, (some condition on $E$ only) if and only if there exists full-rank $D$ such that $D \odot E$ has full rank", then the condition on $E$ that makes it hold is "contains a nonzero transversal".

For the following questions, I don't know the answer, and I hope some expert here will contribute answers:

  1. Given a matrix $E$, (some condition on $E$ only) if and only if for every full-rank $D$, $D \odot E$ has full rank.

This holds for the all $1$s matrix $E$ and scalar multiples of it. At a wild guess, those might be the only ones. (But I haven't thought about it.)

  1. Given a matrix $E$ and a full-rank $D$, (some condition on $D$ and $E$) if and only if $D \odot E$ has full rank.

If $D \odot E$ has full rank, then both $D$ and $E$ contain nonzero transversals in the same positions, this is a necessary condition for $D \odot E$ to have full rank. But it's not sufficient.

Of course these questions can just be the "if" or "only if" parts instead of "if and only if", and there can be other questions.

$\endgroup$
2
  • $\begingroup$ Thanks so much. It is indeed hard to find a general condition for $E$ to make sure $D\odot E$ is a full-column rank when $D$ is a full-column. In my problem, I have a full-column rank matrix $D\in \mathbb{C}^{d\times p}$ and a matrix $E\in \mathbb{C}^{d\times p}$ with $e^{-j\omega_d\delta_p}$ as its $(d,p)$th element. I have found that $D\odot E$ is a also full-column rank, which was confirmed by the numerical simulations (I selected $\omega_d>0$ and $\delta_p>0$ randomly). However, I have no idea to prove this. Can you please give me some advice? $\endgroup$ – tianhang Apr 26 at 7:29
  • $\begingroup$ Hmm, that $E$ is not rank 1. It would be easy if it was rank 1 (the Hadamard product would just be rescaling rows and columns of $D$, so not changing the rank). I’m not sure how to prove it for this $E$... I’ll think about it. $\endgroup$ – Zach Teitler Apr 26 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.