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Let $\circ$ be the Hadamard product (i.e. element-wise multplication). We know that

$$ \operatorname{rank}(A\circ B) \leq \operatorname{rank}(A)\operatorname{rank}(B) $$

If we are given that $A$ is a rank 1 matrix, then this becomes

$$ \operatorname{rank}(A\circ B) \leq \operatorname{rank}(B) $$

Under what conditions does the equality hold, i.e.

$$ \operatorname{rank}(A\circ B) = \operatorname{rank}(B) $$

?

Follow-up questions:

  1. What can we say about the product rank if $A$ has one or more zero column or rows, but $B$ is full rank?
  2. What can we say about the product rank if $A$ has no zero column or rows, but $B$ is nearly full rank but not quite? e.g. $rank(B) = n-1$ where $B$ is a $nxn$ matrix?
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1 Answer 1

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I'm assuming $\circ$ is the Hadamard, i.e. element-wise, product.

If $A$ is a rank-$1$ matrix, we can write it as $A = u v^T$ where $u$ and $v$ are column vectors of the appropriate dimensions.
Consider an $r \times r$ submatrix $B_{IJ}$ of $B$, corresponding to a set of rows $I$ and a set of columns $J$. Then $$ \det(A_{IJ} \circ B_{IJ}) = \prod_{i \in I} u_i \prod_{j \in J} v_j \det(B_{IJ})$$
If $u$ and $v$ have no entries of $0$, i.e. $A$ has no all-zero rows or columns, then $\det(A_{IJ} \circ B_{IJ})\ne 0$ iff $\det(B_{IJ})\ne 0$, so the rank of $A \circ B$, which is the greatest $r$ such that $A \circ B$ has an $r \times r$ submatrix with nonzero determinant, is the same as the rank of $B$.

On the other hand, if $A$ has an all-zero row or column, then it's easy to find $B$ with rank $1$ such that $A \circ B$ has rank $0$.

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  • $\begingroup$ Do I understand correctly that "If A has no all-zero rows or columns and B is full rank, then $rank(A \circ B) = rank(B)$"? If yes, I have some follow up question (I will edit my original post to include these for readability): 1. What if $A$ has one or more zero column or rows, but $B$ is full rank? 2. What if $A$ has no zero column or rows, but $B$ is nearly full rank but not quite? e.g. $rank(B) = n-1$ where $B$ is a $nxn$ matrix? $\endgroup$ Commented May 1 at 6:47
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    $\begingroup$ If $A$ has rank $1$ and no all-zero rows or columns, then $\text{rank}(A \circ B) = \text{rank}(B)$. If $A$ has one or more zero columns or rows, then $A \circ B$ has those same zero columns or rows. In particular, if $B$ is $n \times n$ with full rank, then that implies $A \circ B$ does not have full rank. $\endgroup$ Commented May 1 at 22:23

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