# Rank of Hadamard product with rank 1 matrix? [closed]

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?

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 $$0$$.

• 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? Commented May 1 at 6:47
• 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. Commented May 1 at 22:23