Related question here: Rank of sum of rank-1 matrices
It looks like my question is a generalisation of the linked question I think.
Suppose I have a $k\times n$ matrix $A$ where $k \geq n$ and suppose that $\mathrm{rank}(A) = m\ (\leq n \leq k)$.
Write $r_1,\dots r_k$ for the rows of $A$ and consider the $n\times n$ matrix
$$
B = r_1^Tr_1 + \cdots + r_k^Tr_k.
$$
Is it true that
$$
\mathrm{rank}(B) = m?
$$
Best Answer
Yes: your matrix $B$ is $A^TA$ and therefore has the same rank as $A$.