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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 $$ \mathrm{rank}(B) = m? $$

SBK
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1 Answers1

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Yes: your matrix $B$ is $A^TA$ and therefore has the same rank as $A$.

Ben Grossmann
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