It's true when $A$ is a matrix over $\mathbb{R}$, see Prove rank $A^TA$ = rank $A$ for any $A_{m \times n}$
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How about $A=\pmatrix{1&1\cr1&1}$ over $\Bbb F_2$?
Angina Seng
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This generalizes to show that the statement doesn't hold for any finite field, or any field of prime characteristic for that matter: The $p \times p$ matrix $A$ with all entries $1$ has rank $1$ but $A^T A = 0$. – Travis Willse Jun 18 '17 at 10:56
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What if $A$ is $m \times n$ and $\operatorname{rank}{A}=n$ ? does it hold ? – Aditya Aug 18 '17 at 14:46