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If $A^TA=cI$ for some scalar $c\neq0$, then $A^T=cA^{-1}$, and thus $AA^T=cI$ as well. (The first equation says that $c^{-1}A^T$ is a left-inverse of $A$. And a left-inverse is also a right-inverse: If $AB = I$ then $BA = I$ .) So what if $c=0$?

Over the real numbers, $A^TA=0$ implies $A=0$, since $A^TA$ contains the sum of squares of elements of each column of $A$.

Over a field with characteristic $p$, the $p\times p$ matrix

$$A=\begin{bmatrix}1&1&\cdots&1\\1&1&\cdots&1\\\vdots&\vdots&\ddots&\vdots\\1&1&\cdots&1\end{bmatrix}$$

(for example) has $A^TA=pA=0$ even though $A\neq0$. But in this case $AA^T=0$ as well. Does that always happen?

mr_e_man
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    your characteristic $p$ example could be improved considerably. First, your $A$ is symmetric, so of course $A^TA =A^2 = AA^T$. Better $B := \mathbf 1_p\mathbf e_1^T$. Then $\text{rank}(B^TB) =0\lt 1 =\text{rank}(BB^T)$ where $BB^T = $ your $A$ – user8675309 Jan 31 '23 at 21:59

1 Answers1

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No.

Suppose $i^2=-1$ for some $i$ in the field (e.g. $i=2$ in the finite field of size $5$), and consider

$$A=\begin{bmatrix}1&0\\i&0\end{bmatrix}.$$

mr_e_man
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