If $A^TA \ge B^TB$ does this imply $AA^T \ge BB^T$?
If not, is there a counter example?
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4$\begingroup$ what is the ordering $\ge$ used here? and are the matrices all square? $\endgroup$– vidyarthiCommented Jun 23, 2020 at 5:54
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10$\begingroup$ A counterexample has already been given, but one can already guess the answer by noting that the two inequalities have different symmetries. The former inequality is invariant with respect to multiplying A,B on the left by arbitrary orthogonal matrices, whilst the latter is invariant with respect to multiplying A,B on the right by arbitrary orthogonal matrices. $\endgroup$– Terry TaoCommented Jun 23, 2020 at 6:36
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1 Answer
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Take $$A = \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}, \quad B = \begin{pmatrix}1 & 0 \\ 0& 0 \end{pmatrix}.$$
