It is a well-known result that when $A$ is a real matrix and $AA^t=A^2$ then $A$ is symmetric. I guess the same proposition is right for a complex matrix, but I can't prove it. Could someone give me some hints on this problem? I will be very appreciated for your help!
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My guess is that this is not true unless you use a conjugate transpose. Try and construct some $2 \times 2$ examples. – Wintermute Mar 23 '16 at 14:23
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Did you inspect the proof of this statement. Does it really rely on the fact that the entries are real? And if yes, what are modifications you could/should assume? – frog Mar 23 '16 at 14:23
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Try $A\in M_2(\mathbb{C})$ with $a_{12}=i$ and $a_{21}=?$ – Damian Reding Mar 23 '16 at 14:24
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At least this is correct for invertible $A$. – Friedrich Philipp Mar 23 '16 at 14:29
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The claim is correct for $2\times 2$ matrices. – Friedrich Philipp Mar 23 '16 at 14:36
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Thanks for your warm replies. I can prove it by elementary algebra when the order of $A$ is 2, or 3. As Friedrich said, this holds for invertible $A$. And we have known this also holds for real A. So, based on some algebraic geometry consideration, I am strongly prone to its rightness. – Marc Lee Mar 23 '16 at 14:40
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Most of the proofs here don't rely on the entries being real. http://math.stackexchange.com/questions/571583/if-matrix-such-aat-a2-then-a-is-symmetric?rq=1 so it is true for all complex matrices. – Wintermute Mar 23 '16 at 14:42
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@ Wintermute, thanks for your link, I will read it right now! – Marc Lee Mar 23 '16 at 14:44
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The entries here are not assumed to be real, in contrast to the matrices in the link. – Friedrich Philipp Mar 23 '16 at 14:46
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oh, i have read that article before. The proofs there are all depending on the condition of $A$ being real. – Marc Lee Mar 23 '16 at 14:47
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At least one can prove that $AA^T = A^TA$. Does this help? – Friedrich Philipp Mar 23 '16 at 14:48
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@FriedrichPhilipp, thanks for your hint. Maybe i need a new try under this. – Marc Lee Mar 23 '16 at 14:50
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@FriedrichPhilipp: Is it immediate that $A A^* = A^* A$? – copper.hat Mar 24 '16 at 14:50
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@copper.hat No, because it's wrong. I don't know what I did yesterday... :-( Daniel's answer gives a counterexample. – Friedrich Philipp Mar 24 '16 at 16:47
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@FriedrichPhilipp: Well, that example shows that $A A^T =A^2$ does not allow us to conclude that $A$ is symmetric, but I imagine that $A A^* =A^2$ would imply that $A$is Hermitian. – copper.hat Mar 24 '16 at 21:54
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@copper.hat Yes. Take the second answer in http://math.stackexchange.com/questions/571583/if-matrix-such-aat-a2-then-a-is-symmetric?rq=1 You can almost copy it. – Friedrich Philipp Mar 24 '16 at 22:27
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This is not true in $M_n(\mathbb{C})$, $n\geq 3$.
A very simple counterxample is the following:
Let $v^t=(0,0,1,0,...,0)$ and $w^t=(1,i,0,...,0)$.
Notice that $w^tw=1+i^2=0$ and $w^tv=0$. Define $A=vw^t$.
Now, $A^2=(w^tv)A=0$ and $AA^t=(w^tw)vv^t=0$, but $A=vw^t\neq w^tv=A^t$.
Daniel
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Hi @Wojowu , $A^2=v(w^tv)w^t=(w^tv)vw^t=(w^tv)A$ and $AA^t=v(w^tw)v^t=(w^tw)vv^t$ – Daniel Mar 23 '16 at 16:15
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Ah, right, $w^tv$ and $w^tw$ are scalars, so they commute with all matrices. – Wojowu Mar 23 '16 at 16:17
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