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Disclaimer: This might be an SE question, but I'm not quite sure...

Thanks in advance!

Setup

So, it is known (see Proposition 5.2) that if $A + A^T$ is positive-definite then $A$ must be a $P$-matrix. Thus this gives a simple way for generating P-matrices (which are otherwise very complex objects, due to the "principal minors" conditions defining them)

Question

  • What (if any ) interesting things can be said about matrices $A$ for which $A + A^T$ is p.d. ?

  • What are some interesting non-trivial ways for generating such matrices (i.e matrices for which $A + A^T$ is p.d) ?

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  • $\begingroup$ isn't it true that for any real square $A$ it holds that $A$ is positive definite iff $A+A^T$ is positive definite? $\endgroup$ – Carlo Beenakker Mar 6 at 16:33
  • $\begingroup$ Hum, I see your point. Indeed, $x^T(A + A^T)x = x^TAx + x^TA^Tx = x^TAx + x^TAx = 2x^TAx \ge 0 \iff x^TAx \ge 0$. But for positive definiteness, there is also the symmetry condition which has to be satisfied. But then, there are non-symmetric matrices $A$ for which $x^TAx \ge 0$ with equality iff $x=0$. forgot the the name of such matrices. Thus your claim is not (very) true, except perhaps you're using a definition of +ve definiteness which is different from the standard one. $\endgroup$ – dohmatob Mar 6 at 17:18
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    $\begingroup$ I am probably missing something, but can't you just generate an arbitrary positive definite real symmetric matrix $B$ and then choose $A_{nm}$ randomly $\in\mathbb{R}$ for $n\leq m$, followed by $A_{mn}=B_{nm}-A_{nm}$. $\endgroup$ – Carlo Beenakker Mar 6 at 18:09
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    $\begingroup$ I deleted my answer because @CarloBeenakker 's comment already said it. To sample a symmetric positive definite matrix, one could e.g. pick a diagonal matrix $\Lambda$ with positive entries, a random orthogonal matrix $Q$, and a random upper triangular matrix $U$ to set $A = Q \Lambda Q^T + U - U^T$. $\endgroup$ – student Mar 7 at 1:09
  • $\begingroup$ @student: This is indeed an interesting sub-family of such matrices. Thanks. Thanks to Carlo too. Indeed, in general, any squared matrix of the form $A=B+C-C^T$ where $B$ is p.d. (no conditions on $C$) should do the job since $A+A^T=B+C-C^T + B + C^T-C = 2B$ which is p.d. BTW, if think your comment should be an answer. $\endgroup$ – dohmatob Mar 7 at 7:44

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