# Non-trivial ways for generating matrices $A$ for which $A + A^T$ is positive-definite?

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

# 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) ?

• isn't it true that for any real square $A$ it holds that $A$ is positive definite iff $A+A^T$ is positive definite? – Carlo Beenakker Mar 6 at 16:33
• 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. – dohmatob Mar 6 at 17:18
• 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 $A_{mn}=B_{nm}-A_{nm}$. – Carlo Beenakker Mar 6 at 18:09
• 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 $A = Q \Lambda Q^T + U - U^T$. – student Mar 7 at 1:09
• @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. – dohmatob Mar 7 at 7:44