$\det(A + A^T) > 0$

Question

I have a real square matrix $A$ such that all its eigenvalues are positive. And I need to prove that $\det(A + A^T) > 0$

For symmetric or diagonalizable matrices it's easy, but I don't know what to do in general. I would be grateful fo any hints.

score 7 · Accepted Answer · answered Jan 18 '18 at 14:36

7

Your claim is wrong. The statement is not true for every diagonalisable matrix. Counterexample: let $0<s<1,\ t\ge2$ and $$ A=\pmatrix{1&t\\ 0&s},\ A+A^T=\pmatrix{2&t\\ t&2s}. $$ Then $\det(A+A^T)=4s-t^2<0$.

answered Jan 18 '18 at 14:36

user1551

139,064

Oh, really. Thanks a lot. – D F Jan 18 '18 at 14:37
Why is $A$ diagonalizable? – WeakestTopology Jan 18 '18 at 14:39
4

@DoughnutPump All eigenvalues of $A$ are distinct and real. So, it is diagonalisable over $\mathbb R$. – user1551 Jan 18 '18 at 14:40

$\det(A + A^T) > 0$

1 Answers1