My question is: if we know that $M>0$ for some matrix $M$ and $MA^T + AM <0$ for another matrix $A$, can we say that $A+A^T<0$? The matrices $A$ and $M$ are both real.
I am using the notation $M>0$ to mean $\boldsymbol{v}^\dagger M \boldsymbol{v} >0 \, \forall \boldsymbol{v}$, so $M$ is positive definite, real, and symmetric. The notation $^\dagger$ indicates conjugate transpose. The question could be rephrased as 'if $M$ is positive definite and $MA^T + AM$ is negative definite, then is $A+A^T$ negative definite?'
So far I have tried by writing the eigenequations for $M$ which are $M \boldsymbol{v}_j = m_j \boldsymbol{v}_j$ where $m_j>0$ are the eigenvalues of $M$ and $\boldsymbol{v}_j$ are its eigenvectors. Since $M$ is symmetric, its eigenvectors form a basis: $\{\boldsymbol{v}_j\}$. This allows us to write an arbitrary vector $\boldsymbol{w} = \sum_j w_j \boldsymbol{v}_j$ in this basis. We can then write the equation $\boldsymbol{w}^\dagger (MA^T + AM)\boldsymbol{w}<0$ in terms of the basis $\{\boldsymbol{v}_j\}$ to get:
$$ \sum_{j,k}w^*_jw_k (m_j \boldsymbol{v}^\dagger_jA^T\boldsymbol{v}_k + m_k \boldsymbol{v}^{\dagger}_jA\boldsymbol{v}_k) <0 $$
Since we have $m_j>0$, I feel like it should be possible to prove $\boldsymbol{w}^\dagger(A+A^T)\boldsymbol{w}<0$ from here, but I am having trouble 'seeing' what to do next.
Extra question: if $A+A^T<0$ is not true for a general $A$ with $MA^T + AM <0$ , is it true when we have the extra condtion that the real parts of the eigenvalues of $A$ are all negative as well as $MA^T + AM <0$?
Any help appreciated! Thanks!
The answer to your first question is no. Let $m>0$, $$ H=-\pmatrix{1&1\\ 1&2}<0, \quad M=\pmatrix{1&0\\ 0&m}>0, \quad A=HM^{-1}. $$ Then $AM+(AM)^T=2H<0$. However, as $\lim_{m\to+\infty}(A+A^T)=-\pmatrix{2&1\\ 1&0}$ is indefinite, $A+A^T$ is indefinite when $m$ is sufficiently large.