Prove the matrix is symmetric negative definite Given a curve of symmetric positive definite matrices $t\mapsto P(t)\in\mathbb{R}^{n\times n}$ and a constant symmetric negative definite matrix $A\in\mathbb{R}^{n\times n}$, how would you prove that the symmetric matrix $Q(t)=AP(t)+P(t)A$ is negative definite? Or, if it is not true, what property would you impose on $A$ so that this is true?
It is quite easy to prove that both $AP(t)$ and $P(t)A$ have negative eigenvalues, however, the sum of the two matrices I think needs additional assumptions on $A$ in order to have negative eigenvalues. I say this because for a given matrix $S$ with all negative eigenvalues, we can not say anything about the Eigenvalues of symmetric and antisimmetric part
It is clear that for some choice of $A$ it works, for example $A=-c^2Id$ but in general it does not hold for any $A$ negative definite.
An example where this does not work is $$A = \begin{bmatrix} -34 &amp; 8 \\ 8 &amp; -2\end{bmatrix}$$ and $$P=\begin{bmatrix} 26 &amp; -4 \\ -4 &amp; 2\end{bmatrix}$$ where the matrix $AP+PA$ has a one eigenvalue which is positive and one which is negative. I know this is not a nice example, but I've found it playing a little bit with some matrices.
Do you understand what could be an additional property to impose onto $A$ to guarantee that $AP+PA$ has negative eigenvalues?
Hints:
All the matrices here are real symmetric:

Any symmetric matrix is diagonalizable, so we can assume its form is
$$\;\begin{pmatrix}\lambda_1&amp;0&amp;\ldots&amp;0\\
0&amp;\lambda_2&amp;\ldots&amp;0\\
\ldots&amp;\ldots&amp;\ldots&amp;\ldots\\
0&amp;0&amp;\ldots&amp;\lambda_n\end{pmatrix}\;$$

Any symmetric matrix is positive definite (=PD) iff all its eigenvalues are positive, and it is negative definite (=ND) iff all its eigenvalues are negative.

The product of a PD matrix and a ND definite matrix is a ND matrix

The sum of two PD (ND) matrices is a PD (ND) matrix.


Now put the above together.
Hints:
All the matrices here are real symmetric:

Any symmetric matrix is diagonalizable, so we can assume its form is
$$\;\begin{pmatrix}\lambda_1&amp;0&amp;\ldots&amp;0\\
0&amp;\lambda_2&amp;\ldots&amp;0\\
\ldots&amp;\ldots&amp;\ldots&amp;\ldots\\
0&amp;0&amp;\ldots&amp;\lambda_n\end{pmatrix}\;$$

Any symmetric matrix is positive definite (=PD) iff all its eigenvalues are positive, and it is negative definite (=ND) iff all its eigenvalues are negative.

The product of a PD matrix and a ND definite matrix is a ND matrix

The sum of two PD (ND) matrices is a PD (ND) matrix.


Now put the above together.