# 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 $$A = \begin{bmatrix} -34 & 8 \\ 8 & -2\end{bmatrix}$$ and $$P=\begin{bmatrix} 26 & -4 \\ -4 & 2\end{bmatrix}$$ 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:

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

2. 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.

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

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

Now put the above together.

• Ok, thank you very much, these were my computations too, I just was concerned about assuming that they are diagonal matrices, but this is just a matter of checking it via diagonalization. Thank you again Mar 30 at 8:50
• Thinking a bit more about your answer, it is not clear to me how you can pass from 3 to 4, since the single matrices are not indeed negative definite, but they just have negative eigenvalues and are not symmetric in general. The symmetric part of a negative definite matrix I think is not negative definite for free... @DonAntonio please see the edit in the question Mar 30 at 10:46
• @Dadeslam There is no meaning, at least known to me, of "positive/negative definite" for general matrices, only for symmetric ones...and that's what I wrote at the beginning of my answer. Mar 30 at 11:25
• Perfect, so you are assuming that $AP$ commute? Otherwise, how can $AP$ be symmetric and hence negative definite as you are saying? Mar 30 at 11:27
• I'm sorry but I really don't get it, ok $AP+PA$ is symmetric but the single $AP$ and $PA$ are not in general, so all the results in that PDF do not apply, as far as I understand, since they hold for symmetric matrices. Playing a little bit with matrices I've found a counter example which you can find added in the question..do you understand why this case does not work? Mar 30 at 11:55