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?


1 Answer 1



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.

  • 1
    $\begingroup$ 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 $\endgroup$
    – Dadeslam
    Commented Mar 30, 2021 at 8:50
  • 1
    $\begingroup$ 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 $\endgroup$
    – Dadeslam
    Commented Mar 30, 2021 at 10:46
  • $\begingroup$ @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. $\endgroup$
    – DonAntonio
    Commented Mar 30, 2021 at 11:25
  • $\begingroup$ Perfect, so you are assuming that $AP$ commute? Otherwise, how can $AP$ be symmetric and hence negative definite as you are saying? $\endgroup$
    – Dadeslam
    Commented Mar 30, 2021 at 11:27
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    $\begingroup$ 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? $\endgroup$
    – Dadeslam
    Commented Mar 30, 2021 at 11:55

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