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?



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
    Mar 30 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
    Mar 30 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
    Mar 30 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
    Mar 30 at 11:27
  • 1
    $\begingroup$ 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
    Mar 30 at 11:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.