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Suppose I have two real-valued diagonalizable matrices $A$ and $B$ and that all of the eigenvalues of each matrix have strictly negative real components. Moreover, assume:

  1. $A$ is symmetric (which means that all of its eigenvalues must be real)
  2. The diagonal entries of $B$ are negative
  3. The off-diagonal entries of $B$ are positive
  4. The sum of any column of $B$ is nonpositive ($\le 0$).

I have a few questions I would like to answer:

  1. Are the real components of the eigenvalues of $A+B$ guaranteed to be negative?
  2. Is $A+B$ diagonalizable?
  3. What else can be said about the eigenvalues of $A+B$?

If it is difficult to answer the above questions, what if we assume that all of the eigenvalues of $B$ are real? Or that the "leading" eigenvalue (eigenvalue with the largest/least negative real component) of $B$ is real?

If $A$ and $B$ are both symmetric, I think we could say that they are both negative definite operators and that their sum must also be a negative definite operator, so the answers to 1 and 2 in that case would be yes. (Correct me if I'm wrong!) However, I'm interested in the more general case where $B$ may not be symmetric. Moreover, $B$, in my physical problem of interest, is generally not a negative definite operator; i.e., $\exists x \ne 0$ such that $x^T B x > 0$.

Any thoughts on this would be greatly appreciated!

EDIT1: Added further restrictions on $B$ based on the physics of my problem.

nukeguy
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  • Question 1 might have an interesting answer. For question 2 the answer is no. 3 is related to 1; you might be able to get some useful inequalities. – Ben Grossmann Feb 27 '17 at 15:36
  • To your additional statements: $B$ having real eigenvalues probably won't make a difference. The leading eigenvalue doesn't make much of a difference here either. Your reasoning in the symmetric case is correct. – Ben Grossmann Feb 27 '17 at 15:38
  • Something nice can be said in the case that $B + B^T$ has negative eigenvalues (which in turn guarantees that the eigenvalues of $B$ have negative real part). This doesn't lead to a general answer, though. – Ben Grossmann Feb 27 '17 at 15:39
  • "Something nice can be said in the case that $B+B^T$ has negative eigenvalues" -- could you elaborate more on what that "something nice" is? – nukeguy Feb 27 '17 at 15:43
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    in particular, we could guarantee that $A + B$ will have eigenvalues with negative real part since its symmetric part $\frac{(A + B) + (A + B)^T}{2}$ is negative definite. – Ben Grossmann Feb 27 '17 at 15:52
  • Hm, so it sounds like you're claiming that, if $(C+C^T)/2$ is negative definite, then $C$ will have eigenvalues with negative real part. I see that $\frac{(A+B)+(A+B)^T}{2}$ would be negative definite but I don't see why this implies $A+B$ will have eigenvalues with negative real part. Could you explain this a bit more? Thanks! – nukeguy Feb 27 '17 at 16:31
  • I am indeed claiming that if $(C + C^T)/2$ is negative definite, then $C$ will have eigenvalues with negative real part. – Ben Grossmann Feb 27 '17 at 16:33
  • Er sorry, I should've been more clear in my question -- why is that claim true? – nukeguy Feb 27 '17 at 16:58
  • Doing so would constitute an answer in itself, and that's not the primary question you're asking. I'm trying to find a reference – Ben Grossmann Feb 27 '17 at 16:59
  • here it is. Took 30 minutes to find that one. – Ben Grossmann Feb 27 '17 at 17:02
  • Let us continue this discussion in chat. – nukeguy Feb 27 '17 at 17:08

0 Answers0