I've been working on this proof from a test review for about an hour and a half trying to figure out what to do. I have also scoured the internet in an attempt to find a similar problem to hint at where I should be going. I'm in an elementary level linear algebra course.

The proof reads: Let A and B be symmetric real n × n matrices with all negative eigenvalues. Prove that the matrix C = A + B also has negative eigenvalues.

I have tried looking at the matrices A, B, and C as an equation in spectral decomposition form and in a form of orthogonal diagonalization and I must be missing something obvious or going about it in the wrong way. In my head, I can kinda get the picture of what's going on, but I can't find the theorems and the words to describe it. Help of any form appreciated. Thanks!

  • $\begingroup$ Do you know Weyl's inequality? Because that is what I would use here. $\endgroup$ – Cameron Williams Apr 19 '16 at 0:02
  • $\begingroup$ We've not gone over it in class... I can look it up though :) Thanks $\endgroup$ – nickroll Apr 19 '16 at 0:03
  • $\begingroup$ I guess the matrices are real ones, right? $\endgroup$ – thanasissdr Apr 19 '16 at 0:25
  • $\begingroup$ Yes. Sorry. I'll edit it now. $\endgroup$ – nickroll Apr 19 '16 at 0:26
  • $\begingroup$ Ok, then the answer holds, otherwise we would need $A,B$ to be Hermitian ones. $\endgroup$ – thanasissdr Apr 19 '16 at 0:27


Since $A,B$ are real symmetric matrices, this means they are negative definite, since all their eigenvalues are negative. We claim that $A+B$ is a negative definite matrix, as well. We can do so by following the definition. For every vector $x$...

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