Let A a normal matrix that sets $ A^9+3A^5=3A^7+A^3 $, prove that A is self-adjoint.

I can use this auxiliary argument - "A is self-adjoint if and only if A is normal and all the roots of it's characteristic polynomial are real", so I just need to prove the last part.

Now, since $ A^9-3A^7+3A^5-A^3=0 $ we know that the minimal polynomial of A divides the polynomial $ t^9-3t^7+3t^5-t^3= t^3(t^6-3t^4+3t^2-1)$ But how am I suppose to solve the polynomial $t^6-3t^4+3t^2-1$ at a test time, there is some kind of trick? Or perhaps there is another way to solve this, without having to find the roots of a 6-deg polynomial?

Thanks in advance!

  • 3
    $\begingroup$ There is a trick here, just substitute $t^2=u$ to get a cubic equation $\endgroup$ – Peter Sep 23 '17 at 19:24
  • 7
    $\begingroup$ Come on, it is just $(t^2-1)^3=0$ $\endgroup$ – Aqua Sep 23 '17 at 19:28

Your Answer

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

Browse other questions tagged or ask your own question.