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

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!

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