@CarloBeenakker; even if I go one order higher and prove it, how will I show that it is true for all other dimensions? I am confused, can you please help

@CarloBeenakker; how would we get an accurate expression of $\alpha_1,\alpha_2$? I thought giving an upper bound would be easier than finding the exact eigenvalues, so I insisted on the bound

please help me out, thanks in advance

@CarloBeenakker; I had one question, how can we prove that $\alpha_1-\alpha_2<-3$ when we are in higher order. You did here the approximation for $0$ and $1$ order, isn't it? or am I missing something

If we could successfully prove it, then we won't require to use any complex calculations, so I asked you if it can be proved as is done in pure Mathematics, but I really appreciate the hard work you are putting in for me and I thank you for that.

Yes, but finding the eigenvalues requires an approximation whereas finding the bound is quite sharp, and frankly speaking I never did perturbation theory, so I am not aware of this concept

Is there any other way we can prove this fact that $\alpha_1+\alpha_2<-3$ without using numerical mathematics

I was just wondering if this can be done with any bounds that exist on smallest and second smallest eigenvalue of a matrix, it just came to my mind so I thought of sharing with you

I am thankful for such a detailed explanation, I would be grateful if you could explain how you obtained the two matrices $M_0,M_1$

@user64494; I understand that but the question has to be done without numeric, it needs to be proved.

@CarloBeenakker; is there any trick by which it can be proved mathematically without using any numerical computation

@user64494; thanks for your efforts, but the problem needs to be proved mathematically and not through any software like Mathematica

@gmvh; I have edited it