Query about Newton's method for two variables - example problem

[Sowmiya_R]

Hi, I liked the explanation and understood most of the part except when I tried to find the eigenvalue of the practice problem: f(x,y) = x^4 + 0.8y^4 + 4x^2 + 2y^2 - xy - 0.2x^2y.
After you solved it with newton’s method and concluded that [x* y*] = [0 0] , I tried to find the eigen value of the initial Hessian matrix in two ways:

1. taking the initial Hessian matrix as such where I tried to solve: det(H-Lambda * I ) = 115.2x^2y^2 + 47.84x^2 - 0.8x - 12x^2z + 76.8y^2 + 31 - 8z - 3.84y^3 - 1.6y + 0.4yz - 9.6y^2 - 4z + z^2 = 0 (where z = lambda) but I ended up doubting if I am going right and thus not solving

2. by substituting (0,0) in the above equation i.e., in det(H-Lambda * I ). Now I got Hessian matrix [[8 -1] [-1 4]]. On solving which I got two roots: 14.1854 and -2.1854. Here one of the roots is positive while the other is negative, so shouldn’t the concavity/convexity of the problem be inconclusive.

How is that we got the info about the curvature as concave up? Could you kindly explain where I went wrong? I was trying to find the concavity/convexity of the curvature from solving the Hessian after you arrived (x*,y*) = (0,0). Kindly explain me, sir/someone.

[lucas.coutinho]

Hi @Sowmiya_R,

Sorry for the late response.

The first method you did can be quite troublesome, since finding roots of polynomials is usually a very hard task.

However, you did right in the second way, but I think you messed up the eigenvalues calculations, because both are positive. Let me show:

The matrix is
M = \left[ \begin{array}{cc} 8 & -1 \\ -1 & 4 \end{array} \right]

So to find the eigenvalues we must find \lambda such that \text{det}(M - \lambda I) = 0.

Now, M - \lambda I = \left[ \begin{array}{cc} 8 - \lambda & -1 \\ -1 & 4 - \lambda \end{array} \right] and \text{det}(M - \lambda I) = (8-\lambda)(4-\lambda) - 1 = \lambda^2 - 12x + 31 which has roots \lambda_1 = 6 - \sqrt{5} and \lambda_2 = 6 + \sqrt{5}, which are all positive.