Computing eigenspectrum of a Riemannian Hessian matrix

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Rajendra

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Mar 1, 2017, 4:30:15 AM3/1/17
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Dear sir,

I have two doubts,

1. I get different eigenspectrum of  a Riemannian Hessian matrix when I use the 'hessianspectrum(problem, X)' function and use eigs function on the matrix getHessian(problem, X, Z), where X is a point on the manifold and the point of the convergence of the RTR method and Z is a tangent vector in the tangent space at the point X.

2. How to analytically prove  that a Riemannian Hessian matrix (Hess f(X) [Z] ) is a positive definite matrix or not? Will the definiteness of the Riemannian Hessian matrix depend on the choice of the tangent vector Z at the point X on the manifold?

with regards,
Rajendra

BM

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Mar 1, 2017, 12:20:51 PM3/1/17
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Hello Rajendra, 

Thanks for the questions. 

1. I get different eigenspectrum of  a Riemannian Hessian matrix when I use the 'hessianspectrum(problem, X)' function and use eigs function on the matrix getHessian(problem, X, Z), where X is a point on the manifold and the point of the convergence of the RTR method and Z is a tangent vector in the tangent space at the point X.

I should mention that
getHessian(problem, X, Z) is not the Hessian, but it is "Hessian along Z". For example, if X were a vector (so, Z is a vector too), then getHessian(problem, X, Z) is "Hessian at X multiplied with Z". Hence, doing eigs on getHessian(problem, X, Z) does not make sense. 

The correct spectrum is indeed given by hessianspectrum(problem, X). 


2. How to analytically prove  that a Riemannian Hessian matrix (Hess f(X) [Z] ) is a positive definite matrix or not? Will the definiteness of the Riemannian Hessian matrix depend on the choice of the tangent vector Z at the point X on the manifold?

No, positiveness does not depend on Z. We look at hessianspectrum(problem, X) and see whether they are positive. 

Regards,
Bamdev

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