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In http://apmath.spbu.ru/cnsa/pdf/monograf/Numerical_Optimization2006.pdf on page 151, we are told that approximations of the Hessian matrix with the BFGS formula are always symmetric and positive definite if the initial matrix is symmetric and positive definite - if the Wolfe condition is met in the choosing of the step length. My question: What if the initial matrix is symmetric and negative definite? (The reason: I want to use BFGS for finding the maximum instead of the minimum.) Are all subsequent matrices calculated with the BFGS formula symmetric and negative definite? How to proof this?

EDIT: For future readers, the following questions might be usefull:

(I am still trying to get around this ;-).)

Make42
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    Just multiply the objective function by $-1$ and minimize. – Rodrigo de Azevedo Sep 19 '19 at 17:11
  • @Rodrigo de Azevedo: I wasn't asking for a way to solve a practical programming problem (I would have been able to figure that out ;-)). I was asking for a mathematical proof. – Make42 Sep 19 '19 at 17:37
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    You could modify the BFGS method to work on maximization problems, with an approximate Hessian that was always negative definite, but the update formula would change- this is a simple exercise. – Brian Borchers Sep 19 '19 at 17:41
  • @BrianBorchers: "A simple exercise"? Check out the derivation of the BFGS formula - this does not look like a simple exercise to me. I understand the derivation from Eq. (6.1) up to (6.9), but I am lost afterwards. I am able to pick up at Eq. (6.15), but again am lost with all this "weighted Frobenius norm" and "average Hessian" business. I guess I get the very general idea, but have no clue how to arrive at the presented formulas. – Make42 Sep 20 '19 at 09:38
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    It would be logical for you to start by understanding the derivation of the BFGS method for minimization problems. It would be appropriate to ask a question of this group if you can't follow the discussion in Nocedal and Wright. – Brian Borchers Sep 20 '19 at 14:09

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