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I want to know the geometrical significance of Hessian Matrix. Please could anyone have any idea about it?

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marked as duplicate by Giuseppe Negro, azimut, Rick Decker, TZakrevskiy, Davide Giraudo Sep 16 '13 at 13:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It is unclear to me what the question is. Are you aware of the relationship between the Hessian and convexity? Can you please be more specific? – Jonas Meyer Jul 12 '12 at 6:40
A sci-fi film set in the 18th century with German mercenaries. – copper.hat Jul 12 '12 at 6:41
@copper, no, it's what I'd use to trap Hessian flies... nasty buggers. – J. M. Jul 12 '12 at 6:48
'Sisal'ing observation! Hessian flies were supposedly imported by the Hessians (it says so on wiki, so it must be true...). – copper.hat Jul 12 '12 at 6:51
I must add, copper.hat, that comment is probably the best I've seen on this site. – Jeff Jul 12 '12 at 8:21

Basically, it's a symmetric matrix (Young's theorem) used to describe curvature for functions of a vector variable. For a real valued function of a vector variable, $f:\mathbb{R^n}\rightarrow\mathbb{R}$ it's $n\times n$. The results of $uHu$ are of interest for optimization problems because the Hessian serves to describe local behavior of the function at those points (much like the second derivative test works for $\mathbb{R}\rightarrow\mathbb{R}$). Traits of the eigenvalues of the Hessian also do this (i.e. "positive definite").

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@loganecolss I think this is not my answer at all. I just edited it some minutes after user12345121212 wrote it. – Siminore Sep 6 '13 at 14:25
as said in your answer, Hessian serves to describe local behavior of function, Hessian is just the quadratic term in Taylor expansion, so how could it alone describe the local behavior without taking 1st order derivative into account? – loganecolss Sep 6 '13 at 14:36

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