Hessian as "Square" of Jacobian?

  • #1

WWGD

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Hi,
Is there a way of representing the Laplacian ( Say for 2 variables, to start simple) ##\partial^2(f):= f_{xx}+f_{yy} ## as a "square of Jacobians" ( More precisely, as ##JJ^T ; J^T ## is the transpose of J, for dimension reasons)? I am ultimately trying to use this to show that the Laplacian is rotationally-invariant, using a rotation matrix and manipulating the product.
 
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  • #2
Don't understand. ##\Delta \equiv \nabla^2## but even for 2 variables the Hessian is not as you write it !? it is the matrix product of ##\nabla## and ##\nabla##: ##H_{ij} = \nabla_i\nabla_j##

Ah, my bad: Hessian = det##H##
 

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