Graduate Hessian as "Square" of Jacobian?

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SUMMARY

The discussion centers on the representation of the Laplacian operator as a "square of Jacobians," specifically in the context of two variables. The Laplacian, denoted as ##\Delta \equiv \nabla^2##, is expressed as the sum of second partial derivatives, ##\partial^2(f) = f_{xx} + f_{yy}##. The user seeks to demonstrate the rotational invariance of the Laplacian using a rotation matrix and the relationship between the Hessian and the gradient. The Hessian is clarified as the matrix product of the gradient, represented as ##H_{ij} = \nabla_i \nabla_j##, and its determinant is denoted as det##H##.

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  • Understanding of Laplacian operator in multivariable calculus
  • Familiarity with Jacobian and Hessian matrices
  • Knowledge of gradient operations and their properties
  • Basic concepts of rotational invariance in mathematics
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  • Investigate the relationship between Jacobians and Hessians in multivariable calculus
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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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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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