The Gradient and the Hessian of a Function of Two Vectors

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Discussion Overview

The discussion revolves around finding the gradient and Hessian of a function defined by two n-dimensional vectors, f(𝑥,𝑦). Participants explore the mathematical treatment of this function in terms of its derivatives.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant suggests treating the function as a function of 2n variables to derive the gradient as a 1-by-2n vector and the Hessian as a 2n-by-2n matrix.
  • Another participant elaborates on the interpretation of the gradient as a linear map and the Hessian as a bilinear map, providing a more detailed mathematical framework.
  • There is a mention of breaking down the source space into two n-dimensional vectors, leading to the concept of "partial" gradients and a block structure for the Hessian.

Areas of Agreement / Disagreement

Participants generally agree on the approach to deriving the gradient and Hessian, but the discussion includes varying levels of detail and interpretation regarding the mathematical structures involved.

Contextual Notes

Some assumptions about the definitions of the gradient and Hessian in this context may not be explicitly stated, and the discussion does not resolve all nuances regarding the treatment of the function's derivatives.

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Hi,

Suppose we have a function of two n-dimensional vectors [tex]f(\mathbf{x},\mathbf{y})[/tex]. How can we find the gradient and Hessian of this function?

Regards
 
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i would just treat it as a function of 2n variables and take the vector of 1st, then the matrix of 2nd derivatives.
 
mathwonk said:
i would just treat it as a function of 2n variables and take the vector of 1st, then the matrix of 2nd derivatives.

So, the gradient will be 1-by-2n vector, and the Hessian will be 2n-by-2n matrix?
 
yep. basically the gradient is a linear map approximating the original map. so you could also view it as a linear map from R^n x R^n -->R, and the Hessian I suppose as a bilinear such map.

I.e. if R^n = V, the 1st derivative is a linear map VxV-->R, and the second derivative is a symmetric bilinear map (VxV)x(VxV)-->R.

So if you really want to break up your source space into a pair of n vectors, then
you get two "partial" gradients, each a 1byn matrix, and you get a 2 by 2 Hessian matrix, where each block is an nbyn matrix, i.e. 4 nbyn matrices of vector second derivatives
 
mathwonk said:
yep. basically the gradient is a linear map approximating the original map. so you could also view it as a linear map from R^n x R^n -->R, and the Hessian I suppose as a bilinear such map.

I.e. if R^n = V, the 1st derivative is a linear map VxV-->R, and the second derivative is a symmetric bilinear map (VxV)x(VxV)-->R.

So if you really want to break up your source space into a pair of n vectors, then
you get two "partial" gradients, each a 1byn matrix, and you get a 2 by 2 Hessian matrix, where each block is an nbyn matrix, i.e. 4 nbyn matrices of vector second derivatives

Ok, I see. Thank you.

Regards
 

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