Understanding Matrix Calculus: Laplacian, Hessian, and Jacobian Explained

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SUMMARY

This discussion focuses on matrix calculus, specifically the definitions and relationships between the Jacobian, Hessian, and Laplacian. The Jacobian is defined as the matrix of first derivatives, while the Hessian is a 2x2 matrix representing second derivatives. The Laplacian, defined as the divergence of the gradient, is equivalent to the second derivative in one dimension and is described as the trace of the Hessian. The confusion arises from the overlapping definitions of the Hessian and Laplacian, which both involve second derivatives.

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  • Understanding of matrix calculus concepts
  • Familiarity with derivatives and their notation
  • Knowledge of matrix operations
  • Basic grasp of vector calculus
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Jhenrique
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Hellow!

I was studying matrix calculus and learned new things as:
[tex]\frac{d\vec{y}}{d\vec{x}}=\begin{bmatrix} \frac{dy_1}{dx_1} & \frac{dy_1}{dx_2} \\ \frac{dy_2}{dx_1} & \frac{dy_2}{dx_2} \\ \end{bmatrix}[/tex]
[tex]\frac{d}{d\vec{r}}\frac{d}{d\vec{r}} = \frac{d^2}{d\vec{r}^2} = \begin{bmatrix} \frac{d^2}{dxdx} & \frac{d^2}{dydx}\\ \frac{d^2}{dxdy} & \frac{d^2}{dydy}\\ \end{bmatrix}[/tex]
Those are the real definition for Jacobian and Hessian. However, the definition for Laplacian is ##\triangledown \cdot \triangledown = \triangledown^2##, that corresponds to ##\frac{d}{d\vec{r}} \cdot \frac{d}{d\vec{r}} = \frac{d^2}{d\vec{r}^2}##, but this definition conflicts with the definition for Hessian that is ##\frac{d^2}{d\vec{r}^2}## too. So, where is the mistake with respect to these definitions? I learned something wrong?
 
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Hello Jhenrique! :smile:

The Hessian is a 2x2 matrix, (column-vector)(row-vector).

The Laplacian is a 1x1 matrix, (row-vector)(column-vector). :wink:
 
The laplacian is the trace of the hessian.
 

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