Gradient and Jacobian

[simple fox]

Why is the hessian the jacobian of the gradient, rather than the jacobian of the jacobian? I know that the jacobian is the generalization of the derivative for functions R^n to R^m, but for multivariate functions R^n to R^1 for some reason we use the gradient instead (the transpose of the jacobian). I am confused generally about why this transpose is necessary.

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[true grotto]

The hessian is the jacobian of grad(f) cause you compute the jacobian of grad(f) and get the matrix composed of all the 2nd partials of f

[simple fox]

Yes, that's the definition. So I guess it's just a dimensionality thing?

[true grotto]

https://math.stackexchange.com/questions/2053229/the-connection-between-the-jacobian-hessian-and-the-gradient?noredirect=1&lq=1

This seems to say the transpose just arises due to the indexing, but ill let you read it yourself

[simple fox]

thank you for finding this

[dense nest]

@simple fox has given 1 rep to @true grotto

[true grotto]

you can google stuff too lol