Let's say we have a gradient vector:

$$\nabla (x) \in R^d$$

Since the gradient is also a function (just like how in 1D, the derivative of a function is also a function), then we can say:

$$\nabla: R^d\longrightarrow R^d$$

That is, the gradient transforms inputs of $R^d$ to yield the gradient at that point.

Can we represent the gradient as a matrix $R^d \times R^d $ in this case? (Since the gradient is a linear operator). Is it useful/used in practice?

Note: A follow-up is that the gradient "matrix" may not be square, since we may need an extra term to represent affine transforms. What about this case?


  • $\begingroup$ The input of the gradient is a scalar field, not a vector. The vector is the input of the scalar field itself. $\endgroup$ – Rodrigo de Azevedo Jan 30 at 8:44

The gradient is a vector of multivariate functions (partial derivatives), not a matrix.

Except for the pure quadrics, it has no reason to be a linear function of the coordinates and cannot be represented as a product of a constant matrix by the position vector.

You can indeed extend to the general quadrics by adding a constant vector. But these are special cases.

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  • $\begingroup$ Thanks for the answer. I understand that the gradient is a vector of partial derivatives, for a real-valued multivariable function. But you could represent it with a non-constant matrix (i.e. matrix with variables for the elements) and a position vector. For instance, if the gradient is wrt $x,y,z$ then the matrix can also contain non-linear functions of $x, y, z$ But I think you are saying this is not useful? $\endgroup$ – information_interchange Jan 30 at 1:33
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    $\begingroup$ @information_interchange: It is worse than useless, it does not make sense. Try it on an example and you'll understand. $\endgroup$ – Yves Daoust Jan 30 at 8:59
  • $\begingroup$ Thanks you are right. But in general, we can still view the gradient as a mapping from $R^n \rightarrow R^n$ $\endgroup$ – information_interchange Mar 28 at 1:25

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