# Derivative of $max(0, \mathbf{x})$ for the vector $\mathbf{x} \in R^n$

I'm reading the tutorial The Matrix Calculus You Need For Deep Learning: https://arxiv.org/abs/1802.01528. In Page 25, the derivative of the ReLu function $$\text{max}(0, \mathbf{x})$$, where the variable $$\mathbf{x}$$ is a vector $$\in R^n$$, is given as follows: My question is, why is the derivative a vector instead of a diagnol matrix as follows?

\begin{align*} \frac{\partial}{\partial \mathbf{x}}max(0, \mathbf{x}) &= diag( \frac{\partial}{\partial x_1}max(0, x_1), \frac{\partial}{\partial x_2}max(0, x_2), \dotsc, \frac{\partial}{\partial x_n}max(0, x_n) ) \\ \end{align*}

The result of the ReLu function $$max(0, \mathbf{x})$$ is a vector, and the derivative of a vector with respect to a vector variable is a Jacobian matrix. In this case, though, the Jacobian matrix happens to be diagonal too.

Page 7 of the same tutorial presents a general rule as below. I'm not sure how this does not apply to the derivative of ReLu function. You are correct, by definition the form should be a matrix. However, in this case all terms of the off-diagonal evaluate to zero. Thus, when applying the gradient $$H$$ to an arbitrary vector $$v$$, it holds
$$Hv = diag(H) \odot v = h \odot v$$
Therefore, it is often simpler/more efficient to calculate only the diagonal terms ($$h$$) and employ the Hadamard/element-wise ($$\odot$$) product instead of doing the full matrix product. This is probably what your reference does.