TITLE: Hessian matrix as derivative of gradient
QUESTION [10 upvotes]: From a text:

For a real-valued differentiable function $f:\mathbb{R}^n\rightarrow\mathbb{R}$, the Hessian matrix $D^2f(x)$ is the derivative matrix of the vector-valued gradient function $\nabla f(x)$; i.e., $D^2f(x)=D[\nabla f(x)]$.

$\nabla f(x)$ is just an $n\times 1$ matrix consisting of $\partial f/\partial x_1,\partial f/\partial x_2,\ldots,\partial f/\partial x_n$.
Then $D[\nabla f(x)]$ must be a $1\times n$ matrix.
But I know that the Hessian matrix is an $n\times n$ matrix consisting of $\partial ^2f/\partial x_i\partial x_j$. How can the given definition be consistent with this?

REPLY [4 votes]: This is an old discussion and I would not intent to correct anything.
Just in mind that this had caused myself long-time confusion so that I wish to accept some simple rules to make it clear.
If one allows to define the operators of the function $f(x):\mathbb{R}^n\rightarrow\mathbb{R}$:

Gradient operator $\nabla$: defined as $n\times1$ column vector.
Derivative operator $\nabla^T$: defined as a row vector (i.e., $1\times n$),
Hessian operator $\mathbf{H}$: defined as the gradient of the derivative of $f(x)$.
$$\mathbf{H}=\left(\begin{matrix}\frac{\partial}{\partial x_1}\\
\vdots\\\frac{\partial}{\partial x_n}\end{matrix}\right)\left(\begin{matrix}\frac{\partial}{\partial x_1}&
\ldots&\frac{\partial}{\partial x_n}\end{matrix}\right)=\left(\begin{matrix}\frac{\partial}{\partial x_1^2} & \ldots & \frac{\partial^2}{\partial x_1\partial x_n}\\\vdots &\ddots&\vdots \\
\frac{\partial^2}{\partial x_n\partial x_1}& \ldots & \frac{\partial^2}{\partial x_n^2}\end{matrix}\right).$$