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Hi I have a doubt: What is the matrix associated at the second differential? Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$, differiantable and let $df: \mathbb{R}^n \rightarrow \mathbb{R}$, differentiable in $x_0 \in \mathbb{R}^n$.

After some calculations I find the relation: $d^2 f(x_0)(e_k)=\sum_{j=1}^n \frac{\partial}{\partial x_k}(\frac{\partial}{\partial x_k} f(x_0) dx_j)=\sum_{j=1}^n \frac{\partial^2}{\partial x_k \partial x_j} f(x_0) dx_j$

Where ${(e_j)}$ is a basis of $\mathbb{R}^n$ and ${(dx_j)}$ is a basis of the applications from $\mathbb{R}^n$ to $\mathbb{R}$ linear and continuous.

Hence we have also $d^2 f(x_0)(e_k)(e_l)=\frac{\partial^2}{\partial x_k \partial x_l} f(x_0)$

Therefore the matrix associate to $d^2 f(x_0)$ is:

$\begin{bmatrix} \frac{\partial^2}{\partial x_1 \partial x_1} f(x_0) & \cdots & \frac{\partial^2}{\partial x_n \partial x_1} f(x_0) \\ \vdots &\ddots & \vdots \\ \frac{\partial^2}{\partial x_1 \partial x_n} f(x_0) & \cdots & \frac{\partial^2}{\partial x_n \partial x_n} f(x_0)\end{bmatrix}$

That is the transpose of the Hessian of $f$. It should not be the Hessian?

Second fast question: the hessian of $f$ is the transpose of the Jacbian of the gradient of $f$, right?

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Since the partial derivatives commute, the Hessian matrix is symmetric, so you can stop worrying about transpose. (I'm ignoring some pathological and not really relevant examples; in the sense of distributions the partial derivatives always commute.) – user53153 Jan 1 '13 at 17:33
up vote 1 down vote accepted

The matrix of $d^2 f(x_0)$ is indeed the Hessian of $f$ at $x_0$; the point is that for $A \in M_n(\mathbb{R})$ and $X,Y \in \mathbb{R}^n$, $\sum_{k,l=1}^n A_{kl} X_k Y_l = \left\langle X, AY \right\rangle = X^T A Y$ and not $\left\langle AX, Y \right\rangle = X^T A^T Y$.

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