where A is an nxn symmetric matrix, b is one n-dimensional vector, and also alpha a scalar. Show that

$igtriangledown _xf(x) = Ax + b$


$H = igtriangledown ^2_xf(x) = A$

Is this just a issue of acquisition a derivative v respect to X, how would you attack this one?



$ abla f = (partial f/partial x_1, ldots, partial f/partial x_n)^t$ denotes the vector the partial derivatives of $f$ and also is a completely standard notation.

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On the other hand, $ abla^2 f$ appears to be used here in an unusual way, namely to denote the Hessian (the matrix of all second order partial derivatives), $(partial^2 f/partial x_i partial x_j)_i,j=1^n$.

(The usual an interpretation of $ abla^2 f$ is the Laplacian, $partial^2 f/partial x_1^2 + ldots + partial^2 f/partial x_n^2$.)


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