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Write $H_{f}$ for the Hessian of a real function $f:\mathbb{R}^n\mapsto \mathbb{R}$, and define the bordered Hessian as

$$ H_{f} = \left(\begin{matrix}0 & \nabla f' \\ \nabla f & H \end{matrix}\right),\quad H := \left[ \frac{\partial f}{\partial x_ix_j}\right]_{i,j} \quad i,j=\overline{1,n}$$

and consider the composition $g = h \circ f$. What is the best way to show the relationship of $H_g$ and $H_f$?

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    I didn't change the title, but why does it contain the word "determinant"?2012-02-06
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    Oh, some authors refer to the determinant of the $H_f$ matrix as the bordered Hessian. I find that a lot. It's a good point though.2012-02-06

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We have $\frac{\partial g}{\partial x_i}=h'(f)\cdot\frac{\partial f}{\partial x_i}$ so $\nabla g=h'(f)\nabla f$ and $$\frac{\partial^2g}{\partial x_i\partial x_j}=h''(f)\cdot \frac{\partial f}{\partial x_i}\cdot \frac{\partial f}{\partial x_j}+h'(f)\frac{\partial ^2f}{\partial x_i\partial x_j},$$ hence $$H_g=h'(f)H_f+h''(f)\pmatrix{0\\\ \nabla F}\pmatrix{0\\\ \nabla F}^T.$$

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    The two matrices are not conformal. $H_f$ is $n\times n$, whereas $$ \left(\begin{matrix} 0 \\ \nabla f \end{matrix}\right)\left(\begin{matrix} 0 \\ \nabla f \end{matrix}\right)' $$ is $n+1\times n+1$.2012-02-06
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    In the definition $H_f$ seems to be $(n+1)\times (n+1)$ since $H$ is $n\times n$.2012-02-06
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    Right, sorry, my notation is confusing even to me. Thank you!2012-02-06