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I'm having trouble understanding the relationship between the Hessian and the directional derivative.

If given $\bf H$ as the Hessian of some function $f(\bf x)$, and $u$ some unit direction vector, from this answer, the second derivative of $f$ in the direction of $u$ is $\partial_{uu}^2f({\bf x})=u^T{\bf H}u$. I understand that this is essentially the rate of change of the directional derivative $\partial_u f({\bf x})=\nabla f(x) \cdot u $ in the direction of $u$.

$u^T{\bf H}u$ appears say that $\partial_{uu}^2f({\bf x})$ does not change if $u$ is replaced with $-u$. Why is this the case intuitively? I had thought that if the directional derivative increases as we move in the direction of $u$, that the directional derivative would decrease as we move in the opposite direction?

Yandle
  • 903
  • $x^T{\bf H}y$ is the expression for a quadratic form. So the hessian matrix can be seen as the matrix associated to a inner product $\Phi: \mathbb{R^n}\rightarrow \mathbb{R^n}$. Now it is quite intuitive to understand why you get the same result if you consider a vector $u$ or its opposite: the second derivative is the $\Phi$-norm of the vector $u$ which has always a positive value because the norm is positive definite. – Vajra Sep 09 '20 at 14:46

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