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Suppose that we are trying to minimize a function $f$ on $\mathbb{R}^n$ and we apply Newton's method, updating: \begin{align} \mathbf{x}_{n+1} = \mathbf{x}_n - [\nabla^2 f(\mathbf{x}_n)]^{-1} \nabla f(\mathbf{x}_n). \end{align} If the Hessian $\nabla^2 f$ is not positive semidefinite, Newton's method may update $x$ against the gradient because \begin{align} \mathbf{x}_{n+1} - \mathbf{x}_n &= - [\nabla^2 f(\mathbf{x}_n)]^{-1} \nabla f(\mathbf{x}_n) \\ \implies \nabla f(\mathbf{x}_n)^T(\mathbf{x}_{n+1} - \mathbf{x}_n) &= - \nabla f(\mathbf{x}_n)^T[\nabla^2 f(\mathbf{x}_n)]^{-1} \nabla f(\mathbf{x}_n) \end{align} could have either sign. Is there a graphical way to interpret what's going on?

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up vote 1 down vote accepted

You say you're trying to minimize, but Newton's method actually looks for a critical point (a solution of $\nabla f(x) = 0$). A critical point $x_0$ where the Hessian matrix is not positive semidefinite is not a local minimum but a saddle point or a local maximum: if $v$ is an eigenvector of $\nabla^2 f(x_0)$ for a negative eigenvalue $\lambda$ then $f(x_0 + t v) = f(x_0) + \lambda t^2 |v|^2/2 + O(|t|^3)$ as $t \to 0$.

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Thanks for taking the time to answer. You're right that Newton's method will only give a local minimum (and will stop at a local maximum) since it's looking for a point where the gradient is zero. I am still having trouble visualizing the second part of your answer. – Neil G Mar 21 '13 at 0:45
For the purposes of visualization, it may be helpful to consider a case where the Hessian matrix is diagonal. – Robert Israel Mar 21 '13 at 1:22
Thanks, I think I see it now. Essentially, Newton's method is looking for points with zero slope, and decides for each eigenvector whether to go towards a local maximum or minimum based on the sign of the eigenvalue. Is that right? – Neil G Mar 21 '13 at 4:55
If you're close enough to a critical point it will go to that critical point, whether it's a local max, local min, or saddle. – Robert Israel Mar 21 '13 at 7:08

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