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Stationary points classification using definiteness of the Lagrangian
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[QUOTE="pasmith, post: 6635678, member: 415692"] You are constrained to look at the behaviour of [itex]f[/itex] restricted to the one-dimensional hyperbola [itex]g(x,y) = x^2 - 2y^2 - 1 = 0[/itex]. If [itex]f[/itex] increases or decreases as you move off this curve, then that does not concern you. To remain on the curve, your direction of travel must be orthogonal to [itex]\nabla g[/itex]. In this case, the eigenvector corresponding to the negative eigenvalue is parallel to [itex]\nabla g[/itex] and the eigenvector corresponding to the positive eigenvalue is orthogonal to it, so as far as you are concerned the negative eigenvalue is irrelevant: this critical point is a minimum. In general, [itex]\nabla g[/itex] will not be an eigenvector of the hessian of [itex]f[/itex]. The text therefore defines a vector [itex]\mathbf{w}[/itex] orthogonal to [itex]\nabla g[/itex]) and looks at [tex] f(\mathbf{x} + \alpha\mathbf{w}) \approx f(\mathbf{x}) + \tfrac12\alpha^2 \mathbf{w}^T H \mathbf{w}[/tex] to determine whether a critical point of [itex]f[/itex] subject to this constraint is a minimum ([itex]\mathbf{w}^T H\mathbf{w} > 0[/itex]) or a maximum ([itex]\mathbf{w}^T H\mathbf{w} < 0[/itex]). [/QUOTE]
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Stationary points classification using definiteness of the Lagrangian
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