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The method of Lagrange multipliers finds critical points (including maxima and minima) of a differentiable function subject to differentiable constraints.

1 vote

What would happen to the solution of primal SVM problem we had 0 in constraint instead of 1

Then $\mathbf{w} = \mathbf{0}$, $b = 0$ satisfies every constraint, but that's not a useful solution. Stanford Prof. Stephen Boyd has an excellent discussion in lecture 13 of his Convex Optimization …
Matthew Gunn's user avatar
8 votes

KKT in a nutshell graphically

The basic idea of the KKT conditions as necessary conditions for an optimum is that if they don't hold at a feasible point $\mathbf{x}$, then there exists a direction $\boldsymbol{\delta}$ that will i …
Matthew Gunn's user avatar
13 votes

The proof of equivalent formulas of ridge regression

It's perhaps worth reading about Lagrangian duality and a broader relation (at times equivalence) between: optimization subject to hard (i.e. inviolable) constraints optimization with penalties for …
Matthew Gunn's user avatar