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I am reading this question: shortest distance from point to hyperplane lagrange method

And in the answer, it directly says "Consider the Lagrangian function" $$L(\mathbf x,\lambda)=\|\mathbf x-\mathbf x_0\|^2+2\lambda(\mathbf w^T\mathbf x+b)$$

My question is: $$\|\mathbf x-\mathbf x_0\|^2$$

Why is this term squared without squared root? and how to understand the whole Lagrangian function? Any help is appreciated.

user3692015
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  • This is simply the definition of the Lagrangian function. The relevant proofs work out when the Lagrangian is defined like this. – NicNic8 Aug 14 '18 at 14:00
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    In this case, minimizing something is the same as minimizing its square. The square makes things much simpler. – Claude Leibovici Aug 14 '18 at 14:10
  • @NicNic8 thank you for your reply but I'm still confused. In this question https://math.stackexchange.com/questions/450370/lagrange-multipliers-to-find-shortest-distance The first term of the Lagrangian function is not squared, so why it is squared here? – user3692015 Aug 14 '18 at 14:12
  • @ClaudeLeibovici Thanks for your reply. But the derivative will become different if you square the first term of the Lagrangian function... – user3692015 Aug 14 '18 at 14:53
  • It does not change anything to the problem. Think about minimizing a distance under any constraint. For sure, the derivative will be different but the )problem is really simpler. Why are using least-square methods ? Just for that. – Claude Leibovici Aug 14 '18 at 15:17

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