Should the lagrange multipliers have a fixed sign for equality constraints?

Question

I Have implemented the augmented lagrangian optimization with equality constraints, with a problem formulation as follows. $$\min f(x)$$

s.t.

$$\sum_{i=1}^{n}c_{i}(x)=0$$ I write the lagrangian formulation as $$ \mathcal{L}(x, \lambda) = f(x) - \sum_{i=1}^{n} \lambda_i c_i(x) + \frac{1}{2}(\tau c_i^2(x)) $$ What I've been getting very confused about is about the sign of $\lambda_i$s here. I've read answers to other questions (q1, q2, q3) and I see that for equality constraints the sign does not matter, but does it still mean that we have

either $(\lambda_i \geq 0 \;\forall i)$ or $(\lambda_i \leq 0 \;\forall i)$

In other words, should $\lambda$ always have the same sign in case of equality constraints, or is it unrestricted and can take any value?

Answer 1

Strictly speaking, the Lagrange multipliers can take any value. In some economic applications (and others, I suppose) there are reasons to fix signs in a certain way. This is related with the fact that, when you use the minus sign in the Lagrangian function, the value of the Lagrange multiplier is the derivative of the optimum with respect to the (constant) rhs in the equality constraint.

Answer 2

The problem

$$ \min f(x)\ \ \text{s.t}\ \ \sum_{i=1}^n c_i(x)=0 $$

can be handled making a due formulation according to the Lagrange multipliers paradigm.

The lagrangian

$$ L(x,\lambda) = f(x)+\lambda\sum_{i=1}^n c_i(x) $$

has the stationary points located at the solutions for

$$ \cases{ \nabla f(x) + \lambda \nabla \sum_{i=1}^n c_i(x) = 0\\ \sum_{i=1}^n c_i(x) = 0 } $$

Giving $\{x^*,\lambda^*\}$ as a stationary point, we have

$$ \nabla f(x^*) + \lambda^* \nabla \sum_{i=1}^n c_i(x^*) = 0 $$

which affirms that

$$ \nabla f(x^*) = -\lambda^* \nabla \sum_{i=1}^n c_i(x^*) $$

There, two vectors are related $\nabla f(x^*)$ and $\nabla \sum_{i=1}^n c_i(x^*)$ so the sign for $\lambda^*$ is determined by the relative orientation of those two vectors.