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Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a convex differentiable function and let $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^m$. I need to show that for a feasible solution $x^*$ to the convex problem: $$\text{min}\ f(x)\\Ax = b$$ such a solution is optimal if and only if $\exists \lambda \in \mathbb{R}^m$ s.t. $\nabla f(x^*) = A^T\lambda$


I'm not completely sure how to go about this. I can start with the condition that a solution $x^*$ minimises $f$ if and only if: $$\nabla f(x^*)^T (x - x^*) \geq 0, \ \ \ \forall x \in \{x \in \mathbb{R}^n | Ax = b\}$$ But then I'm not sure how to go from here (is the above right? I'm still a bit of a newbie in optimisation). For example, can I say that since $x^*$ is feasible that if we take another $x$ in the domain of $f$ as above: $$Ax = b\\ Ax^* = b\\ \implies A(x - x^*) = 0$$ and do something with that?

Any hints in the right direction would help. Thanks a lot!

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    Hint: That $\lambda$ might be the eigenvalue associated to a certain matrix.2017-02-09
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    I'm sorry it has been quite a while since I studied linear algebra. Are you saying it is an eigenvalue for $A$ so $A \underline{v} = \lambda \underline{v}$?2017-02-09

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