Maximize $\frac{\sum_i \lambda_i x_i}{\sum_i |x_i|}$

Question

Suppose $x=(x_1,x_2,\ldots,x_n) \in \mathbb{R}^n$ under the constraint that $Ax = 0$. $A$ is a $m(

My intuition is that we first write out a basis of $ker(A)$ as $\{\lambda,\alpha,\ldots,\beta\}$ and then we try adding $\{\alpha,\ldots,\beta\}$ to $\lambda$ so that $x=\lambda+c_1\alpha+c_2\beta+\cdots$ has the same order of elements as $\lambda$ and also majorizes $\lambda$. But I don't know if it is correct. Also I hope to find an analytical expression of $x$ as a function of the basis of $ker(A)$.

Answer 1

I started solving this problem, but did not end up with a final answer. I'm still posting this so others may continue (or not make the same derivations to arrive at the same conclusion).

Note that if $x$ is optimal, then so is $cx$ for $c>0$. We can therefore rewrite the problem to: $$\max_x \left\{ \sum_i \lambda_i x_i : \sum_i |x_i| \leq 1, Ax=0\right\}$$ Note that this is a convex optimization problem (and can be written as a linear optimization problem and solved by free solvers like Clp). Its Lagrangian is: $$L(x,y,z) = \sum_i \lambda_i x_i - y\left(\sum_i |x_i| - 1\right) + \sum_i (A_i^Tz)x_i$$ The dual is obtained by taking the maximum over $x$. The maximum is $\infty$ unless $|\lambda_i + A_i^Tz| \leq y \; \forall i$. The dual is therefore: $$\min_{y\geq 0,z} \left\{ y : |\lambda_i + A_i^Tz| \leq y \; \forall i \right\}$$ or: $$\min_{z} \max_i |\lambda_i + A_i^Tz|$$ I don't see if there is an analytical solution to the dual. If there is, you can easily obtain a solution to the original problem.