# An equality with vectors

Here is a problem with its solution.

Let $$\mathbf{s}_{n\times1}$$ is a vector with elements $$s_{j}\in\left[-1,1\right]$$ for $$j=1,2,\ldots,n$$. I have the equality $$\lambda\alpha\mathbf{s}=\mathbf{a}$$ where $$\lambda$$ is an unknown parameter, $$\alpha\in\left(0,1\right)$$ is a fixed constant and $$\mathbf{a}$$ is an $$n\times1$$ vector. We can show that the minimum value of $$\lambda$$ that satisfies this equality for some $$s_{j}\in\left[-1,1\right]$$ is $$\lambda_{min}=\frac{1}{\alpha}\left\Vert \mathbf{a}\right\Vert _{\infty}$$ where $$\left\Vert \mathbf{a}\right\Vert _{\infty}=\max_{j}\left|a_{j}\right|$$. I would like to extend this equality with a fixed vector $$\mathbf{b}$$ like this: $$\lambda(\alpha\mathbf{s}+\left(1-\alpha\right)\mathbf{b})=\mathbf{a}$$ The problem is the same: What is the minimum value of $$\lambda$$ (if exists) that satisfies this equality for some $$s_{j}\in\left[-1,1\right]$$?

I tried to use inequalities with absolute values but could not get a solution.

• is $b$ fixed? do you mean 'for some s_j' (instead of 'for all')? Commented Dec 27, 2018 at 21:02
• Yes, thanks. I edited the question.
– mert
Commented Dec 28, 2018 at 20:47
• did you appreciate my answer? Commented Jan 8, 2019 at 15:03

Your problem can be summarized as: $$\min_{x \in \mathbb{R},y \in [-\alpha,\alpha]^n} \{ x : xy + x(1-\alpha)b = a\}$$ Substitute $$xy=z$$: $$\min_{x \in \mathbb{R},z \in [-\alpha x,\alpha x ]^n} \{ x : z + x(1-\alpha)b = a\}$$ The dual problem is: $$\max_{v \in \mathbb{R}^n,w_1 \in \mathbb{R}_+^n,w_2 \in \mathbb{R}_+^n} \{ a^Tv : (1-\alpha)b^Tv + \alpha e^T(w_1+w_2) = 1, \; e^T(v + w_1-w_2) = 0\}$$ I do not see an immediate solution to any of these problems, but the last two problems you can just feed to a linear optimization solver.