Suppose $A \in \mathbb{R}^{n \times n}$ is invertible and we are looking for the solution of $Ax = b$. Given an initial guess $x_1 \in \mathbb{R}$, we show that by reflecting through hyperplanes generated by the rows of $A$, we can generate an infinite sequence $(x_k)_{k=1}^{\infty}$ such that all elements have the same distance to the solution, i.e. $\|x_k - x\| = \|x_1 - x\|$. If the hyperplanes are chosen at random, averages over the sequence converge and $$ \mathbb{E} \left\| x - \frac{1}{m} \sum_{k=1}^{m}{ x_k} \right\| \leq \frac{1 + \|A\|_F \|A^{-1}\|}{\sqrt{m}} \cdot\|x-x_1\|.$$ The bound does not depend on the dimension of the matrix. This introduces a purely geometric way of attacking the problem: are there fast ways of estimating the location of the center of a sphere from knowing many points on the sphere? Our convergence rate (coinciding with that of the Random Kaczmarz method) comes from averaging, can one do better?
翻译：假设$A\ in\ mathbb{R ⁇ n\ times n} 美元是不可忽略的, 我们正在寻找 $Ax = b$ 的解决方案。 根据最初的猜测 $x_ 1\ in\ mathbb{R} 美元, 我们显示, 通过由 $A 生成的超高平面, 我们可以通过 $A 生成一个无限的序列$( x_ k)\\ k= 1\\\\ incnfty} 美元, 这样所有元素都具有与解决方案相同的距离, 即 $x_ k - x * * * * * * * * * x_ k - x * * = * x_ x_ x_ x\ x $ 。 如果高平面是随机选择的, 则在序列趋同和 $\ mall x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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