Richardson method optimal parameter The question I am about to ask came to my mind when I was analyzing Richardson method with symmetric and positive definite matrices. But it is really about simple math I somehow can't defeat.
Given real numbers: $\lambda_1\ge \lambda_2 \ge ...\ge \lambda_n>0$ find parameter $\alpha$ such that $\max_k|1-\alpha\lambda_k|$ is minimal. The answer is that $\alpha=\frac{2}{\lambda_1+\lambda_n}$. Why?
 A: This can be proved by induction on $n$. (Side note: This is a nice example of a time when the geometric median in a metric space has a nice formula!) I have to run, however, and so I only sketch an argument below.
First, prove that this holds when $n = 1$ and $n = 2$.
Inductive Hypothesis: Assume that for all $x_1\geq \ldots \geq x_n > 0$ the function $a\mapsto \max_k|1-ax_k|$ is minimized at $\frac{2}{x_1+x_n}$. Given $\lambda_1 \geq \ldots \geq \lambda_n \geq \lambda_{n+1} > 0$, break the argument into three cases:


*

*If $\lambda_1 = \lambda_2 = \ldots = \lambda_{n+1}$, the result follows from the $n=1$ case.

*Similarly, if $\lambda_1 = \lambda_2 = \ldots = \lambda_n > \lambda_{n+1}$, then the result follows from the $n=2$ case.

*Now for the interesting case. Assume that at least one of the first $n-1$ inequalities is strict, i.e., that $\lambda_1 > \lambda_n \geq \lambda_{n+1}$. For $k = 1,\ldots,n$, define


$$
x_k = \begin{cases}
          \lambda_1 - \lambda_n &\text{if } k = 1\\
          \lambda_k             &\text{if } 2\leq k \leq n-1\\
          \lambda_n + \lambda_{n+1} &\text{if } k = n.
      \end{cases}
$$
By the inductive hypothesis
$$
a\mapsto \max_k|1-ax_k|
$$
is minimized at
$$
a = \frac{2}{x_1+x_n} = \frac{2}{\lambda_1+\lambda_{n+1}}.
$$
You can then do some algebra to infer from this that $\min_k|1-a\lambda_k|$ is also minimal at this $a$.
A: since ‎$‎\lambda_1‎\geq ‎\lambda‎_2‎\geq ‎\ldots ‎\geq \lambda_n‎‎‎$ ‎and‎‎ ‎‎$‎‎\alpha‎$ ‎is ‎fixed there, so ‎
‎$‎|1-\alpha \lambda_1|\leq |1-\alpha \lambda_2|‎\leq \ldots ‎\leq‎ |1-\alpha \lambda_n|‎.‎ ‎‎$‎‎
‎$‎|1-\alpha \lambda_n|‎$ ‎is ‎minimum ‎when ‎‎$‎|1-\alpha \lambda_1|=|1-\alpha \lambda_n|$ ‎or ‎‎$‎1-\alpha \lambda_1=\alpha ‎\lambda_n‎-1$, that is, for ‎$‎\alpha‎=‎\frac{2}{\lambda_1 +\lambda_n}‎‎$‎.‎
