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?
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:
$$ x_k = \begin{cases} \lambda_1 - \lambda_n &\text{if } k = 1\\ \lambda_k &\text{if } 2\leq k \leq n-1\\ \lambda_n + \lambda_{n+1} &\text{if } k = n. \end{cases} $$
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 $a$.
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 $\alpha=\frac{2}{\lambda_1 +\lambda_n}$.
