Help explain an equality why $\max_i \{ 1-\alpha \lambda_i\} = |\max \{1-\alpha \lambda_1, 1-\alpha\lambda_n\}| $, circled in the following and why $\alpha = \frac{2}{\lambda_n + \lambda_1}$. Thanks!
I summarize it as the following. $A$ is symmetric, and $\lambda_1 \le ... \le \lambda_n$ are its eigenvalues (all real) in ascending order. We want to minimize $\|I-\alpha A\|_2$, since it is spectral norm of a symmetric matrix, then $\|I-\alpha A\|_2 = \max_i |1-\alpha \lambda_i| $.
What I can see is $\max_i |1-\alpha \lambda_i|=\max \{ |1-\alpha \lambda_1|,|1-\alpha \lambda_n|\}$. I cannot see why the following circled part holds. Am I missing something here?
For each $1\leq i\leq n,$ let $z_{i}=1-\alpha\lambda_{i}.$ Note that when $\alpha<0,$ $\lambda_{i}\mapsto z_{i}$ is order-preserving, and when $\alpha\geq 0,$ $\lambda_{i}\mapsto z_{i}$ is order-reversing. In either case, this leaves $\min\{z_{1},z_{n}\}\leq z_{i}\leq\max\{z_{1},z_{n}\}$ for all $1\leq i\leq n,$ so $$-\max\{|z_{1}|,|z_{n}|\}\leq\min\{z_{1},z_{n}\}\leq z_{i}\leq \max\{z_{1},z_{n}\}\leq\max\{|z_{1}|,|z_{n}|\},$$ which gives $|z_{i}|\leq\max\{|z_{1}|,|z_{n}|\}$ for all $1\leq i\leq n,$ and thus for the maximum over $i$ in this range.
However, the statement $\max_{i}|z_{i}|\leq|\max\{z_{1},z_{n}\}|$ is false generally: Let $\{\lambda_{1},\lambda_{2},\lambda_{3}\}=\{1,2,3\}.$ Then $\max\{1-(1)1,1-(1)2,1-(1)3\}=\max\{0,-1,-2\}=0,$ but $\max_{i}|z_{i}|=2$ for this example, with $\alpha=1.$
The minimizer of $\max\{|1-\alpha\lambda_{1}|,|1-\alpha\lambda_{n}|\}$ with respect to $\alpha$ is then given by $2/(\lambda_{1}+\lambda_{n}),$ as claimed.