Minimum of inner products in unitary sphere

Question

Let $A$ be a real $n\times m$ matrix and $x$ a real $m\times 1$ vector. Define the sphere $$ S^{n-1} = \{u\in\mathbb{R}^n:\|u\|_2=1\}. $$ Is it true that $$ -\|Ax\|_2 = \min_{u\in S^{n-1}}u^{T}Ax\quad ? $$ This part of the next equality found in "High-dimensional statistics" by MJ Wainwright (p. 188): $$ -\min_{v\in V}\|Ax\|_2 = \max_{v\in V}\{-\|Ax\|_2\}=\max_{v\in V}\min_{u\in S^{n-1}}u^{T}Ax. $$ Is this true?

Answer 1

Yes, it is true.

By the Cauchy-Schwartz inequality we get $$|u^TAx|=|\langle u, \, Ax\rangle|\le \|u\|\cdot \|Ax\|=\|Ax\|\,,$$ so $u^TAx\ge -\|Ax\|$ for every $u\in S^{n-1}$,
and if $Ax\ne 0$, we reach $-\|Ax\|$ by setting $u=-\frac{Ax}{\|Ax\|}\in S^{n-1}$.