So, I am trying to prove that given $c \in \mathbb{R}^d $, we have to prove that spectral norm of $c^T$ equals the Euclidean norm of $c$, meaning that:
$$\max_{x \neq 0} \frac{|c^Tx|}{\|x\|} = \|c\|$$
Where $\|c\|$ is the Euclidean norm of $c$.
So far, I've managed to prove that $\|c\| \ge \frac{|c^Tx|}{\|x\|}$ through Cauchy-Schwartz inequality, but I need to prove the other side.
In solutions they say we use that $x = c$ and from there $\frac{|c^Tx|}{\|x\|} = \frac{|c^Tc|}{\|c\|} = \frac{\|c\|^2}{\|c\|} = \|c\|$ so the spectral norm is at least the Euclidean norm. My question is: how comes this side even works and why do we choose $x = c$ for solving?
Best Answer
So I finally understand what I was supposed to be doing; first, we know that: $$\max_{x \neq 0} \frac{|c^Tx|}{\|x\|} \ge \frac{|c^Tx|}{\|x\|}, \forall x \neq 0$$ So from here it doesn't matter which $x$ we use (unless it is 0), so for the sake of proving that spectral norm is equal to Euclidean norm, we use $x = c \neq 0$ and insert it to the right side of the above inequality, and from where we get that: $$\max_{x \neq 0} \frac{|c^Tx|}{\|x\|} \ge \frac{|c^Tc|}{\|c\|} = \frac{\|c\|^2}{\|c\|} = \|c\|$$
Thus proving for both inequalities and finally getting that : $$\max_{x \neq 0} \frac{|c^Tx|}{\|x\|} = \|c\|$$