How is it true that if $||v||_2=1$ then we can always find a vector $w$ satisfying $w^{*}v=1$ with $||w||_2=1$? How is it true that if $||v||_2=1$ then we can always find a vector $w$ satisfying $w^{*}v=1$ with $||w||_2=1$ ?  I am seeing my instructor write that on the board as if we should know how to prove that, but I am missing its understanding it!!
Here is an example of where it appears: https://www.cs.ox.ac.uk/projects/pseudospectra/thms/thm1.pdf  Look halfway into the proof for the first theorem.
I am not able to understand why this is the case.  Can someone help me understand this please ?
 A: As stated in the file that you linked, this is a corollary of the Hahn-Banach theorem.
There are many forms of the Hahn-Banach theorem; the Wikipedia page states it in much generality, and specializes to the case of normed vector spaces only in passing (under the section "Important consequences"). A statement of Hahn-Banach for normed vector spaces can be found here (Theorem 1), reproduced here:

Let $X$ be a normed vector space, and let $Y$ be a subspace of $X$. Then any continuous linear functional $f$ on $Y$ can be extended to a linear functional $\tilde{f}$ on $X$ with the same operator norm; thus $\tilde{f}$ agrees with ${f}$ on $Y$ and $\|\tilde{f}\| = \|f\|$.

Let us apply this form of the theorem to your question. $Y:=\text{span}\{v\}$ is a subspace of your vector space, whose elements are of the form $cv$ for scalars $c$. You can check that $f(cv)=c\|v\|$ is a linear functional on this subspace $Y$ that has norm $\|f\|=1$. By the Hahn-Banach theorem, this functional can be extended to a functional $\tilde{f}$ on your entire vector space such that $\tilde{f}=f$ on $Y$, and $\|\tilde{f}\|=1$. In your situation you are in $\mathbb{C}^n$ (more generally, the following is true whenever the Riesz representation theorem applies), so there exists some vector $w$ such that this linear functional $\tilde{f}$ can be written as $\tilde{f}(x) = w^* x$. Transferring the properties of $\tilde{f}$, we have $w^*v =\tilde{f}(v) = f(v) = \|v\|$ and $\|w\| = \|\tilde{f}\|=1$.
