If two normalized vectors have inner product 1, are they equal (up to a normalized constant)? I am trying to use this fact in adapting a no-cloning theorem in quantum mechanics. Thus consider the following question phrased more precisely.
Let $V$ be an inner product space (over the field $\mathbb{C}$ -- does this matter?) and $v_1, v_2 \in V$. If $\langle v_j,v_j \rangle=1$ for each $j$ and $\langle v_1,v_2 \rangle=1$, does it follow that $v_1 = cv_2$ for some $c \in \mathbb{C}$ with $|c| = 1$?
I think I can argue as follows: from the full statement/the standard proof of the Cauchy-Schwarz inequality I know that $|\langle v_1,v_2 \rangle| = \sqrt{\langle v_1,v_1 \rangle \langle v_2,v_2 \rangle}$ (as is the case here) if and only if $v_1 = cv_2$ for some $c \in \mathbb{C}$. $|c| = 1$ then follows from the given normalization and the positive homogeneity of the norm. Is this correct?