One of the proofs I am working on (Cauchy-Schwarz Inequality) requires me to simplify $\Bigl\lVert\frac{\langle u,v \rangle}{\lVert v \rVert} v \Bigr\rVert ^2$ into the form $\frac{\lvert \langle u,v \rangle \rvert ^2}{\lVert v \rVert ^2}$ where $u,v \in V$ over field $\mathbb F$
I have no clue where I went wrong...but here is what I have so far.
Firstly, by definition:
$\lVert v \lVert = \sqrt{\langle v,v \rangle}$ where $\lVert v \rVert \in \mathbb R$
Further, note that $\langle u,v \rangle$ (the inner product) is a map between a vector space $V$ and a field $\mathbb F$. Therefore, $\langle u,v \rangle \in \mathbb F$.
Although my textbook (Linear Algebra as an Introduction to Abstract Mathematics) has not explicitly mentioned it, based on some of the things I've read on this site, I believe the inner product can only map a vector to either $\mathbb F = \mathbb R$ or $\mathbb F=\mathbb C$.
So with that being said, the expression $\frac{\langle u,v \rangle}{\lVert v \rVert}$ is simply a scalar belonging to $\mathbb F$, which means they can be pulled out of the inner product. Continuing:
$\Bigl\lVert\frac{\langle u,v \rangle}{\lVert v \rVert} v \Bigr\rVert ^2 = \sqrt{\langle \frac{\langle u,v \rangle}{\lVert v \rVert} v, \frac{\langle u,v \rangle}{\lVert v \rVert} v \rangle }^2 = \langle \frac{\langle u,v \rangle}{\lVert v \rVert} v, \frac{\langle u,v \rangle}{\lVert v \rVert} v \rangle $.
Applying the properties of linearity and conjugate linearity on the first and second "slots" (term the author uses), respectively, of the inner product:
$\langle \frac{\langle u,v \rangle}{\lVert v \rVert} v, \frac{\langle u,v \rangle}{\lVert v \rVert} v \rangle = \frac{\langle u,v \rangle}{\lVert v \rVert} \overline{\Big(\frac{\langle u,v \rangle}{\lVert v \rVert}\Big)}\langle v , v \rangle $.
Looking at $\frac{\langle u,v \rangle}{\lVert v \rVert} \overline{\Big(\frac{\langle u,v \rangle}{\lVert v \rVert}\Big)}$, let's assume the more general case that $\frac{\langle u,v \rangle}{\lVert v \rVert} \in \mathbb C$...specifically, let it equal (in its trigonemtric form) some arbitrary $z = r\big(\cos(\theta), \sin(\theta)\big)$. Correspondingly, $\bar z = r\big(\cos(\theta), -\sin(\theta)\big)$.
From trigonometric identities, $r\big(\cos(\theta), -\sin(\theta)\big) = r\big(\cos(-\theta), \sin(-\theta)\big)$. Following the rules of complex multiplication, we get:
$\frac{\langle u,v \rangle}{\lVert v \rVert} \overline{\Big(\frac{\langle u,v \rangle}{\lVert v \rVert}\Big)} = r\big(\cos(\theta), \sin(\theta)\big)*r\big(\cos(-\theta), \sin(-\theta)\big) = r^2\big(\cos(0),\sin(0)\big)=r^2 \in \mathbb R$.
From the definition of the modulus of a complex number, recall that $r=\lvert z \rvert$. Therefore, $r^2 = \lvert z \rvert^2 = \Big\lvert \frac{\langle u,v \rangle}{\lVert v \rVert} \Big\rvert^2$.
Therefore:
$\frac{\langle u,v \rangle}{\lVert v \rVert} \overline{\Big(\frac{\langle u,v \rangle}{\lVert v \rVert}\Big)}\langle v , v \rangle = \Big\lvert \frac{\langle u,v \rangle}{\lVert v \rVert} \Big\rvert^2 \langle v, v \rangle$.
Note that: $\langle v , v \rangle = \lVert v \rVert^2$ thus:
$\Big\lvert \frac{\langle u,v \rangle}{\lVert v \rVert} \Big\rvert^2 \langle v, v \rangle = \Big\lvert \frac{\langle u,v \rangle}{\lVert v \rVert} \Big\rvert^2 \lVert v \rVert ^2$
I get the sense that I am close...but I really cannot see the misstep. Any help is greatly appreciated! Thank you.
Edit: Whoops. Typo on my part. The author actually wrote:
$\Bigl\lVert\frac{\langle u,v \rangle}{\lVert v \rVert^2} v \Bigr\rVert ^2$
Given everyone's comments...this makes perfect sense now.
