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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.

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    $\begingroup$ When I simplify that, I get $|\langle{u,v}\rangle|^2$. $\endgroup$ – Angina Seng Sep 21 '20 at 2:31
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As you said, we can pull out the inner product as a scalar: $$\left\| \frac{\langle u,v \rangle}{\|v\|}v\right\|^2 = |\langle u,v \rangle|^2 \left\| \frac{v}{\|v\|}\right\|^2$$ But notice that $ \frac{v}{\|v\|}$ is a unit vector. Therefore, $ \|\frac{v}{\|v\|}\| =1$. So we have: $$|\langle u,v \rangle|^2 \left\| \frac{v}{\|v\|}\right\|^2=|\langle u,v \rangle|^2 .$$

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  • $\begingroup$ I'm not quite sure I understand how you arrived at your very first statement. Could you explain, please? $\endgroup$ – S.Cramer Sep 21 '20 at 4:45
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    $\begingroup$ The first line is due to the property that for a normed space and any $\alpha$ scalar, we have $\|\alpha x\| = |\alpha| \| x \|$. In particular, the inner product of two vectors is a scalar. $\endgroup$ – travvytree Sep 21 '20 at 13:21

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