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, $r^2 = \lvert z \rvert^2 = \Big\lvert \frac{\langle u,v \rangle}{\lVert v \rVert} \Big\rvert^2$.


$\frac{\langle u,v \rangle}{\lVert v \rVert} \overline{\Big(\frac{\langle u,v \rangle}{\lVert v \rVert}\Big)}\langle v , v \rangle = \Big\lvert \frac{\langle u,v \rangle}{\lVert v \rVert} \Big\rvert^2 \langle v, v \rangle$.

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.

  • 3
    $\begingroup$ When I simplify that, I get $|\langle{u,v}\rangle|^2$. $\endgroup$ – Angina Seng Sep 21 '20 at 2:31

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: $$|\langle u,v \rangle|^2 \left\| \frac{v}{\|v\|}\right\|^2=|\langle u,v \rangle|^2 .$$

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.