# Trouble simplifying the follow expression of a squared norm: $\Bigl\lVert\frac{\langle u,v \rangle}{\lVert v \rVert} v \Bigr\rVert ^2$

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.

• When I simplify that, I get $|\langle{u,v}\rangle|^2$. – 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: $$|\langle u,v \rangle|^2 \left\| \frac{v}{\|v\|}\right\|^2=|\langle u,v \rangle|^2 .$$
• 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. – travvytree Sep 21 '20 at 13:21