What is the motivation for setting $v \cdot v = \|v\|^2$ for $v \in \mathbb{R}^n$?

Question

Recently, we discussed inner products in my Linear Algebra course. We state:

$v \cdot v = ||v||^2$ for $v \in \mathbb{R}^n$ Why can we say this?

We later define the norm as $||v|| = \sqrt{\langle v, v}$ for $v \in V$. How can we say that (1) $||v|| \ge 0$ when we only know that $||v||^2 \ge 0$ from the axioms, and how do we known $\langle v, v \rangle = ||v||^2$ in the first place?

We also state that $\langle v, w \rangle = 0$ iff $v, w$ are "orthogonal". What enables us to say this statement?

Answer 1

Given a vector space $V$ equipped with an inner product $\langle\cdot,\cdot\rangle:V\times V\to\textbf{F}$ we can define a norm as you have noticed. More precisely, one has that $\|v\|^{2} = \langle v,v\rangle$ for every $v\in V$. But how do we prove it behaves as a norm indeed? In order to do so, let us remember the properties satisfied by an inner product:

• $\langle u,u\rangle \geq 0$ for every $u\in V$ and $\langle u,u\rangle = 0$ iff $u = 0\in V$,
• $\langle \alpha u,v\rangle = \alpha\langle u,v\rangle$
• $\overline{\langle u,v\rangle} = \langle v,u\rangle$,
• $\langle u + v,w\rangle = \langle u,w\rangle + \langle v,w\rangle$.

for every $u,v,w\in V$ and $\alpha\in\textbf{F}$. Therefore we must prove that:

• $\|u\|\geq 0$ and $\|u\| = 0$ iff $u = 0$,
• $\|\alpha u\| = |\alpha|\|u\|$,
• $\|u + v\| \leq \|u\| + \|v\|$.

for every $u,v\in V$ and $\alpha\in\textbf{F}$.

Indeed, the first properties holds because $\langle u,u\rangle \geq 0$ and $\langle u,u\rangle = 0$ iff $u = 0$.

The second property does also holds because $\langle \alpha u,\alpha u \rangle = \alpha\overline{\alpha}\langle u,u\rangle = |\alpha|^{2}\langle u,u\rangle$.

And finally the last property is true because of the Cauchy-Schwarz inequality: \begin{align*} \|u + v\|^{2} & = \langle u + v, u + v\rangle\\\\ & = \langle u,u\rangle + \langle u,v\rangle + \langle v,u\rangle + \langle v,v\rangle\\\\ & = \|u\|^{2} + 2\text{Re}\langle u,v\rangle + \|v\|^{2}\\\\ & \leq \|u\|^{2} + 2|\langle u,v\rangle| + \|v\|^{2}\\\\ & \leq \|u\|^{2} + 2\|u\|\|v\| + \|v\|^{2}\\\\ & = (\|u\| + \|v\|)^{2} \end{align*}

and we are done.

Given that you have a pre-Hilbert space, you can define orthogonality as you did: \begin{align*} u\perp v & \Longleftrightarrow \langle u,v\rangle = 0. \end{align*}

EDIT

Consider that $x = (x_{1},x_{2},x_{3})$ and $y = (y_{1}, y_{2}, y_{3})$ are to vectors in $\mathbb{R}^{3}$. According to the cosine's law, one has that the following relations holds: \begin{align*} \|x - y\|^{2} = \|x\|^{2} + \|y\|^{2} - 2\|x\|\|y\|\cos(\theta) & \end{align*} where $\theta$ is the angle between $x$ and $y$.

Simplifying the calculations, it results the following expression for the inner product in $\mathbb{R}^{3}$: \begin{align*} \langle x,y\rangle := x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} = \|x\|\|y\|\cos(\theta) \end{align*} Consequently, if we take $x = y$, we conclude that $\theta = 0$ and \begin{align*} \langle x,x\rangle = x^{2}_{1} + x^{2}_{2} + x^{2}_{3} = \|x\|^{2} \end{align*}

Hopefully this helps!