# pre-Hilbert spaces

Wednesday September 01, 2021

## Logistics

• Self assessment

• Assignment 2 coming

• Graduate teaching assistants + office hours

• Monday 1:00pm - 2:00pm: TSRB 423 (Shi-Yuan)
• Tuesdays 12:00pm - 1:00pm: Tutorial Lab in Van Leer room C449 Cubicle D (Kayla)
• Wednedays 5:30 pm - 6:30pm: virtual (Kayla)
• Thursdays 10am - 11 am: Tutorial Lab in Van Leer room C449 Cubicle B (Kayla)
• Friday 1:00pm - 2:00pm: TSRB 423 (Meng-Che)
• Course slides: https://bloch.ece.gatech.edu/teaching/ece7750fa21/

## What’s on the agenda for today?

• Last time:

• Formal introduction to vector spaces
• Key take-way: understand that vectors are not just arrows in $\bbR^3$ (functions)
• Key concepts: subspaces, spans, linear combinations
• Questions?
• Today: pre-Hilbert spaces (inner product spaces)

• More on vector spaces: linear independence, bases, dimension
• Norm, inner product
• Monday September 06, 2021: Hilbert spaces

## Linear independence

• Let $\set{v_i}_{i=1}^n$ be a set of vectors in a vector space $\calV$

• $\set{v_i}_{i=1}^n$ is linearly independent (or the vectors $\set{v_i}_{i=1}^n$ are linearly independent ) if (and only if) $\sum_{i=1}^na_iv_i = 0\Rightarrow \forall i\in\intseq{1}{n}\,a_i=0$ Otherwise the set is (or the vectors are) linearly dependent.

• $\calV\eqdef\bbR^3\quad \bfv_1=\mat{ccc}{2&1^0}^\intercal, \bfv_2=\mat{ccc}{1&1&0}^\intercal, \bfv_3=\mat{ccc}{1&2&0}^\intercal$
• $\calV\eqdef\calC([0;1])\quad v_1=\cos(2\pi t), v_2=\sin(2\pi t), v_3=2\cos(2\pi t +\frac{\pi}{3})$
• Any set of linearly dependent vectors contains a subset of linearly independent vectors with the same span.

## Bases

• A basis of vector subspace $\calW$ of a vector space $\calV$ is a countable set of vectors $\calB$ such that:

1. $\text{span}(\calB)=\calW$
2. $\calB$ is linearly independent
• If a non vector space $\calV\neq \set{0}$ as a finite basis with $n\in\bbN^*$ elements, $n$ is called the dimension of $\calV$, denoted $\dim{\calV}$. If the basis has an infinite number of elements, the dimension is infinite
• Any two bases for the same finite dimensional vector space contain the same number of elements.
• You should be somewhat familiar with bases (at least in $\bbR^n$):

• the representation of a vector on a basis is unique
• every subspace has a basis
• having a basis reduces the operations on vectors to operations on their components
• Things sort of work in infinite dimensions, but we have to be bit more careful

• $\calV=\bbR^n$, $\calV=\set{f:f\text{ is periodic with period }2\pi}$

## Norm

• The properties of vector space seen thus far provide an algebraic structure

• We are missing a topological structure to measure length and distance

• A norm on a vector space $\calV$ over $\bbR$ is a function $\norm{\cdot}:\calV\to\bbR$ that satisfies:
• Positive definiteness: $\forall x\in\calV$ $\norm{x}\geq 0$ with equality iff $x=0$
• Homogeneity: $\forall x\in\calV$ $\forall\alpha\in\bbR$ $\norm{\alpha x}=\abs{\alpha}\norm{x}$
• Subadditivity: $\forall x,y\in\calV$ $\norm{x+y}\leq \norm{x}+\norm{y}$
• $\norm{x}$ measures a length, $\norm{x-y}$ measures a distance

• ? $\bfx\in\bbR^d\qquad\norm[0]{\bfx}\eqdef\card{\set{i:x_i\neq 0}}\quad\norm[1]{\bfx}\eqdef\sum_{i=1}^d\abs{x_i}\quad \norm[2]{\bfx}\eqdef\sqrt{\sum_{i=1}^d x_i^2}$

• See board

## Inner product

• In addition to a topological and algebraic strucure, what if we want to do geometry?

• An inner product space over $\bbR$ is a vector space $\calV$ equipped with a positive definite symmetric bilinear form $\dotp{\cdot}{\cdot}:\calV\times\calV\to\bbR$ called an inner product

• An inner product space is also called a pre-Hilbert space

• See board

• Unless stated otherwise, we will only deal with vector spaces on $\bbR$ (things don’t change too much on $\bbC$ but one should be a bit more careful) ## Induced norm and orthogonality

• In an inner product space, an inner product induces a norm $\norm{x} \eqdef \sqrt{\dotp{x}{x}}$

• A norm $\norm{\cdot}$ is induced by an inner product on $\calV$ iff $\forall x,y\in\calV$ $\norm{x}^2+\norm{y}^2 = \frac{1}{2}\left(\norm{x+y}^2+\norm{x-y}^2\right)$

If this is the case, the inner product is given by the polarization identity $\dotp{x}{y}=\frac{1}{2}\left(\norm{x}^2+\norm{y}^2-\norm{x-y}^2\right)$

• An inner product satisfies $\forall x,y\in\calV$ $\dotp{x}{y}^2\leq\dotp{x}{x}\dotp{y}{y}$

• Two vectors $x,y\in\calV$ are orthogonal if $\dotp{x}{y}=0$. We write $x\perp y$ for simplicity.

A vector $x\in\calV$ is orthogonal to a set $\calS\subset\calV$ if $\forall s\in\calS$ $\dotp{x}{s}=0$. We write $x\perp \calS$ for simplicity.
• If $x\perp y$ then $\norm{x+y}^2=\norm{x}^2+\norm{y}^2$