Dr. Matthieu R Bloch

Wednesday September 01, 2021

**Self assessment**- Due today (firm deadline)
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**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/

**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

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.

A basis of vector subspace \(\calW\) of a vector space \(\calV\) is a

*countable*set of vectors \(\calB\) such that:- \(\text{span}(\calB)=\calW\)
- \(\calB\) 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

- \(\calV=\bbR^n\), \(\calV=\set{f:f\text{ is periodic with period }2\pi}\)

The properties of vector space seen thus far provide an

*algebraic*structureWe 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

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

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.*orthogonal*if \(\dotp{x}{y}=0\). We write \(x\perp y\) for simplicity.- If \(x\perp y\) then \(\norm{x+y}^2=\norm{x}^2+\norm{y}^2\)