# september 2

$$\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}} \def\Field{\mathbb{F}}$$

#### Fields, vector spaces, subspaces, linear operators, range space, null space

A field $$(F,+,·)$$ is a set with two commutative associative operations with $$0$$ and$$1$$, additive and multiplicative inverses and subject to the distributive law.

##### Examples (Common fields)

$$(\Real,+,·)$$ is a field. Rational, Complex numbers. Matrices (when they are fields?)

Vector spaces – modules over a field. (Real definition: abelian group with an action of $$\Field$$ and several natural axioms.)

##### Examples

Fun($$A,\Field$$), for any $$A$$ – vector space over $$\Field$$. (For example, sequences…)

The standard $$n$$-dimensional vector space over $$\Field$$.

Kronecker delta’s. Finite sums of delta’s.

What if the set $$A$$ is infinite? Finite sums are not enough: many more functions.

Subclasses of functions: continuous, integrable,…

Linear subspace (of a linear space)- subset that is a linear space (over the same field, of course). Easiest way to fix a linear subspace: by a linear equation.

##### Examples:

Set of functions vanishing on $$B\subset A$$ (fixing other values, non-zeros, would it work?).
All but finite values vanish. (What about vanishing ni at least one point?)
Functions with finite number of discontinuities?

#### Linear functions

Also form a linear space! Examples.

#### Linear independence.\

Nice criterion of linear independence: if $$e_1,\ldots,e_n\in V$$ is a set of vectors, and $$f_1,\ldots,f_n\in V^*$$ a set of linear functions, and
$\left| \begin{array}{ccc} f_1(e_1)&\ldots &f_1(e_n)\\ \vdots &\ddots &\vdots \\ f_n(e_1) & \ldots&f_n(e_n)\\ \end{array} \right|\neq 0,$
then both sets, $$e$$’s and $$f$$’s are linear independent.

Bases (a set of vectors is a basis if it is linearly independent and spans the space).

In particular, there is a unique representation of any vector in a basis.
If there exists a finite basis, the space is finite-dimensional.

Representing a vector in a basis involves a duel basis!

Dimension (the size of a basis) is well-defined.

##### Theorem. Any linearly independent set can be completed to a basis.

Change of basis:
$x=\sum_k a(k) e^k\times \sum_l f_l f^l=\sum_l (\sum_k P(k,l)a(k)) f^l.$
Here $$P(k,l)=f_l(e^k)$$.