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Added vector spaces to functional analysis.

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Luc Bijl 2024-08-05 18:39:01 +02:00
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- 'Tensor transformations': mathematics/linear-algebra/tensors/tensor-transformations.md
- 'Functional analysis':
- 'Metric spaces':
- 'Definition': mathematics/functional-analysis/metric-spaces/definition.md
- 'Metric spaces': mathematics/functional-analysis/metric-spaces/metric-spaces.md
- 'Topological notions': mathematics/functional-analysis/metric-spaces/topological-notions.md
- 'Convergence': mathematics/functional-analysis/metric-spaces/convergence.md
- 'Completeness': mathematics/functional-analysis/metric-spaces/completeness.md
- 'Completion': mathematics/functional-analysis/metric-spaces/completion.md
- 'Normed spaces':
- 'Vector spaces': mathematics/functional-analysis/normed-spaces/vector-spaces.md
- 'Definition': mathematics/functional-analysis/normed-spaces/definition.md
- 'Normed spaces': mathematics/functional-analysis/normed-spaces/normed-spaces.md
- 'Compactness': mathematics/functional-analysis/normed-spaces/compactness.md
- 'Linear operators': mathematics/functional-analysis/normed-spaces/linear-operators.md
- 'Linear functionals': mathematics/functional-analysis/normed-spaces/linear-functionals.md

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# Definition of a metric space
# Metric spaces
> *Definition 1*: a **metric space** is a pair $(X,d)$, where $X$ is a set and $d$ is a metric on $X$, which is a function on $X \times X$ such that
>

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# Vector spaces
> *Definition 1*: a **vector space** $X$ over a **scalar field** $F$ is a non-empty set, on which two algebraic operations are defined; vector addition and scalar multiplication. Such that
>
> 1. $(X, +)$ is a commutative group with neutral element 0.
> 2. the scalar multiplication satisfies $\forall x, y \in X$ and $\lambda, \mu \in F$
> * $\lambda (x + y) = \lambda x + \lambda y$,
> * $(\lambda + \mu) x = \lambda x + \mu x$,
> * $\lambda (\mu x) = (\lambda \mu) x$,
> * $1 x = x$.
When $F = \mathbb{R}$ we have a real vector space while when $F = \mathbb{C}$ we have a complex vector space.
We have that the metric spaces $\mathbb{R}^n$, $C$, $l^p$ and $l^\infty$ are also vector spaces.
??? note "*Proof*:"
I am too lazy to add this trivial proof. Maybe some time in the future, if I do not forget.
> *Definition 2*: a **subspace** of a vector space $X$ is a non-empty subset $M$ of $X$, such that $\forall x, y \in M$ and $\lambda, \mu \in F$:
>
> $$
> \lambda x + \mu y \in M,
> $$
>
> with $M$ itself a vector space.
A special subspace $M$ of a vector space $X$ is the *improper subspace* $M = X$. Every other subspace of $X$ is a *proper subspace*.
## Linear combinations
> *Definition 3*: a **linear combination** of the vectors $\{x_i\}_{i=1}^n$ with $n \in \mathbb{N}$ is vector of the form
>
> $$
> \alpha_1 x_1 + \dots + \alpha_n x_n = \sum_{i=1}^n \alpha_i x_i,
> $$
>
> with $\{\alpha_i\}_{i=1}^n \in F$.
The set of all linear combinations of a set of vectors is defined as follows.
> *Definition 4*: the **span** of a subset $M \subset X$ of a vector space $X$, denoted by $\mathrm{span}(M)$, is the set of all linear combinations of vectors from $M$.
It follows that $\mathrm{span}(M)$ is a subspace of $X$.
## Linear independence
> *Definition 5*: a finite subset of vectors $M = \{x_i\}_{i=1}^n$ is **linearly independent** if
>
> $$
> \sum_{i=1}^n \alpha_i x_i = 0 \implies \forall i \in \{1, \dots, n\}: \alpha_i = 0.
> $$
The converse may also be defined.
> *Definition 6*: a finite subset of vectors $M = \{x_i\}_{i=1}^n$ is **linearly dependent** if $\exists \{\alpha_i\}_{i=1}^n \in F$ not all zero such that
>
> $$
> \sum_{i=1}^n \alpha_i x_i = 0.
> $$
The notions of linear dependence and independence may also be extended to infinite subsets.
> *Definition 7*: a subset $M$ of a vector space $X$ is **linearly independent** if every non-empty finite subset of $M$ is linearly independent.
While the converse in this case is defined by the contradiction.
> *Definition 8*: a subset $M$ of a vector space $X$ is **linearly dependent** if $M$ is not linearly independent.
## Dimension and basis
> *Definition 9*: a vector space $X$ is **finite dimensional** if there exists a $n \in \mathbb{N}$, such that $X$ contains a set of $n$ linearly independent vectors, while every set of $n+1$ vectors in $X$ is linearly dependent. In this case $n$ is the dimension of $X$, denoted by $\dim X = n$.
By definition $X = \{0\}$ is finite dimensional and $\dim X = 0$.
> *Definition 10*: if a vector space $X$ is not finite dimensional then $X$ is **infinite dimensional**.
The following definition of a basis is both relevant to finite and infinite dimensional vector spaces.
> *Definition 11*: a **basis** $B$ of a vector space $X$ is a linearly independent subset of $X$, that spans $X$.
Such a set $B$ is also called a *Hamel basis* of $X$.
> *Theorem 1*: every vector space $X$ has a Hamel basis.
??? note "*Proof*:"
Read it again, a proof is not necessary.
> *Theorem 2*: let $X$ be a vector space with $\dim X = n \in \mathbb{N}$. Then any proper subspace $M \subset X$ has dimension less than $n$.
??? note "*Proof*:"
If $n = 0$, then $X = \{0\}$ and $X$ has no proper subspace.
If $\dim M = 0$, then $M = \{0\}$ and $X \neq M \implies \dim X \geq 1$.
If $\dim M = n$ then $M$ would have a basis of $n$ elements, which would also be a basis for $X$ since $\dim X = n$, so that $X = M$.
This shows that any linearly independent set of vectors in $M$ must have fewer than $n$ elements and $\dim M < n$.