Added vector spaces to functional analysis.

config/en/mkdocs.yaml

@@ -99,14 +99,14 @@ nav:
         - '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

docs/en/mathematics/functional-analysis/metric-spaces/metric-spaces.md

@@ -1,4 +1,4 @@
-# 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 
 >

docs/en/mathematics/functional-analysis/normed-spaces/vector-spaces.md

@@ -0,0 +1,100 @@
+# 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 $\dim M < n$.