Added part of vector spaces in linear algebra.

config/en/mkdocs.yaml

@@ -86,6 +86,7 @@ nav:
         - 'Matrix algebra': mathematics/linear-algebra/matrices/matrix-algebra.md
         - 'Elementary matrices': mathematics/linear-algebra/matrices/elementary-matrices.md
       - 'Determinants': mathematics/linear-algebra/determinants.md
+      - 'Vector spaces': mathematics/linear-algebra/vector-spaces.md
     - 'Calculus': 
       - 'Limits': mathematics/calculus/limits.md
       - 'Continuity': mathematics/calculus/continuity.md

config/nl/mkdocs.yaml

@@ -71,26 +71,26 @@ nav:
     
   - 'Scheikunde':
     - scheikunde/index.md
-    - 'Basis':
+#    - 'Basis':
-      - 'Materialen en stoffen': scheikunde/basis/materialen-en-stoffen.md
+#      - 'Materialen en stoffen': scheikunde/basis/materialen-en-stoffen.md
-      - 'Scheidingsmethoden': scheikunde/basis/scheidingsmethoden.md
+#      - 'Scheidingsmethoden': scheikunde/basis/scheidingsmethoden.md
-      - 'Reacties': scheikunde/basis/reacties.md
+#      - 'Reacties': scheikunde/basis/reacties.md
-      - 'Chemisch rekenen': scheikunde/basis/chemisch-rekenen.md
+#      - 'Chemisch rekenen': scheikunde/basis/chemisch-rekenen.md
-      - 'Energie omzetting': scheikunde/basis/energie-omzetting.md
+#      - 'Energie omzetting': scheikunde/basis/energie-omzetting.md
-      - 'Bindingstypen': scheikunde/basis/bindingstypen.md
+#      - 'Bindingstypen': scheikunde/basis/bindingstypen.md
-      - 'Evenwichten': scheikunde/basis/evenwichten.md
+#      - 'Evenwichten': scheikunde/basis/evenwichten.md
-    - 'Organische chemie':
+#    - 'Organische chemie':
-      - 'Organische verbindingen': scheikunde/organische-chemie/organische-verbindingen.md
+#      - 'Organische verbindingen': scheikunde/organische-chemie/organische-verbindingen.md
-      - 'Lewis theorie': scheikunde/organische-chemie/lewis-theorie.md
+#      - 'Lewis theorie': scheikunde/organische-chemie/lewis-theorie.md
-      - 'VSEPR theorie': scheikunde/organische-chemie/vsepr-theorie.md
+#     - 'VSEPR theorie': scheikunde/organische-chemie/vsepr-theorie.md
-      - 'Isomerie': scheikunde/organische-chemie/isomerie.md
+#      - 'Isomerie': scheikunde/organische-chemie/isomerie.md
-      - 'Cis-trans isomerie': scheikunde/organische-chemie/cis-trans-isomerie.md
+#      - 'Cis-trans isomerie': scheikunde/organische-chemie/cis-trans-isomerie.md
-      - 'Spiegelbeeld isomerie': scheikunde/organische-chemie/spiegelbeeld-isomerie.md
+#      - 'Spiegelbeeld isomerie': scheikunde/organische-chemie/spiegelbeeld-isomerie.md
-      - 'Reacties': scheikunde/organische-chemie/reacties.md
+#      - 'Reacties': scheikunde/organische-chemie/reacties.md
-    - 'Reactie kinetiek': scheikunde/reactie-kinetiek.md
+#    - 'Reactie kinetiek': scheikunde/reactie-kinetiek.md
-    - 'Zuren en basen': scheikunde/zuren-basen.md
+#    - 'Zuren en basen': scheikunde/zuren-basen.md
-    - 'Redox chemie': scheikunde/redox-chemie.md
+#    - 'Redox chemie': scheikunde/redox-chemie.md
-    - 'Analyse methoden': scheikunde/analyse-methoden.md
+#    - 'Analyse methoden': scheikunde/analyse-methoden.md
     
   - 'Filosofie':
     - filosofie/index.md

docs/en/mathematics/linear-algebra/vector-spaces.md

@@ -0,0 +1,335 @@
+# Vector spaces
+
+## Definition
+
+> *Definition*: a **vector space** $V$ is a set on which the operations of addition and scalar multiplication are defined. Such that for all vectors $\mathbf{u}$ and $\mathbf{v}$ in $V$ the vectors $\mathbf{u} + \mathbf{v}$ are in $V$ and for each scalar $a$ the vector $a\mathbf{v}$ is in $V$. With the following axioms satisfied.
+> 
+> 1. Associativity of vector addition: $\mathbf{u} + (\mathbf{v} + \mathbf{w}) = (\mathbf{u} + \mathbf{v}) + \mathbf{w}$ for any $\mathbf{u},\mathbf{v}, \mathbf{w} \in V$.
+> 2. Commutativity of vector addition: $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$ for any $\mathbf{u},\mathbf{v} \in V$.
+> 3. Identity element of vector addition: $\exists \mathbf{0} \in V$ such that $\mathbf{v} + \mathbf{0}$ for all $\mathbf{v} \in V$. 
+> 4. Inverse element of vector addition: $\forall \mathbf{v} \in V \exists (-\mathbf{v}) \in V$ such that $\mathbf{v} + (-\mathbf{v}) = \mathbf{0}$. 
+> 5. Distributivity of scalar multiplication with respect to vector addition: $a(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v}$ for any scalar $a$ and any $\mathbf{u}, \mathbf{v} \in V$.
+> 6. Distributivity of scalar multiplication with respect to field addition: $(a + b) \mathbf{v} = a \mathbf{v} + b \mathbf{v}$ for any scalars $a$ and $b$ and any $\mathbf{v} \in V$.
+> 7. Compatibility of scalar multiplication with field multiplication: $a(b\mathbf{v}) = (ab) \mathbf{v}$ for any scalars $a$ and $b$ and any $\mathbf{v} \in V$. 
+> 8. Identity element of scalar multiplication: $1 \mathbf{v} = \mathbf{v}$ for all $\mathbf{v} \in V$. 
+
+Some important properties of a vector space can be derived from this definition in the following proposition a few have been listed. 
+
+> *Proposition*: if $V$ is a vector space and $\mathbf{u}$, $\mathbf{v}$ is in $V$, then
+>
+> 1. $0 \mathbf{v} = \mathbf{0}$.
+> 2. $\mathbf{u} + \mathbf{v} = \mathbf{0} \implies \mathbf{u} = - \mathbf{v}$.
+> 3. $(-1)\mathbf{v} = - \mathbf{v}$. 
+
+??? note "*Proof*:"
+
+    For 1, suppose $\mathbf{v} \in V$ then it follows from axioms 3, 6 and 8
+
+    $$
+        \mathbf{v} = 1 \mathbf{v} = (1 + 0)\mathbf{v} = 1 \mathbf{v} + 0 \mathbf{v} = \mathbf{v} + 0\mathbf{v},
+    $$
+
+    therefore
+
+    $$
+    \begin{align*}
+        -\mathbf{v} + \mathbf{v} &= - \mathbf{v} + (\mathbf{v} + 0\mathbf{v}) = (-\mathbf{v} + \mathbf{v}) + 0\mathbf{v}, \\
+        \mathbf{0} &= \mathbf{0} + 0\mathbf{v} = 0\mathbf{v}.
+    \end{align*}
+    $$
+
+    For 2, suppose for $\mathbf{u}, \mathbf{v} \in V$ that $\mathbf{u} + \mathbf{v} = \mathbf{0}$ then it follows from axioms 1, 3 and 4
+
+    $$
+        - \mathbf{v} = - \mathbf{v} + \mathbf{0} = - \mathbf{v} + (\mathbf{v} + \mathbf{u}),
+    $$
+
+    therefore
+
+    $$
+        -\mathbf{v} = (-\mathbf{v} + \mathbf{v}) + \mathbf{u} = \mathbf{0} + \mathbf{u} = \mathbf{u}.
+    $$
+
+    For 3, suppose $\mathbf{v} \in V$ then it follows from 1 and axioms 4 and 6
+
+    $$
+        \mathbf{0} = 0 \mathbf{v} = (1 + (-1))\mathbf{v} = 1\mathbf{v} + (-1)\mathbf{v},
+    $$
+
+    therefore
+
+    $$
+        \mathbf{v} + (-1)\mathbf{v} = \mathbf{0},
+    $$
+
+    from 2 it follows then that
+
+    $$
+        (-1)\mathbf{v} = -\mathbf{v}.
+    $$
+
+### Euclidean spaces
+
+Perhaps the most elementary vector spaces are the Euclidean vector spaces $V = \mathbb{R}^n$ with $n \in \mathbb{N}$. Given a nonzero vector $\mathbf{u}  \in \mathbb{R}^n$ defined by 
+
+$$
+    \mathbf{u} = \begin{pmatrix}u_1 \\ \vdots \\ u_n\end{pmatrix},
+$$ 
+
+it may be associated with the directed line segment from $(0, \dots, 0)$ to $(u_1, \dots, u_n)$. Or more generally line segments that have the same length and direction can be represented by any line segment from $(a_1, \dots, a_n)$ to $(a_1 + u_1, \dots, a_n + u_n)$. Vector addition and scalar multiplication in $\mathbb{R}^n$ are respectively defined by
+
+$$
+    \mathbf{u} + \mathbf{v} = \begin{pmatrix} u_1 + v_1 \\ \vdots \\ u_n + v_n \end{pmatrix} \quad \text{ and } \quad a \mathbf{u} = \begin{pmatrix} a u_1 \\ \vdots \\ a u_n \end{pmatrix},
+$$
+
+for any $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$ and any scalar $a$. 
+
+This can be extended to matrices with $V = \mathbb{R}^{m \times n}$ with $m,n \in \mathbb{N}$, the set of all matrices. Given a nonzero matrix $A \in \mathbb{R}^{m \times n}$ defined by $A = (a_{ij})$. Matrix addition and scalar multiplication in $\mathbb{R}^{m \times n}$ are respectively defined by
+
+$$
+    A + B = C \iff a_{ij} + b_{ij} = c_{ij} \quad \text{ and } \quad \alpha A = C \iff \alpha a_{ij} = c_{ij},
+$$
+
+for any $A, B, C \in \mathbb{R}^{m \times n}$ and any scalar $\alpha$. 
+
+### Function spaces
+
+Let $V$ be a vector space over a field $F$ and let $X$ be any set. The functions $X \to F$ can be given the structure of a vector space over $F$ where the operations are defined by
+
+$$
+\begin{align*}
+    (f + g)(x) = f(x) + g(x), \\
+    (af)(x) = af(x),
+\end{align*}
+$$
+
+for any $f,g: X \to F$, any $x \in X$ and any $a \in F$. 
+
+### Polynomial spaces
+
+Let $P_n$ denote the set of all polynomials of degree less than $n \in \mathbb{N}$ where the operations are defined by
+
+$$
+\begin{align*}
+    (p+q)(x) = p(x) + q(x), \\
+    (ap)(x) = ap(x),
+\end{align*}
+$$
+
+for any $p,q: X \to P_n$, any $x \in X$ and any $a \in P_n$. 
+
+## Vector subspaces
+
+> *Definition*: if $S$ is a nonempty subset of a vector space $V$ and $S$ satisfies the conditions
+>
+> 1. $a \mathbf{u} \in S$ whenever $\mathbf{u} \in S$ for any scalar $a$.
+> 2. $\mathbf{u} + \mathbf{v} \in S$ whenever $\mathbf{u}, \mathbf{v} \in S$. 
+>
+> then $S$ is said to be a **subspace** of $V$.
+
+In a vector space $V$ it can be readily verified that $\{\mathbf{0}\}$ and $V$ are subspaces of $V$. All other subspaces are referred to as *proper subspaces* and $\{\mathbf{0}\}$ is referred to as the *zero subspace*.
+
+> *Theorem*: Every subspace of a vector space is a vector space. 
+
+??? note "*Proof*:"
+
+    May be proved by testing if all axioms remain valid for the definition of a subspace. 
+
+
+### The null space of a matrix
+
+> *Definition*: let $A \in \mathbb{R}^{m \times n}$, $\mathbf{x} \in \mathbb{R}^n$ and let $N(A)$ denote the set of all solutions of the homogeneous system $A\mathbf{x} = \mathbf{0}$. Therefore
+>
+> $$
+>   N(A) = \{\mathbf{x} \in \mathbb{R}^n \;|\; A \mathbf{x} = \mathbf{0}\},
+> $$
+>
+> referred to as the null space of $A$. 
+
+Claiming that $N(A)$ is a subspace of $\mathbb{R}^n$. Clearly $\mathbf{0} \in N(A)$ so $N(A)$ is nonempty. If $\mathbf{x} \in N(A)$ and $\alpha$ is a scalar then
+
+$$
+    A(\alpha \mathbf{x}) = \alpha A\mathbf{x} = \alpha \mathbf{0} = \mathbf{0}
+$$
+
+and hence $\alpha \mathbf{x} \in N(A)$. If $\mathbf{x}, \mathbf{y} \in N(A)$ then 
+
+$$
+    A(\mathbf{x} + \mathbf{y}) = A\mathbf{x} + A\mathbf{y} = \mathbf{0} + \mathbf{0} = \mathbf{0}
+$$
+
+therefore $\mathbf{x} + \mathbf{y} \in N(A)$ and it follows that $N(A)$ is a subspace of $\mathbb{R}^n$. 
+
+
+### The span of a set of vectors
+
+> *Definition*: let $\mathbf{v}_1, \dots, \mathbf{v}_n$ be vectors in a vector space $V$ with $n \in \mathbb{N}$. A sum of the form 
+>
+> $$
+>   a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n,
+> $$
+>
+> with scalars $a_1, \dots, a_n$ is called a **linear combination** of $\mathbf{v}_1, \dots, \mathbf{v}_n$.
+>
+> The set of all linear combinations of $\mathbf{v}_1, \dots, \mathbf{v}_n$ is called the **span** of $\mathbf{v}_1, \dots, \mathbf{v}_n$ which is denoted by $\text{Span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$. 
+
+The nullspace can be for example defined by a span of vectors.
+
+> *Theorem*: if $\mathbf{v}_1, \dots, \mathbf{v}_n$ are vectors in a vector space $V$ with $n \in \mathbb{N}$ then $\text{Span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$ is a subspace of $V$. 
+
+??? note "*Proof*:"
+
+    Let $b$ be a scalar and $\mathbf{u} \in \text{Span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$ given by
+
+    $$
+        a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n,
+    $$
+
+    with scalars $a_1, \dots, a_n$. Since
+
+    $$
+        b \mathbf{u} = (b a_1)\mathbf{v}_1 + \dots + (b a_n)\mathbf{v}_n,
+    $$
+
+    it follows that $b \mathbf{u} \in \text{Span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$. 
+
+    If we also have $\mathbf{w} \in \text{Span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$ given by
+
+    $$
+        b_1 \mathbf{v}_1 + \dots + b_n \mathbf{v}_n,
+    $$
+
+    with scalars $b_1, \dots, b_n$. Then
+
+    $$
+        \mathbf{u} + \mathbf{w} = (a_1 + b_1) \mathbf{v}_1 + \dots + (a_n + b_n)\mathbf{v}_n, 
+    $$
+
+    it follows that $\mathbf{u} + \mathbf{w} \in \text{Span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$ is a subspace of $V$.
+
+For example, a vector $\mathbf{x} \in \mathbb{R}^3$ is in $\text{Span}(\mathbf{e}_1, \mathbf{e}_2)$ if and only if it lies in the $x_1 x_2$-plane in 3-space. Thus we can think of the $x_1 x_2$-plane as the geometrical representation of the subspace $\text{Span}(\mathbf{e}_1, \mathbf{e}_2)$. 
+
+> *Definition*: the set $\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$ with $n \in \mathbb{N}$ is a spanning set for $V$ if and only if every vector $V$ can be written as a linear combination of $\mathbf{v}_1, \dots, \mathbf{v}_n$. 
+
+## Linear independence
+
+We have the following observations.
+
+> *Proposition*: if $\mathbf{v}_1, \dots, \mathbf{v}_n$ with $n \in \mathbb{N}$ span a vector space $V$ and one of these vectors can be written as a linear combination of the other $n-1$ vectors then thoses $n-1$ vectors span $V$. 
+
+??? note "*Proof*:"
+
+    suppose $\mathbf{v}_n$ with $n \in \mathbb{N}$ can be written as a linear combination of the vectors $\mathbf{v}_1, \dots, \mathbf{v}_{n-1}$ given by
+
+    $$
+        \mathbf{v}_n = a_1 \mathbf{v}_1 + \dots + a_{n-1} \mathbf{v}_{n-1}.
+    $$
+
+    Let $\mathbf{v}$ be any element of $V$. Since we have
+
+    $$
+    \begin{align*}
+        \mathbf{v} &= b_1 \mathbf{v}_1 + \dots + b_{n-1} \mathbf{v}_{n-1} + b_n \mathbf{v}_n, \\
+                &= b_1 \mathbf{v}_1 + \dots + b_{n-1} \mathbf{v}_{n-1} + b_n (a_1 \mathbf{v}_1 + \dots + a_{n-1} \mathbf{v}_{n-1}), \\
+                &= (b_1 + b_n a_1)\mathbf{v}_1 + \dots + (b_{n-1} + b_n a_{n-1}) \mathbf{v}_{n-1},
+    \end{align*}
+    $$
+
+    we can write any vector $\mathbf{v} \in V$ as a linear combination of $\mathbf{v}_1, \dots, \mathbf{v}_{n-1}$ and hence these vectors span $V$. 
+
+> *Proposition*: given $n$ vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ with $n \in \mathbb{N}$, it is possible to write one of the vectors as a linear combination of the other $n-1$ vectors if and only if there exist scalars $a_1, \dots, a_n$ not all zero such that
+>
+> $$
+>   a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n = \mathbf{0}.
+> $$
+
+??? note "*Proof*:"
+
+    Suppose that one of the vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ with $n \in \mathbb{N}$ can be written as a linear combination of the others
+
+    $$
+        \mathbf{v}_n = a_1 \mathbf{v}_1 + \dots + a_{n-1} \mathbf{v}_{n-1}.
+    $$
+
+    Subtracting $\mathbf{v}_n$ from both sides obtains
+
+    $$
+        a_1 \mathbf{v}_1 + \dots + a_{n-1} \mathbf{v}_{n-1} - \mathbf{v}_n = \mathbf{0},
+    $$
+
+    we have $a_n = -1$ and 
+
+    $$
+        a_1 \mathbf{v}_1 + \dots + a_n\mathbf{v}_n = \mathbf{0}.
+    $$
+
+We may use these oberservations to state the following definitions.
+
+> *Definition*: the vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ in a vector space $V$ with $n \in \mathbb{N}$ are said to be **linearly independent** if 
+>
+> $$
+>   a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n = \mathbf{0} \implies \forall i \in \{1, \dots, n\} [c_i = 0].
+> $$
+
+It follows from the above propositions that if $\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$ is a minimal spanning set of a vector space $V$ then $\mathbf{v}_1, \dots, \mathbf{v}_n$ are linearly independent. A minimal spanning set is called a basis of the vector space. 
+
+> *Definition*: the vectors $\mathbf{v}_1, \dots, \mathbf{v}_n$ in a vector space $V$ with $n \in \mathbb{N}$ are said to be **linearly dependent** if there exists scalars $a_1, \dots, a_n$ not all zero such that 
+>
+> $$
+>   a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n = \mathbf{0}.
+> $$
+
+It follows from the above propositions that if a set of vectors is linearly dependent then at least one vector is a linear combination of the other vectors. 
+
+> *Theorem*: let $\mathbf{x}_1, \dots, \mathbf{x}_n$ be vectors in $\mathbb{R}^n$ with $n \in \mathbb{N}$ and let $X = (\mathbf{x}_1, \dots, \mathbf{x}_n)$. The vectors $\mathbf{x}_1, \dots, \mathbf{x}_n$ will be linearly dependent if and only if $X$ is singular.
+
+??? note "*Proof*:"
+
+    Let $\mathbf{x}_1, \dots, \mathbf{x}_n$ be vectors in $\mathbb{R}^n$ with $n \in \mathbb{N}$ and let $X = (\mathbf{x}_1, \dots, \mathbf{x}_n)$. Suppose we have the linear combination given by
+
+    $$
+        a_1 \mathbf{x}_1 + \dots + a_n \mathbf{x}_n = \mathbf{0},
+    $$
+
+    can be rewritten as a matrix equation by
+
+    $$
+        X\mathbf{a} = \mathbf{0},
+    $$
+
+    with $\mathbf{a} = (a_1, \dots, a_n)^T$. This equation will have a nontrivial solution if and only if $X$ is singular. Therefore $\mathbf{x}_1, \dots, \mathbf{x}_n$ will be linearly dependent if and only if $X$ is singular.
+
+This result can be used to test whether $n$ vectors are linearly independent in $\mathbb{R}^n$ for $n \in \mathbb{N}$.
+
+> *Theorem*: let $\mathbf{v}_1, \dots, \mathbf{v}_n$ be vectors in a vector space $V$ with $n \in \mathbb{N}$. A vector $\mathbf{v} \in \text{Span}(\mathbf{v}_1, \dots, \mathbf{v}_n)$ can be written uniquely as a linear combination of $\mathbf{v}_1, \dots, \mathbf{v}_n$ if and only if $\mathbf{v}_1, \dots, \mathbf{v}_n$ are linearly independent.
+
+??? note "*Proof*:"
+
+    If $\mathbf{v} \in \text{Span}(\mathbf{v}_1, \dots \mathbf{v}_n)$ with $n \in \mathbb{N}$ then $\mathbf{v}$ can be written as a linear combination 
+
+    $$
+        \mathbf{v} = a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n.
+    $$
+
+    Suppose that $\mathbf{v}$ can also be expressed as a linear combination 
+
+    $$
+        \mathbf{v} = b_1 \mathbf{v}_1 + \dots + b_n \mathbf{v}_n.
+    $$
+
+    If $\mathbf{v}_1, \dots \mathbf{v}_n$ are linearly independent then subtracting both expressions yields
+
+    $$
+        (a_1 - b_1)\mathbf{v}_1 + \dots + (a_n - b_n)\mathbf{v}_n = \mathbf{0}.
+    $$
+
+    By the linear independence of $\mathbf{v}_1, \dots \mathbf{v}_n$, the coefficients must all be 0, hence
+
+    $$
+        a_1 = b_1,\; \dots \;, a_n = b_n 
+    $$
+
+    therefore the representation of $\mathbf{v}$ is unique when $\mathbf{v}_1, \dots \mathbf{v}_n$ are linearly independent. 
+
+    On the other hand if $\mathbf{v}_1, \dots \mathbf{v}_n$ are linearly dependent then the coefficients must not all be 0 and $a_i \neq b_i$ for some $i \in \{1, \dots, n\}$. Therefore the representation of $\mathbf{v}$ is not unique when $\mathbf{v}_1, \dots \mathbf{v}_n$ are linearly dependent.
+

docs/nl/scheikunde/index.md

@@ -1,3 +1,3 @@
 # Scheikunde
 
-Welkom op de scheikunde pagina, hier bewaar ik mijn aantekeningen van scheikunde. Dit zijn voornamelijk aantekeningen van vwo scheikunde en aantekeningen van een paar boeken die ik na die tijd gelezen heb. Tijdens mijn verdere studies ben ik niet heel veel meer scheikunde tegengekomen helaas.
+Welkom op de scheikunde pagina.