linear-algebra: update

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

@@ -1,8 +1,8 @@
 # Dual vector spaces
 
-We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n.$ In the following sections we make use of the Einstein summation convention introduced and $\mathbb{K} = \mathbb{R} \lor\mathbb{K} = \mathbb{C}$. 
+We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n$ and a field $F$. In the following sections we make use of the Einstein summation convention introduced. 
 
-> *Definition 1*: let $\mathbf{\hat f}: V \to \mathbb{K}$ be a **covector** or **linear functional** on $V$ if for all $\mathbf{v}_{1,2} \in V$ and $\lambda, \mu \in \mathbb{K}$ we have
+> *Definition 1*: let $\mathbf{\hat f}: V \to F$ be a **covector** or **linear functional** on $V$ if for all $\mathbf{v}_{1,2} \in V$ and $\lambda, \mu \in F$ we have
 >
 > $$
 >   \mathbf{\hat f}(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2) = \lambda \mathbf{\hat f}(\mathbf{v}_1) + \mu \mathbf{\hat f}(\mathbf{v}_2).
@@ -10,7 +10,7 @@ We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim
 
 Throughout this section covectors will be denoted by hats to increase clarity. 
 
-> *Definition 2*: let the the dual space $V^* \overset{\text{def}} = \mathscr{L}(V, \mathbb{K})$ denote the vector space of covectors on the vector space $V$.
+> *Definition 2*: let the the dual space $V^* \overset{\text{def}} = \mathscr{L}(V, F)$ denote the vector space of covectors on the vector space $V$.
 
 Each basis $\{\mathbf{e}_i\}$ of $V$ therefore induces a basis $\{\mathbf{\hat e}^i\}$ of $V^*$ by 
 

docs/mathematics/linear-algebra/tensors/tensor-formalism.md

@@ -1,13 +1,13 @@
 # Tensor formalism
 
-We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n$ and a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}.$ In the following sections we make use of the Einstein summation convention and $\mathbb{K} = \mathbb{R} \lor\mathbb{K} = \mathbb{C}$.
+We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n$, a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}$ and a field $F$. In the following sections we make use of the Einstein summation convention.
 
 ## Definition
 
 > *Definition 1*: a **tensor** is a multilinear mapping of the type 
 >
 > $$
->   \mathbf{T}: \underbrace{V^* \times \dots \times V^*}_p \times \underbrace{V \times \dots \times V}_q \to \mathbb{K},
+>   \mathbf{T}: \underbrace{V^* \times \dots \times V^*}_p \times \underbrace{V \times \dots \times V}_q \to F,
 > $$
 >
 > with $p, q \in \mathbb{N}$. Tensors are collectively denoted as 
@@ -16,7 +16,7 @@ We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim
 >   \mathbf{T} \in \underbrace{V \otimes \dots \otimes V}_p \otimes \underbrace{V^* \otimes \dots \otimes V^*}_q = \mathscr{T}_q^p(V),
 > $$
 >
-> with $\mathscr{T}_0^0(V) = \mathbb{K}$. 
+> with $\mathscr{T}_0^0(V) = F$. 
 
 We refer to $\mathbf{T} \in \mathscr{T}_q^p(V)$ as a $(p, q)$-tensor; a mixed tensor of **contravariant rank** $p$ and **covariant rank** $q.$ It may be observed that we have $\dim \mathscr{T}_q^p (V) = n^{p+q}$ with $\dim V = n \in \mathbb{N}$.  
 
@@ -45,7 +45,7 @@ $$
 
 ## Outer product
 
-> *Definition 3*: the outer product $f \otimes g: X \times Y \to \mathbb{K}$ of two scalar functions $f: X \to \mathbb{K}$ and $g: Y \to \mathbb{K}$ is defined as
+> *Definition 3*: the outer product $f \otimes g: X \times Y \to F$ of two scalar functions $f: X \to F$ and $g: Y \to F$ is defined as
 >
 > $$
 >   (f \otimes g)(x,y) = f(x) g(y),
@@ -69,7 +69,7 @@ The following statements are given with $p=q=r=s=1$ without loss of generality.
 
 From this definition the subsequent theorem follows naturally. 
 
-> *Theorem 1*: let $\mathbf{T} \in \mathscr{T}_q^p(V)$ be a tensor, then there exists **holors** $T_j^i \in \mathbb{K}$ such that
+> *Theorem 1*: let $\mathbf{T} \in \mathscr{T}_q^p(V)$ be a tensor, then there exists **holors** $T_j^i \in F$ such that
 >
 > $$
 >   \mathbf{T} = T^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j,
@@ -90,11 +90,11 @@ From this definition the subsequent theorem follows naturally.
     \end{align*}
     $$
 
-For $\mathbf{T} \in \mathscr{T}^0_q(V)$ it follows that there exists holors $T_i \in \mathbb{K}$ such that $\mathbf{T} = T_i \mathbf{\hat e}^i$ with $T_i = \mathbf{T}(\mathbf{e}_i)$, are referred to as the **covariant components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$. 
+For $\mathbf{T} \in \mathscr{T}^0_q(V)$ it follows that there exists holors $T_i \in F$ such that $\mathbf{T} = T_i \mathbf{\hat e}^i$ with $T_i = \mathbf{T}(\mathbf{e}_i)$, are referred to as the **covariant components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$. 
 
-For $\mathbf{T} \in \mathscr{T}^p_0(V)$ it follows that there exists holors $T^i \in \mathbb{K}$ such that $\mathbf{T} = T^i \mathbf{e}_i$ with $T^i = \mathbf{T}(\mathbf{\hat e}^i)$, are referred to as the **contravariant components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$. 
+For $\mathbf{T} \in \mathscr{T}^p_0(V)$ it follows that there exists holors $T^i \in F$ such that $\mathbf{T} = T^i \mathbf{e}_i$ with $T^i = \mathbf{T}(\mathbf{\hat e}^i)$, are referred to as the **contravariant components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$. 
 
-If $\mathbf{T} \in \mathscr{T}^p_q(V)$, it follows that there exists holors $T^i_j \in \mathbb{K}$ are coined the **mixed components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$. 
+If $\mathbf{T} \in \mathscr{T}^p_q(V)$, it follows that there exists holors $T^i_j \in F$ are coined the **mixed components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$. 
 
 By definition tensors are basis independent. Holors are basis dependent.
 
@@ -135,10 +135,10 @@ We have from theorem 2 that the outer product of two tensors yields another tens
 
 ## Inner product
 
-> *Definition 5*: an **inner product** on $V$ is a bilinear mapping $\bm{g}: V \times V \to \mathbb{K}$ which satisfies
+> *Definition 5*: an **inner product** on $V$ is a bilinear mapping $\bm{g}: V \times V \to F$ which satisfies
 >
 > 1. for all $\mathbf{u}, \mathbf{v} \in V: \; \bm{g}(\mathbf{u}, \mathbf{v}) = \overline{\bm{g}}(\mathbf{v}, \mathbf{u}),$ 
-> 2. for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and $\lambda, \mu \in \mathbb{K}: \;\bm{g}(\mathbf{u}, \lambda \mathbf{v} + \mu \mathbf{w}) = \lambda \bm{g}(\mathbf{u}, \mathbf{v}) + \mu \bm{g}(\mathbf{u}, \mathbf{w}),$
+> 2. for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and $\lambda, \mu \in F: \;\bm{g}(\mathbf{u}, \lambda \mathbf{v} + \mu \mathbf{w}) = \lambda \bm{g}(\mathbf{u}, \mathbf{v}) + \mu \bm{g}(\mathbf{u}, \mathbf{w}),$
 > 3. for all $\mathbf{u} \in V\backslash \{\mathbf{0}\}: \bm{g}(\mathbf{u},\mathbf{u}) > 0,$
 > 4. for $\mathbf{u} = \mathbf{0} \iff \bm{g}(\mathbf{u},\mathbf{u}) = 0.$
 
@@ -209,7 +209,7 @@ $$
         \mathbf{k}(\mathbf{g}(\mathbf{u}), \mathbf{v}) = \text{g}_{ij} u^i v^k\mathbf{k}(\mathbf{\hat e}^j, \mathbf{e}_k) = \text{g}_{ij} u^i v^k \delta^j_k = \text{g}_{ij} u^i v^j. 
     $$
 
-    Since $u^i, v^j \in \mathbb{K}$ are arbitrary it follows that $\text{g}_{ij} = g_{ij}$. 
+    Since $u^i, v^j \in F$ are arbitrary it follows that $\text{g}_{ij} = g_{ij}$. 
 
 Consequently, the inverse $\mathbf{g}^{-1}: V^* \to V$ has the property $\mathbf{g}^{-1}(\mathbf{\hat u}) = G^{-1} \mathbf{\hat u}$ for all $\mathbf{\hat u} \in V^*$. The bijective linear map $\mathbf{g}$ is commonly known as the **metric** and $\mathbf{g}^{-1}$ as the **dual metric**. 
 

docs/mathematics/linear-algebra/tensors/tensor-symmetries.md

@@ -1,6 +1,6 @@
 # Tensor symmetries
 
-We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n,$ a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}$ and a pseudo inner product $\bm{g}$ on $V.$
+We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n$, a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}$, a field F and a pseudo inner product $\bm{g}$ on $V$.
 
 ## Symmetric tensors
 
@@ -53,7 +53,7 @@ Alternatively one may write $\bigwedge_q(V) = V^* \otimes_a \cdots \otimes_a V^*
 It follows from the definitions of symmetric and antisymmetric tensors that for $0$-tensors we have
 
 $$
-    {\bigvee}_0(V) = {\bigvee}^0(V) = {\bigwedge}_0(V) = {\bigwedge}^0(V) = \mathbb{K}.
+    {\bigvee}_0(V) = {\bigvee}^0(V) = {\bigwedge}_0(V) = {\bigwedge}^0(V) = F.
 $$
 
 Furthermore, for $1$-tensors we have
@@ -188,6 +188,42 @@ An interesting result of the definition of the symmetric and antisymmetric produ
 
 ??? note "*Proof*:"
 
-    Will be added later.
+    We have for the symmetric case
 
-In some literature theorem 2 is used as definition for the symmetric and antisymmetric product from which the relation with the symmetrisation maps can be proven. Either method is valid, however it has been chosen that defining the products in terms of the symmetrisation maps is more general.
+    $$
+        \begin{align*}
+            \mathbf{\hat u}_1 \vee \mathbf{\hat u}_2 &= 2 \mathscr{S}(\mathbf{\hat u}_1 \otimes \mathbf{\hat u}_2),\\
+            &= u^1_{(i} u^2_{j)} \mathbf{\hat e}_i \otimes \mathbf{\hat e}^j,\\
+            &= u^1_i u^2_j \mathbf{\hat e}^i \otimes \mathbf{\hat e}^j + u^2_i u^1_j \mathbf{\hat e}^i \otimes \mathbf{\hat e}^j,
+        \end{align*}
+    $$
+
+    such that 
+
+    $$
+        \begin{align*}
+            \mathbf{\hat u}_1 \vee \mathbf{\hat u}_2 (\mathbf{v}_1, \mathbf{v}_2) &= u^1_i u^2_j v_1^i v_2^j + u^2_i u^1_j v_1^i v_2^j,\\
+            &= \mathrm{perm}\big(\mathbf{k}(\mathbf{\hat u}_i, \mathbf{v}_j)\big).
+        \end{align*}
+    $$
+
+    And we have for the antisymmetric case
+
+    $$
+        \begin{align*}
+            \mathbf{\hat u}_1 \wedge \mathbf{\hat u}_2 &= 2 \mathscr{A}(\mathbf{\hat u}_1 \otimes \mathbf{\hat u}_2),\\
+            &= u^1_{[i} u^2_{j]} \mathbf{\hat e}_i \otimes \mathbf{\hat e}^j,\\
+            &= u^1_i u^2_j \mathbf{\hat e}^i \otimes \mathbf{\hat e}^j - u^2_i u^1_j \mathbf{\hat e}^i \otimes \mathbf{\hat e}^j,
+        \end{align*}
+    $$
+
+    such that 
+
+    $$
+        \begin{align*}
+            \mathbf{\hat u}_1 \wedge \mathbf{\hat u}_2 (\mathbf{v}_1, \mathbf{v}_2) &= u^1_i u^2_j v_1^i v_2^j - u^2_i u^1_j v_1^i v_2^j,\\
+            &= \det\big(\mathbf{k}(\mathbf{\hat u}_i, \mathbf{v}_j)\big).
+        \end{align*}
+    $$
+
+In some literature theorem 2 is used as definition for the symmetric and antisymmetric product from which the relation with the symmetrisation maps can be proven. Either method is valid, however it has been chosen that defining the products in terms of the symmetrisation maps is more general.