Added dual vector spaces and layout for tensors.

config/en/mkdocs.yaml

@@ -91,6 +91,11 @@ nav:
       - 'Inner product spaces': mathematics/linear-algebra/inner-product-spaces.md
       - 'Orthogonality': mathematics/linear-algebra/orthogonality.md
       - 'Eigenspaces': mathematics/linear-algebra/eigenspaces.md
+      - 'Dual vector spaces': mathematics/linear-algebra/dual-vector-spaces.md
+      - 'Tensors':
+        - 'Tensor formalism': mathematics/linear-algebra/tensors/tensor-formalism.md
+        - 'Tensor symmetries': mathematics/linear-algebra/tensors/tensor-symmetries.md
+        - 'Volume forms': mathematics/linear-algebra/tensors/volume-forms.md
     - 'Calculus': 
       - 'Limits': mathematics/calculus/limits.md
       - 'Continuity': mathematics/calculus/continuity.md

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

@@ -0,0 +1,57 @@
+# 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 in [vector analysis](/en/physics/mathematical-physics/vector-analysis/curvilinear-coordinates/) and $\mathbb{K} = \mathbb{R} \lor\mathbb{K} = \mathbb{C}$. 
+
+> *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 $\lambda, \mu \in \mathbb{K}$ 
+>
+> $$
+>   \mathbf{\hat f}(\lambda \mathbf{v}_1 + \mu \mathbf{v}_2) = \lambda \mathbf{\hat f}(\mathbf{v}_1) + \mu \mathbf{\hat f}(\mathbf{v}_2).
+> $$
+
+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 $V$.
+
+Each basis $\{\mathbf{e}_i\}$ of $V$ therefore induces a basis $\{\mathbf{\hat e}^i\}$ of $V^*$ by 
+
+$$
+    \mathbf{\hat e}^i(\mathbf{v}) = v^i,
+$$
+
+for all $\mathbf{v} = v^i \mathbf{e}_i \in V$. 
+
+> *Theorem 1*: the dual basis $\{\mathbf{\hat e}^i\}$ of $V^*$ is uniquely determined by
+>
+> $$
+>   \mathbf{\hat e}^i(\mathbf{e}_j) = \delta_j^i,
+> $$
+>
+> for each basis $\{\mathbf{e}_i\}$ of $V$. 
+
+??? note "*Proof*:"
+
+    Let $\mathbf{\hat f} = f_i \mathbf{\hat e}^i \in V^*$ and let $\mathbf{v} = v^i \mathbf{e}_i \in V$, then we have
+
+    $$
+        \mathbf{\hat f}(\mathbf{v}) = \mathbf{\hat f}(v^i \mathbf{e}_i) = \mathbf{\hat f}(\mathbf{e}_i) v^i = \mathbf{\hat f}(\mathbf{e}_i) \mathbf{\hat e}^i(\mathbf{v}) = f_i \mathbf{\hat e}^i (\mathbf{v}), 
+    $$
+
+    therefore $\{\mathbf{\hat e}^i\}$ spans $V^*$. 
+
+    Suppose $\mathbf{\hat e}^i(\mathbf{e}_j) = \delta_j^i$ and $\lambda_i \mathbf{\hat e}^i = \mathbf{0} \in V^*$, then
+
+    $$
+        \lambda_i = \lambda_j \delta_i^j = \lambda_j \mathbf{\hat e}^j(\mathbf{e}_i) = (\lambda_j \mathbf{\hat e}^j)(\mathbf{e}_i) = \mathbf{0},
+    $$
+
+    for all $i \in \mathbb{N}[i \leq n]$. Showing that $\{\mathbf{\hat e}^i\}$ is a linearly independent set.
+
+Obtaining a vector and consequent covector space having the same dimension $n$. 
+
+From theorem 1 it follows that for each covector basis $\{\mathbf{\hat e}^i\}$ of $V^*$ and each $\mathbf{\hat f} \in V^*$ there exists a unique collection of numbers $\{f_i\}$ such that $\mathbf{\hat f} = f_i \mathbf{\hat e}^i$. 
+
+> *Theorem 2*: the dual of the covector space $(V^*)^* \overset{\text{def}} = V^{**}$ is isomorphic to $V$. 
+
+??? note "*Proof*:"
+
+    Will be added later.

docs/en/mathematics/linear-algebra/eigenspaces.md

@@ -42,7 +42,7 @@ Furthermore it follows from the definition that any linear combination of eigenv
         A \mathbf{x} - \lambda \mathbf{x} = (A - \lambda I) \mathbf{x} = \mathbf{0},
     $$
 
-    which implies that $(A - \lambda I)$ is singular and $\det(A - \lambda I) = 0$ by [definition](..//determinants/#properties-of-determinants). 
+    which implies that $(A - \lambda I)$ is singular and $\det(A - \lambda I) = 0$ by [definition](../determinants/#properties-of-determinants). 
 
 The eigenvalues $\lambda$ may thus be determined from the **characteristic polynomial** of degree $n$ that is obtained from $\det (A - \lambda I) = 0$. In particular, the eigenvalues are the roots of this polynomial.
 

docs/en/mathematics/linear-algebra/linear-transformations.md

@@ -104,6 +104,7 @@ With these definitions the following theorem may be posed.
 It has therefore been established that each linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ can be represented in terms of an $m \times n$ matrix.
 
 > *Theorem*: let $E = \{\mathbf{e}_1, \dots, \mathbf{e}_n\}$ and $F = \{\mathbf{f}_1, \dots, \mathbf{f}_n\}$ be two ordered bases for a vector space $V$, and let $L: V \to V$ be a linear operator on $V$, $\dim V = n \in \mathbb{N}$. Let $S$ be the $n \times n$ transition matrix representing the change from $F$ to $E$, 
+> 
 > $$
 > \mathbf{e}_i = S \mathbf{f}_i,
 > $$