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docs/mathematics/linear-algebra/tensors/tensor-formalism.md
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docs/mathematics/linear-algebra/tensors/tensor-formalism.md
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# Tensor formalism
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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 introduced in [vector analysis](/en/physics/mathematical-physics/vector-analysis/curvilinear-coordinates/) and $\mathbb{K} = \mathbb{R} \lor\mathbb{K} = \mathbb{C}.$
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## Definition
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> *Definition 1*: a **tensor** is a multilinear mapping of the type
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>
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> $$
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> \mathbf{T}: \underbrace{V^* \times \dots \times V^*}_p \times \underbrace{V \times \dots \times V}_q \to \mathbb{K},
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> $$
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>
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> with $p, q \in \mathbb{N}$. Tensors are collectively denoted as
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>
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> $$
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> \mathbf{T} \in \underbrace{V \otimes \dots \otimes V}_p \otimes \underbrace{V^* \otimes \dots \otimes V^*}_q = \mathscr{T}_q^p(V),
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> $$
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>
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> with $\mathscr{T}_0^0(V) = \mathbb{K}$.
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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 $\dim V = n \in \mathbb{N}$.
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It follows from definition 1 and by virtue of the isomorphism between $V^{**}$ and $V$ that $\mathbf{T} \in \mathscr{T}_1^0(V) = V^*$ is a covector and $\mathbf{T} \in \mathscr{T}_0^1(V) = V$ is a vector.
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## Kronecker tensor
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> *Definition 2*: let $\mathbf{k} \in \mathscr{T}_1^1(V)$ be the **Kronecker tensor** be defined such that
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>
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> $$
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> \mathbf{k}(\mathbf{\hat e}^i, \mathbf{e}_j) = \delta^i_j,
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> $$
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>
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> with $\delta_j^i$ the Kronecker symbol.
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Let $\mathbf{\hat u} = u_i \mathbf{\hat e}^i \in V^*$ and $\mathbf{v} = v^j \mathbf{e}_j \in V$ then the tensor properties and the definition of the Kronecker tensor imply that
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$$
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\begin{align*}
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\mathbf{k}(\mathbf{\hat u}, \mathbf{v}) &= \mathbf{k}(u_i \mathbf{\hat e}^i, v^j \mathbf{e}_j), \\
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&= u_i v^j \mathbf{k}(\mathbf{\hat e}^i, \mathbf{e}_j), \\
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&= u_i v^j \delta^i_j, \\
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&= u_i v^i.
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\end{align*}
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$$
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## Outer product
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> *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
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>
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> $$
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> (f \otimes g)(x,y) = f(x) g(y),
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> $$
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>
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> for all $(x,y) \in X \times Y$.
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The outer product is associative and distributive with respect to addition and scalar multiplication, but not commutative.
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Note that although the same symbol is used for the outer product and the denotation of a tensor space, these are not equivalent.
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The following statements are given with $p=q=r=s=1$ without loss of generality.
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> *Definition 4*: the mixed $(p, q)$-tensor $\mathbf{e}_i \otimes \mathbf{\hat e}^j \in \mathscr{T}_q^p(V)$ is defined as
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>
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> $$
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> (\mathbf{e}_i \otimes \mathbf{\hat e}^j)(\mathbf{\hat u}, \mathbf{v}) = \mathbf{k}(\mathbf{\hat u}, \mathbf{e}_i) \mathbf{k}(\mathbf{\hat e}^j, \mathbf{v}),
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> $$
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>
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> for all $(\mathbf{\hat u}, \mathbf{v}) \in V^* \times V$.
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From this definition the subsequent theorem follows naturally.
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> *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
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>
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> $$
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> \mathbf{T} = T^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j,
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> $$
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>
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> with $T^i_j = \mathbf{T}(\mathbf{\hat e}^i, \mathbf{e}_j)$.
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??? note "*Proof*:"
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Let $\mathbf{T} \in \mathscr{T}_q^p(V)$ such that
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$$
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\begin{align*}
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\mathbf{T}(\mathbf{\hat e}^i, \mathbf{e}_j) &= T^k_l (\mathbf{e}_k \otimes \mathbf{\hat e}^l)(\mathbf{\hat e}^i, \mathbf{e}_j), \\
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&= T^k_l \mathbf{k}(\mathbf{\hat e}^i, \mathbf{e}_k) \mathbf{k}(\mathbf{\hat e}^l,\mathbf{e}_j), \\
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&= T^k_l \delta^i_k \delta^l_j, \\
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&= T^i_j.
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\end{align*}
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$$
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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 $\{\mathbf{e}_i\}$.
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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 $\{\mathbf{e}_i\}$.
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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 $\{\mathbf{e}_i\}$.
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By definition tensors are basis independent. Holors are basis dependent.
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> *Theorem 2*: let $\mathbf{S} \in \mathscr{T}^p_q(V)$ and $\mathbf{T} \in \mathscr{T}^r_s(V)$ be tensors with
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>
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> $$
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> \mathbf{S} = S^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j \quad \land \quad \mathbf{T} = T^r_s \mathbf{e}_r \otimes \mathbf{\hat e}^s,
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> $$
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>
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> then the outer product of $\mathbf{S}$ and $\mathbf{T}$ is given by
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>
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> $$
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> \mathbf{S} \otimes \mathbf{T} = S^i_j T^k_l \mathbf{e}_i \otimes \mathbf{e}_k \otimes \mathbf{\hat e}^j \otimes \mathbf{\hat e}^l,
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> $$
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>
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> with $\mathbf{S} \otimes \mathbf{T} \in \mathscr{T}^{p+r}_{q+s}(V)$.
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??? note "*Proof*:"
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Let $\mathbf{S} \in \mathscr{T}^p_q(V)$ and $\mathbf{T} \in \mathscr{T}^r_s(V)$ with
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$$
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\mathbf{S} = S^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j \quad \land \quad \mathbf{T} = T^r_s \mathbf{e}_r \otimes \mathbf{\hat e}^s,
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$$
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then
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$$
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\begin{align*}
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\mathbf{S} \otimes \mathbf{T} &= S^i_j (\mathbf{e}_i \otimes \mathbf{\hat e}^j) \otimes T^r_s (\mathbf{e}_r \otimes \mathbf{\hat e}^s), \\
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&= S^i_j T^r_s \mathbf{e}_i \otimes \mathbf{e}_r \otimes \mathbf{\hat e}^j \otimes \mathbf{\hat e}^s.
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\end{align*}
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$$
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Which maps two vectors and two covectors, therefore $\mathbf{S} \otimes \mathbf{T} \in \mathscr{T}^{p+r}_{q+s}(V)$.
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We have from theorem 2 that the outer product of two tensors yields another tensor, with ranks adding up.
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## Inner product
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> *Definition 5*: an **inner product** on $V$ is a bilinear mapping $\bm{g}: V \times V \to \mathbb{K}$ which satisfies
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>
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> 1. for all $\mathbf{u}, \mathbf{v} \in V: \; \bm{g}(\mathbf{u}, \mathbf{v}) = \overline{\bm{g}}(\mathbf{v}, \mathbf{u}),$
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> 2. for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and $\lambda, \mu \in \mathbb{K}: \;\bm{g}(\mathbf{u}, \lambda \mathbf{v} + \mu \mathbf{w}) = \lambda \bm{g}(\mathbf{u}, \mathbf{v}) + \mu \bm{g}(\mathbf{u}, \mathbf{w}),$
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> 3. for all $\mathbf{u} \in V\backslash \{\mathbf{0}\}: \bm{g}(\mathbf{u},\mathbf{u}) > 0,$
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> 4. for $\mathbf{u} = \mathbf{0} \iff \bm{g}(\mathbf{u},\mathbf{u}) = 0.$
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It may be observed that $\bm{g} \in \mathscr{T}_2^0$. Unlike the Kronecker tensor, the existence of an inner product is never implied.
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> *Definition 6*: let $G$ be the Gram matrix with its components $G \overset{\text{def}}= (g_{ij})$ defined as
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>
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> $$
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> g_{ij} = \bm{g}(\mathbf{e}_i, \mathbf{e}_j).
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> $$
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For $\mathbf{u} = u^i \mathbf{e}_i, \mathbf{v} = v^j \mathbf{e}_j \in V$ we then have
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$$
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\begin{align*}
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\bm{g}(\mathbf{u}, \mathbf{v}) &= \bm{g}(u^i \mathbf{e}_i, v^j \mathbf{e}_j), \\
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&= u^i v^j \bm{g}(\mathbf{e}_i, \mathbf{e}_j), \\
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&\overset{\text{def}}= u^i v^j g_{ij}.
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\end{align*}
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$$
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> *Proposition 1*: the Gram matrix $G$ is symmetric and nonsingular such that
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>
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> $$
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> g^{ik} g_{kj} = \delta^i_j,
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> $$
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>
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> with $G^{-1} \overset{\text{def}}= (g^{ij})$.
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??? note "*Proof*:"
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Let $G$ be the Gram matrix, symmetry of $G$ follows from defintion 5. Suppose that $G$ is singular, then there exists $\mathbf{u} = u^i \mathbf{e}_i \in V \backslash \{\mathbf{0}\}$ such that $G \mathbf{u} = \mathbf{0} \implies u^i g_{ij} = 0$, as a result we find that
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$$
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\forall \mathbf{v} = v^j \mathbf{e}_j \in V: 0 = u^i g_{ij} v^j = u^i \bm{g}(\mathbf{e}_i, \mathbf{e}_j) v^j = \bm{g}(u^i \mathbf{e}_i, v^j \mathbf{e}_j) = \bm{g}(\mathbf{u}, \mathbf{v}),
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$$
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which contradicts the non-degeneracy of the pseudo inner product in definition 5.
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> *Theorem 3*: there exists a bijective linear map $\mathbf{g}: V \to V^*$ with inverse $\mathbf{g}^{-1}$ such that
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>
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> 1. $\forall \mathbf{u}, \mathbf{v} \in V: \; \bm{g}(\mathbf{u}, \mathbf{v}) = \mathbf{k}(\mathbf{g}(\mathbf{u}), \mathbf{v})$,
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> 2. $\forall \mathbf{\hat u} \in V^*, \mathbf{v} \in V: \; \bm{g}(\mathbf{g}^{-1}(\mathbf{\hat u}), \mathbf{v}) = \mathbf{k}(\mathbf{\hat u}, \mathbf{v})$,
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>
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> with $\mathbf{g}(\mathbf{v}) = G \mathbf{v}$ for all $\mathbf{v} \in V$.
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??? note "*Proof*:"
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Let $\mathbf{u} \in V$ and let $\mathbf{\hat u} \in V^*$, suppose $\mathbf{\hat u}: \mathbf{v} \mapsto \bm{g}(\mathbf{u}, \mathbf{v})$ then we may define $\mathbf{g}: V \to V^*: \mathbf{u} \mapsto \mathbf{g}(\mathbf{u}) \overset{\text{def}} = \mathbf{\hat u}$.
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Let $\mathbf{v} \in V \backslash \{\mathbf{0}\}: \mathbf{g}(\mathbf{v}) = \mathbf{0}$, then
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$$
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0 = \mathbf{k}(\mathbf{g}(\mathbf{v}), \mathbf{w}) \overset{\text{def}} = \bm{g}(\mathbf{v}, \mathbf{w}),
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$$
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for all $\mathbf{w} \in V$, which contradicts the non-degeneracy of the pseude inner product in definition 5. Hence $\mathbf{g}$ is injective, since $\dim V$ is finite $\mathbf{g}$ is also bijective.
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Let $\mathbf{u} = u^i \mathbf{e}_i, \mathbf{v} = v^j \mathbf{e}_j \in V$ and define $\mathbf{g}(\mathbf{e}_i) = \text{g}_{ij} \mathbf{\hat e}^j$ such that
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$$
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\mathbf{k}(\mathbf{g}(\mathbf{u}), \mathbf{v}) \overset{\text{def}} = \bm{g}(\mathbf{u}, \mathbf{v}) = g_{ij} u^i v^j,
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$$
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but also
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$$
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\mathbf{k}(\mathbf{g}(\mathbf{u}), \mathbf{v}) = \text{g}_{ij} u^i v^k\mathbf{k}(\mathbf{\hat e}^j, \mathbf{e}_k) = \text{g}_{ij} u^i v^k \delta^j_k = \text{g}_{ij} u^i v^j.
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$$
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Since $u^i, v^j \in \mathbb{K}$ are arbitrary it follows that $\text{g}_{ij} = g_{ij}$.
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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**.
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It follows from theorem 3 that for $\mathbf{u} = u^i \mathbf{e}_i \in V$ and $\mathbf{\hat u} = u_i \mathbf{\hat e}^i \in V^*$ we have
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$$
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\mathbf{g}(\mathbf{u}) = g_{ij} u^i \mathbf{\hat e}^j = u_j \mathbf{\hat e}^j = \mathbf{\hat u},
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$$
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with $u_j = g_{ij} u^i$ and
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$$
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\mathbf{g}^{-1}(\mathbf{\hat u}) = g^{ij} u_i \mathbf{e}_j = u^j \mathbf{e}_j = \mathbf{u},
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$$
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with $u^j = g^{ij} u_i$.
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> *Definition 7*: the basis $\{\mathbf{e}_i\}$ of $V$ induces a **reciprocal basis** $\{\mathbf{g}^{-1}(\mathbf{\hat e}^i)\}$ of $V$ given by
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>
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> $$
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> \mathbf{g}^{-1}(\mathbf{\hat e}^i) = g^{ij} \mathbf{e}_j.
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> $$
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>
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> Likewise, the basis $\{\mathbf{\hat e}^i\}$ of $V^*$ induces a **reciprocal dual basis** $\{\mathbf{g}(\mathbf{e}_i)\}$ of $V^*$ given by
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>
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> $$
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> \mathbf{g}(\mathbf{e}^i) = g_{ij} \mathbf{\hat e}^j.
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> $$
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So far, a vector space $V$ and its associated dual space $V^*$ have been introduced as a priori independent entities. An inner product provides us with an explicit mechanism to construct a bijective linear mapping associated with each vector by virtue of the metric.
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