The vector space $\mathbb{V}$ with the operation (+) over a field of real number $\mathbb{R}$ is an abelian group with a scalar multiplication. So, suppose $\boldsymbol{x}$, $\boldsymbol{y}$ and $\boldsymbol{z} \in \mathbb{V}$, they satisfy the following conditions:

Abelian group:

• Closure:$\boldsymbol{x} + \boldsymbol{y} \in \mathbb{V}$,
• Associativity: $( \boldsymbol{x} + \boldsymbol{y} )+ \boldsymbol{z}= \boldsymbol{y} + ( \boldsymbol{x}+ \boldsymbol{z} )$,
• Identity: $\exists \boldsymbol{0} \in \mathbb{V}, \forall \boldsymbol{x} \in \mathbb{V}$ such that $\boldsymbol{0} + \boldsymbol{x} = \boldsymbol{x}$ and $\boldsymbol{x} + \boldsymbol{0} = \boldsymbol{x}$,
• Invertibility $\exists ! ( -\boldsymbol{x} ) \in \mathbb{V}, \forall \boldsymbol{x} \in \mathbb{V}$ such that $( -\boldsymbol{x} ) + \boldsymbol{x} = \boldsymbol{0}$ and $\boldsymbol{x} + ( -\boldsymbol{x} ) = \boldsymbol{0}$,
• Commutativity: $\boldsymbol{x} + \boldsymbol{y} = \boldsymbol{y} + \boldsymbol{x}$.

Scalar multiplication:

• $\forall \alpha, \beta \in \mathbb{R}$, $\forall \boldsymbol{x} \in \mathbb{V}$ such that $( \alpha \beta ) \boldsymbol{x} = \alpha ( \beta \boldsymbol{x} )$,
• $\forall \alpha, \beta \in \mathbb{R}$, $\forall \boldsymbol{x}, \boldsymbol{y} \in \mathbb{V}$ such that $\alpha ( \boldsymbol{x} + \boldsymbol{y} ) = \alpha \boldsymbol{x} + \alpha \boldsymbol{y}$ and $( \alpha + \beta ) \boldsymbol{x} = \alpha \boldsymbol{x} + \beta \boldsymbol{x}$,
• $1 \boldsymbol{x} = \boldsymbol{x}$.

This is not a tutorial, please refer to the books for tensor algebra.

### Linearly independent:

For a set of non-zero vectors ${\boldsymbol{x}_1, \boldsymbol{x}_2, \boldsymbol{x}_3, \dots, \boldsymbol{x}_n, }$,  $\exists \alpha_1, \alpha_2, \alpha_3, \dots, \alpha_n \in \mathbb{R}$, such that,

$\displaystyle \sum_{i=1}^n \alpha_i \boldsymbol{x}_i = \boldsymbol{0}$

if and only if  $\alpha_1, \alpha_2, \alpha_3, \dots, \alpha_n = 0$.

Such set of the vectors ${\boldsymbol{x}_1, \boldsymbol{x}_2, \boldsymbol{x}_3, \dots, \boldsymbol{x}_n, }$ is called linearly independent.

### Vector space dimension:

Let $\mathcal{G} = {\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3,\dots,\boldsymbol{g}_n }$ be a subset of a vector space $\mathbb{V}$ and $latex \alpha_i \in \mathbb{R}$, the vector,

$\displaystyle \boldsymbol{x} = \sum_{i=1}^n \alpha_i \boldsymbol{g}_i$

is the linear combination of the vectors ${\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3,\dots,\boldsymbol{g}_n }$.

With the definition of the linear combination, the vector space dimension can be defined: A set $\mathcal{G} \subset \mathbb{V}$ of linearly independent vectors is a basis of a vector space $\mathbb{V}$ if every vector in $\mathbb{V}$ is the linear combination of the elements in $\mathcal{G}$.

Theorem: All the basis of a finite-dimensional vector space $\mathbb{V}$ contains the same number of vectors.

Theorem: For a n-dimension vector space $\mathbb{V}$, every set contains n linearly independent vectors is a basis of $\mathbb{V}$, every set contains more than n vectors is linearly dependent.

### Summation convention:

Let $\mathcal{G} = {\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3,\dots,\boldsymbol{g}_n }$ be the basis of the vector space $\mathbb{V}$, any vector in such space can be represented with the linear combination of the basis,

$\displaystyle \boldsymbol{x} = \sum_{i=1}^{n} x^i \boldsymbol{g}_i = x^i \boldsymbol{g}_i$

The expression is shorten without the summation symbol. This is called the Einstein summation. As shown on the mathworld, there are three rules of this notational convention:

• Repeated indices are implicitly summed over,
• Each index can appear at most twice in any term,
• Each term must contain identical non-repeated indices.

Theorem: The representation of the vector in the vector space $\mathbb{V}$ with respect to a given basis $\mathcal{G}$ is unique.

### Euclidean space:

Suppose $\boldsymbol{x}$, $\boldsymbol{y}$ and $\boldsymbol{z} \in \mathbb{V}$, then the vector space $\mathbb{V}$ with inner product of the vectors satisfying the following conditions is called Euclidean space $\mathbb{E}^n$ :

• Commutativity: $\boldsymbol{x} \cdot \boldsymbol{y} = \boldsymbol{y} \cdot \boldsymbol{x}$,
• Distributivity: $\boldsymbol{x} \cdot ( \boldsymbol{y} + \boldsymbol{z}) = \boldsymbol{x} \cdot \boldsymbol{y} + \boldsymbol{x} \cdot \boldsymbol{z}$,
• Associativity with a scalar: $\alpha (\boldsymbol{x} \cdot \boldsymbol{y}) = (\alpha \boldsymbol{x}) \cdot \boldsymbol{y} = \boldsymbol{x} \cdot (\alpha \boldsymbol{y})$,
• $\boldsymbol{x} \cdot \boldsymbol{x} \ge 0, \forall \boldsymbol{x} \in \mathbb{V}$, $\boldsymbol{x} \cdot \boldsymbol{x} = 0$ if and only if $\boldsymbol{x} = 0$.

Since the inner product of the vectors in $\mathbb{E}^n$ is non-negative, the Euclidean distance or length (Norm) is defined by,

$\displaystyle \|x\|=\sqrt{\boldsymbol{x} \cdot \boldsymbol{x}}$

### Orthonormal basis:

• Orthogonal: $\boldsymbol{x} \cdot \boldsymbol{y} = 0$,
• Normal: $|x|=1$.

Let $\mathcal{E} = {\boldsymbol{e}_1, \boldsymbol{e}_2, \boldsymbol{e}_3, \dots, \boldsymbol{e}_n }$ be a basis of $\mathbb{V}$, if $\forall \boldsymbol{e}_i, \boldsymbol{e}_j \in \mathcal{E}$ satisfy the preceding conditions, then the set $\mathcal{E}^n$ is called orthonormal basis, i.e.

$\displaystyle \boldsymbol{e}_i \cdot \boldsymbol{e}_j = \delta_{ij} ,i,j = 1, 2, 3, \dots, n$

The symbol is called Kronecker delta,

$\displaystyle \delta_{ij}=\delta^{ij}=\delta^i_j=\begin{cases} 1 & \text{if}\ i = j\\ 0 & \text{if}\ i \ne j \end{cases}$

Every set of linearly independent vectors in Euclidean space $\mathbb{E}^n$ can be orthogonalized and normalized with the Gram-Schmidt procedure.

### Dual basis:

The dual basis of vectors is a biorthogonal system of the original basis of vectors in vector space $\mathbb{V}$, the detail information can be found in wikipedia. It is a little bit abstract, the following figure shows the original basis ${ \boldsymbol{g}_1, \boldsymbol{g}_2 }$ and the dual basis ${ \boldsymbol{g}^1, \boldsymbol{g}^2 }$ in 2D Euclidean space $\mathbb{E}^2$.

In this figure, $\boldsymbol{g}_i \cdot \boldsymbol{g}^j = \delta_i^j$. Now, extend the 2D Euclidean space $\mathbb{E}^2$ to nD Euclidean space $\mathbb{E}^n$, the basis $\mathcal{G}^ \prime = {\boldsymbol{g}^1, \boldsymbol{g}^2, \boldsymbol{g}^3, \dots,\boldsymbol{g}^n }$ is the dual basis to the basis $\mathcal{G} = {\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3, \dots,\boldsymbol{g}_n }$, if that,

$\displaystyle \boldsymbol{g}_i \cdot \boldsymbol{g}^j = \delta_i^j \quad i,j = 1, 2, 3, \dots, n$

Theorem: There exists a unique dual basis of each basis in an Euclidean space $\mathbb{E}^n$.

Proof: Let $\mathcal{G}^ \prime = {\boldsymbol{g}^1, \boldsymbol{g}^2, \boldsymbol{g}^3, \dots,\boldsymbol{g}^n }$ and $\mathcal{H}^ \prime = {\boldsymbol{h}^1, \boldsymbol{h}^2, \boldsymbol{h}^3, \dots,\boldsymbol{g}^n }$ be the two dual bases of the original basis $\mathcal{G} = {\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3, \dots,\boldsymbol{g}_n }$, recall the properties of the dual basis, $\displaystyle \boldsymbol{g}_i \cdot \boldsymbol{g}^j = \delta^j_i$ $\displaystyle \boldsymbol{g}_i \cdot \boldsymbol{h}^j = \delta^j_i$ Expand $\boldsymbol{h}^j = h^j_k \boldsymbol{g}^k$ we get, $\displaystyle \delta^j_i = \boldsymbol{g}_i \cdot \boldsymbol{h}^j = \boldsymbol{g}_i \cdot (h^j_k \boldsymbol{g}^k) = h^j_k \delta_i^k = h^j_i$ Then, $\displaystyle \boldsymbol{h}^j = h^j_k \boldsymbol{g}^k = \delta^j_i \boldsymbol{g}^k = \boldsymbol{g}^j$

Now, let $\boldsymbol{g}_i$ be the basis of $\boldsymbol{g}^i$, then,

$\displaystyle \boldsymbol{g}^i = g^{ij} \boldsymbol{g}_j$

Expend the $\boldsymbol{g}_j$ with respect to the $\boldsymbol{g}^j$, then,

$\displaystyle \boldsymbol{g}_j = g_{jk} \boldsymbol{g}^k$

Thus,

$\displaystyle \boldsymbol{g}^i = g^{ij} g_{jk} \boldsymbol{g}^k$ $\displaystyle \boldsymbol{0} = g^{ij} g_{jk} \boldsymbol{g}^k - \boldsymbol{g}^i = (g^{ij} g_{jk} - \delta_k^i) \boldsymbol{g}^k$

we get,

$\displaystyle g^{ij} g_{jk} = \delta_k^i$

Then, the matrices $g^{ij}$ and $g_{jk}$ (i.e. $g_{ji}$) are inverse of each other.

### Dual basis Properties:

For orthonormal basis of $\mathbb{E}^n$, it is a self dual vector space,

$\displaystyle \boldsymbol{e}_i = \boldsymbol{e}^i$ $\displaystyle \boldsymbol{e}_i \cdot \boldsymbol{e}^j = \delta_i^j = \delta_{ij} = \boldsymbol{e}_i \cdot \boldsymbol{e}_j = \delta^{ij} = \boldsymbol{e}^i \cdot \boldsymbol{e}^j$

For non-orthonormal basis of $\mathbb{V}$,

$\displaystyle \boldsymbol{g}^i \cdot \boldsymbol{g}^j = \boldsymbol{g}^j \cdot \boldsymbol{g}^i = g^{ij} = g^{ji}$ $\displaystyle \boldsymbol{g}_i \cdot \boldsymbol{g}_j = \boldsymbol{g}_j \cdot \boldsymbol{g}_i = g_{ij} = g_{ji}$

The inner product of vector $\boldsymbol{x}$ and $\boldsymbol{y}$,

$\displaystyle \boldsymbol{x} \cdot \boldsymbol{y} = x^i \boldsymbol{g}_i \cdot y_j \boldsymbol{y}^j = x^i y_j \delta^j_i = x^i y_i = x^j y_j$

### Levi-Civita symbol:

Levi-Civita symbol is the permutation symbol,

$\displaystyle \varepsilon_{ijk} = \varepsilon^{ijk} = \begin{cases} 1 & \text{ if } ijk \text{ is an even permutation }\\ -1 & \text{ if } ijk \text{ is an odd permutation } \\ 0& \text{ if otherwise } \end{cases}$
• Even permutation: $123, 231, 312$,
• Odd permutation: $132, 321, 213$,
• Otherwise: more than two symbols are the same e.g. $112, 223, 333$.

Let ${\boldsymbol{e}_1, \boldsymbol{e}_2, \boldsymbol{e}_3 }$ be the orthonormal vectors in $\mathbb{E}^3$ and [ ] be the symbol of mixed product of three vectors,

$\displaystyle \varepsilon_{ijk} = [\boldsymbol{e}_i \boldsymbol{e}_j \boldsymbol{e}_k], \quad i, j, k = 1,2,3$

Denote g as the result for mixed product of basis vectors ${\boldsymbol{g}_1, \boldsymbol{g}_2, \boldsymbol{g}_3}$ in $\mathbb{E}^3$,

$\displaystyle g = [ \boldsymbol{g}_1 \boldsymbol{g}_2 \boldsymbol{g}_3] = [ \beta^i_1 \beta^j_2 \beta^k_3 \boldsymbol{e}_i \boldsymbol{e}_j \boldsymbol{e}_k] = \beta^i_1 \beta^j_2 \beta^k_3 [ \boldsymbol{e}_i \boldsymbol{e}_j \boldsymbol{e}_k] = \beta^i_1 \beta^j_2 \beta^k_3 \varepsilon_{ijk}$

The last term $\beta^i_1 \beta^j_2 \beta^k_3 \varepsilon_{ijk}$ is the determinant of the matrix i.e. $g = \textbf{Det}[\beta^i_j]$,

Since we get,

$\displaystyle g_{ij} = \boldsymbol{g}_i \cdot \boldsymbol{g}_j = \beta_i^k \boldsymbol{e}_k \cdot \beta_j^l \boldsymbol{e}_l = \beta_i^k \beta_j^l \delta_{kl} = \beta_i^k \beta_j^k$

Thus,

$\displaystyle |g_{ij}| = \textbf{Det}[g_{ij}] = \|[\beta_i^k][\beta_j^k]\| = \textbf{Det}[\beta^i_j]^2 = g^2$

### Levi-Civita symbol properties:

$\displaystyle \varepsilon_{ijk} \varepsilon^{imn}=\delta_j^{m}\delta_k^n - \delta_j^n\delta_k^m$ $\displaystyle \varepsilon_{jmn} \varepsilon^{imn}=2 \delta_j^i$ $\displaystyle \varepsilon_{ijk} \varepsilon^{ijk}= 6$ $\varepsilon_{ijk} \varepsilon_{lmn} = \delta_{il} ( \delta_{jm}\delta_{kn} - \delta_{jn} \delta_{km}) - \delta_{im} ( \delta_{jl} \delta_{kn} - \delta_{jn} \delta_{kl} ) + \delta_{in} ( \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl})$

### Mixed product (triple product):

With the permutation symbol, the mixed product can be expressed by,

$\displaystyle \varepsilon_{ijk}g = [\boldsymbol{g}_i \boldsymbol{g}_j \boldsymbol{g}_k] = \boldsymbol{g}_i \cdot (\boldsymbol{g}_j \times \boldsymbol{g}_k)$

then multiplying both sides of this relation by the vector $\boldsymbol{g}^i$

$\displaystyle \varepsilon_{ijk}g \boldsymbol{g}^i = \boldsymbol{g}_j \times \boldsymbol{g}_k$

Since $\textbf{Det}[\boldsymbol{g}_i \cdot \boldsymbol{g}^j] = \textbf{Det}[\delta_i^j]=1$,

$\displaystyle 1= \textbf{Det}[\boldsymbol{g}_i \cdot \boldsymbol{g}^j] = \textbf{Det}[\boldsymbol{g}_i \cdot g^{jk} \boldsymbol{g}_k] = \textbf{Det}[g_{jk}] \textbf{Det}[g^{ik}]$

Then we get,

$\displaystyle |g^{ik}| = \frac{1} {|g_{jk}|} \leftrightarrow |g^{ij}| = \frac{1} {|g_{ij}|} = \frac{1}{g^2} = \frac{1}{[\boldsymbol{g}_1 \boldsymbol{g}_2 \boldsymbol{g}_3]^2}$

Thus,

$\displaystyle \frac{\varepsilon_{ijk}}{g} \boldsymbol{g}_i = \boldsymbol{g}^j \times \boldsymbol{g}^k$

Then,

$\displaystyle \boldsymbol{a} \times \boldsymbol{b} = (a^i \boldsymbol{g}_i)\times(a^j \boldsymbol{g}_j) = \varepsilon_{ijk} a^i b^j g \boldsymbol{g}^k=g\begin{vmatrix} a^1 &a^2 &a^3 \\ b^1 &b^2 &b^3 \\ \boldsymbol{g}^1 &\boldsymbol{g}^2 &\boldsymbol{g}^3 \end{vmatrix}$ $\displaystyle \boldsymbol{a} \times \boldsymbol{b} = (a_i \boldsymbol{g}^i)\times(a_j \boldsymbol{g}^j) = \varepsilon^{ijk} a_i b_j \frac{1}{g} \boldsymbol{g}_k=\frac{1}{g} \begin{vmatrix} a_1 &a_2 &a_3 \\ b_1 &b_2 &b_3 \\ \boldsymbol{g}_1 &\boldsymbol{g}_2 &\boldsymbol{g}_3 \end{vmatrix}$

For 3D Euclidean space $\mathbb{E}^3$, $g = 1$, the cross product is,

$\displaystyle \boldsymbol{a} \times \boldsymbol{b} = \begin{vmatrix} a_1 &a_2 &a_3 \\ b_1 &b_2 &b_3 \\ \boldsymbol{e}_1 &\boldsymbol{e}_2 &\boldsymbol{e}_3 \end{vmatrix}$