📂Geometry

Basis Orientation Defined in a Vector Space

Definition ¹

$$U = \left\{ \mathbf{u}_{1}, \cdots, \mathbf{u}_{n} \right\} \\ V = \left\{ \mathbf{v}_{1}, \cdots, \mathbf{v}_{n} \right\}$$ Let $U,V$ be a basis of the vector space $X$, and define the matrix $\left( a_{ij} \right) \in \mathbb{C}^{n \times n}$ such that the following equation is satisfied. $$\mathbf{v}_{j} = \sum_{i=1}^{n} a_{ij} \mathbf{u}_{i}$$ If $\det \left( a_{ij} \right) > 0$ then $U,V$ is said to have the same Orientation. If $\det \left( a_{ij} \right) < 0$ then they have different orientations.

In particular, in the Euclidean space $X = \mathbb{R}^{n}$, orientations have specific names.

• For $\mathbb{R}^{2}$, the basis $\left\{ \mathbf{e}_{2}, \mathbf{e}_{1} \right\}$ is in a clockwise direction.
• For $\mathbb{R}^{3}$, the basis $\left\{ \mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3} \right\}$ is in a right-hand direction.

---

• $\mathbf{e}_{k}$ is a unit vector $\left( 0, \cdots , 0, 1, 0 ,\cdots ,0 \right)$ whose $k$-th component is $1$ and the rest are $0$.

Explanation

Be aware that the two bases are ordered in the definition of orientation. Since orientation is determined by the determinant, changing their order will swap the rows and columns of matrix $\left( a_{ij} \right)$, causing the sign to accurately flip once each time.

The reason for defining orientation geometrically should not be hard to understand. It’s good to think of it as the direction in which the clock hands move on a $2$-dimensional plane $\mathbb{R}^{2}$, and in a $3$-dimensional space $\mathbb{R}^{3}$, it resembles the motion of wrapping the right hand inward while lifting the thumb. From $4$ dimensions onwards, since these concepts do not apply, there’s no specific name, and it is only possible to evaluate whether the two bases are the same or different.

Example

$$\begin{align*} \mathbf{u}_{1} =& \mathbf{e}_{1} & \mathbf{v}_{1} =& \left( 1, 1, 0 \right) \\ \mathbf{u}_{2} =& \mathbf{e}_{2} & \mathbf{v}_{2} =& \left( 1, 0, -1 \right) \\ \mathbf{u}_{3} =& \mathbf{e}_{3} & \mathbf{v}_{3} =& \left( 2, 1, 3 \right) \end{align*}$$

Considering such two bases in $\mathbb{R}^{3}$, $$\begin{align*} \mathbf{v}_{1} = \left( 1, 1, 0 \right) =& 1 \mathbf{e}_{1} + 1 \mathbf{e}_{2} + 0 \mathbf{e}_{3} \\ \mathbf{v}_{2} = \left( 1, 0, -1 \right) =& 1 \mathbf{e}_{1} + 0 \mathbf{e}_{2} - 1 \mathbf{e}_{3} \\ \mathbf{v}_{3} = \left( 2, 1, 3 \right) =& 2 \mathbf{e}_{1} + 1 \mathbf{e}_{2} + 3 \mathbf{e}_{3} \end{align*}$$ the matrix $\left( a_{ij} \right)$ is defined as $$\begin{bmatrix} 1 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & -1 & 3 \end{bmatrix}$$ and its determinant is $$\begin{align*} \det \left( a_{ij} \right) =& 1 \cdot (0 \cdot 3 - (-1)\cdot 1) \\ && - 1 \cdot ( 1 \cdot 3 - (-1) \cdot 2 ) \\ & + 0 \cdot ( 1 \cdot 1 - 0 \cdot 2 ) \\ =& 1 - 5 + 0 \\ =& -4 \\ <& 0 \end{align*}$$ thus, the two bases have different orientations.