Basic Operations on Euclidean n-Space

# Basic Operations on Euclidean n-Space

Recall from the Euclidean n-Space page that for each positive integer $n$, the Euclidean n-space $\mathbb{R}^n$ is the set of all points $\mathbf{x} = (x_1, x_2, ..., x_n)$.

In a moment we will look at some operations defined on Euclidean n-space that the reader should already be familiar with. Before we do though, the reader should note that all of the operations defined below are in compliance to the field axioms of the real numbers in that all of the operations below are all in conjunction with the operations $+$ of addition and $\cdot$ of multiplication of reals.

 Definition: if $\mathbb{x} = (x_1, x_2, ..., x_n), \mathbb{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n$ then we define Equality $\mathbf{x} = \mathbf{y}$ if and only if $x_k = y_k$ for all $k \in \{ 1, 2, ..., n \}$.

For example, if $\mathbf{x} = (1, 4, 7)$ and $\mathbf{y} = (1, 3, 7)$ then $\mathbf{x} \neq \mathbf{y}$ since $4 \neq 3$.

 Definition: If $\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n$ then Addition is defined to be $\mathbf{x} + \mathbf{y} = (x_1 + y_1, x_2 + y_2, ..., x_n + y_n)$ and Subtraction is defined to be $\mathbf{x} - \mathbf{y} = (x_1 - y_1, x_2 - y_2, ..., x_n - y_n)$.

For example, consider the points $\mathbf{x} = (1, 4, 2, 6), \mathbf{y} = (3, -2, 0.5, \pi) \in \mathbb{R}^4$. Then:

(1)
\begin{align} \quad \mathbf{x} + \mathbf{y} = (1+3, 4 +(-2), 2 + 0.5, 6 + \pi) = (4, -2, 2.5, 6 + \pi) \end{align}

And furthermore we have that:

(2)
\begin{align} \quad \mathbf{x} - \mathbf{y} = (1 - 3, 4 - (-2), 2 - 0.5, 6 - \pi) = (-2, 6, 1.5, 6 -\pi) \end{align}

Note that in general $\mathbf{x} + \mathbf{y} \neq \mathbf{y} + \mathbf{x}$ which we are already familiar with in the case when $n = 1$.

 Definition: If $\mathbf{x} = (x_1, x_2, ..., x_n) \in \mathbb{R}^n$ then Scalar Multiplication by the scalar $a \in \mathbb{R}$ is defined to be $a \mathbf{x} = a(x_1, x_2, ..., x_n) = (ax_1, ax_2, ..., ax_n)$.

For example, consider the point $\mathbf{x} = (1, 2, 3, 4, 5) \in \mathbb{R}^5$ and $a = 2 \in \mathbb{R}$. Then:

(3)
\begin{align} \quad a\mathbf{x} = 2(1, 2, 3, 4, 5) = (2, 4, 6, 8, 10) \end{align}