This post will be about the norm of a vector. It is assumed that the reader knows about vectors where a vector in \(\mathbb{R}^{n}\) is of the form \(\textbf{v} = (v_1, v_2, \dots, v_{n})\).
The norm of a vector \(\textbf{v} = (v_1, v_2, \dots, v_{n})\) in \(\mathbb{R}^{n}\) is defined as:
\[ \displaystyle ||\textbf{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2 + \dots + v_{n}^2}\]
Sometimes the norm of a vector \(\textbf{v}\) is referred as the length of \(\textbf{v}\) or the magnitude of \(\textbf{v}.\)
In three dimensions, a vector in \(\mathbb{R}^{3}\) is \(\textbf{v} = (v_1, v_2, v_{3})\) with the norm as:
\[\displaystyle ||\textbf{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2}\]
In two dimensions, we have a vector \(\textbf{v} = (v_1, v_2)\) in \(\mathbb{R}^{2}\). Its norm is:
\[\displaystyle ||\textbf{v}|| = \sqrt{v_1^2 + v_2^2}\]
In one dimension, we have a the vector \(\textbf{v}\) is just \(v\) on the real line \(\mathbb{R}\). The absolute value is a special case of the norm and it is expressed as:
\[\displaystyle ||\textbf{v}|| = \sqrt{v^{2}} = |v|\]
Note that we square the number to ensure a positive number and then take the square root. Doing this is the same as taking the absolute value.
The norm is no longer a vector as it is a scalar/number (with no direction).
Here are some properties of a vector \(\textbf{v}\) in \(\mathbb{R}^{n}\) with a scalar (real number) \(k\).
\[\displaystyle || \textbf{v}|| \geq 0\]
\[\displaystyle ||\textbf{v}|| = 0 \text{ if and only if (iff)} ||\textbf{v}|| = \textbf{0}\]
\[\displaystyle ||k \cdot \textbf{v}|| = |k| \cdot ||\textbf{v}||\]
A vector of a norm of 1 is a unit vector. Unit vectors are of use when length is not relevant. The unit vector \(\textbf{u}\) is defined as:
\[\displaystyle \textbf{u} = \dfrac{1}{|| \textbf{v}||} \cdot \textbf{v}\]
where v is a non-zero vector in \(\mathbb{R}^{n}\).
When we obtain a unit vector u from v, it is called normalizing v.
Example One
Normalize the vector \(\textbf{v} = (-3, 4)\).
Answer:
The norm of \(\textbf{v} = (-3, 4)\) is:
\[\displaystyle \begin{array} {lcl} || \textbf{v}|| & = & \sqrt{ (3)^2 + (-4)^2} \\ & = & \sqrt{9 + 16} \\ & = & \sqrt{25} \\ & = & 5 \\ \end{array} \\ \]
The unit vector u with the same direction as v will be:
\[\displaystyle \textbf{u} = \dfrac{1}{5} \cdot (-3, 4) = (-\dfrac{3}{5}, \dfrac{4}{5})\]
Example Two
Given the vector \(\textbf{v} = (1, 5, -2)\). Find the unit vector u such that it has the same direction as v.
Answer:
The norm of \(\textbf{v} = (1, 5, -2)\) is:
\[\displaystyle \begin{array} {lcl} || \textbf{v}|| & = & \sqrt{ (1)^2 + (5)^2 + (-2)^2} \\ & = & \sqrt{1 + 25 + 4} \\ & = & \sqrt{30} \\ \end{array} \]
Our unit vector u will be:
\[\displaystyle \textbf{u} = \dfrac{1}{\sqrt{30}} \cdot (1, 5, -2) = (\dfrac{1}{\sqrt{30}}, \dfrac{5}{\sqrt{30}}, \dfrac{-2}{\sqrt{30}})\]
You may encounter standard unit vectors (of norm 1) in the form of:
\[\displaystyle \textbf{i} = (1, 0) \text{ and } \textbf{j} = (0, 1)\]
in \(\mathbb{R}^{2}\). For \(\mathbb{R}^{3}\), you may see:
\[\displaystyle \textbf{i} = (1, 0, 0) \text{ , } \textbf{j} = (0, 1, 0) \text{ and } \textbf{j} = (0, 0, 1)\].
For example, we can express the vector (2, 1) as \(2 \cdot \textbf{i} + \textbf{j}\). Likewise, the vector (-3, -1, 5) can be expressed as \(-3 \cdot \textbf{i} - \textbf{j} + 5 \cdot \textbf{k}\).
In the general case in \(\mathbb{R}^{n}\), the standard unit vectors would be:
\[ \displaystyle \textbf{e}_1 = (1, 0, 0, \dots , 0) \]
\[ \displaystyle \textbf{e}_2 = (0, 1, 0, \dots , 0) \]
\[ \displaystyle \textbf{e}_3 = (0, 0, 1, \dots , 0) \]
\[ \displaystyle \vdots\]
\[\displaystyle \textbf{e}_n = (0, 0, 0, \dots , n) \]
and any vector \(\textbf{v} = (v_1, v_2, \dots, v_{n})\) can be expressed as a linear combination as follows:
\[\displaystyle \textbf{v} = v_1 \cdot \textbf{e}_1 + v_2 \cdot \textbf{e}_2 + \dots + v_{n} \cdot \textbf{e}_n \]
Recall that the distance between points \(P_1(x_1, y_1)\) and \(P_2(x_2, y_2)\) in 2-space is:
\[\displaystyle d(\textbf{a}, \textbf{b}) = ||\overrightarrow{P_1 P_2}|| = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2}\]
In a three-dimensional setting, the distance between points \(P_1(x_1, y_1, z_1)\) and \(P_2(x_2, y_2, z_2)\) is:
\[\displaystyle d(\textbf{a}, \textbf{b}) = ||\overrightarrow{P_1 P_2}|| = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]
We can now extend this to n-th dimensional space \(\mathbb{R}^{n}\).
In \(\mathbb{R}^{n}\), the distance between vectors \(\textbf{a} = (a_1, a_2, a_3, \dots, a_n)\) and \(\textbf{b} = (b_1, b_2, b_3, \dots, b_n)\) is
\[\displaystyle d(\textbf{a}, \textbf{b}) = ||\textbf{a} - \textbf{b}|| = \sqrt{ (a_1 - b_1)^2 + (a_2 - b_2)^2 + (a_3 - b_3)^2 + \dots + (a_n - b_n)^2}\]
Example
Calculate the distance between the vectors \(\textbf{v} = (10, 5, -2, -1)\) and \(\textbf{w} = (-1, 0, 2, 1)\) in \(\mathbb{R}^{4}\).
Answer:
We apply the distance formula.
\[\displaystyle \begin{array} {lcl} d(\textbf{v}, \textbf{w}) & = & \sqrt{ (v_1 - w_1)^2 + (v_2 - w_2)^2 + (v_3 - w_3)^2 + (v_4 - w_4)^2} \\ & = & \sqrt{ (10 - (-1))^2 + (5 - 0)^2 + (-2 - 2)^2 + (-1 - 1)^2} \\ & = & \sqrt{ (11)^2 + (5)^2 + (-4)^2 + (-2)^2} \\ & = & \sqrt{ 121 + 25 + 16 + 4} \\ & = & \sqrt{166} \\ & \approx & 12.8841 \text{ units} \\ \end{array} \]
Reference: Elementary Linear Algebra (10th Editon) by Howard Anton
The image is taken from https://www.mathsisfun.com/algebra/images/vector-mag-dir.gif
