The Norm Of A Vector


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}).


Definition Of A Norm

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).


Some Properties Of The Norm

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}||


Unit Vectors

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}})


Standard Unit Vectors

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


Distance Between Two Vectors

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

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