# Vectors in $\mathbb{R}^n$

   

## Definition of Vectors in $\mathbb{R}^n$

A vector in $\mathbb{R}^n$ is an n-tuple (or ordered list of n elements) such as $(x_1, x_2, ...., x_n)$ where each $x_i$ is a real number.
For any positive integer $n$, $\mathbb{R}^n$ is a vector space .

Example 1
a) Examples of vectors in $\mathbb{R}^2$ : $(2,0) , (0,0) , (-1/2 , \sqrt 5)$
b) Examples of vectors in $\mathbb{R}^3$ : $(-7 , 2, 0) , (0,0,0) , (-5 , -1/2 , \sqrt3)$
Note vectors in in $\mathbb{R}^2$ and $\mathbb{R}^3$ are used in physics, engineering and many other fields of studies.
c) Examples of vectors in $\mathbb{R}^5$ : $(2,0,-1,4/5,8) , (0,0,0,0,0) , (-3, -1/2 , \sqrt 5 , 0 , 10)$

A vector $\textbf{u}$ in $\mathbb{R}^n$ may be considered as a column matrix (matrix with one column) with dimension $n \times 1$ and may therefore be written as
$\textbf{u} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ ... \\ x_n \end{bmatrix}$
For a given positive integer $n$, the set of all vectors in $\mathbb{R}^n$ is a space vector and therefore vectors in $\mathbb{R}^n$ have the operations of addition and scalar multiplication defined as follows
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ ... \\ x_n \end{bmatrix} + \begin{bmatrix} x_1' \\ x_2' \\ x_3' \\ ... \\ x_n' \end{bmatrix} = \begin{bmatrix} x_1 +x_1' \\ x_2+x_2'\\ x_3 + x_3'\\ ... \\ x_n + x_n' \end{bmatrix}$
$k \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ ... \\ x_n \end{bmatrix} = \begin{bmatrix} k x_1 \\ k x_2\\ k x_3\\ ... \\ k x_n \end{bmatrix}$ Any vectors $\textbf{u}$, $\textbf{v}$ and $\textbf{w}$ and any real numbers $r$ and $s$, obey the following rules
$\textbf{u} + \textbf{v} = \textbf{v} + \textbf{u}$ , commutativity
$(\textbf{u} + \textbf{v}) + \textbf{w} = \textbf{v} + ( \textbf{u} + \textbf{w})$ , associativity of vectors
$r \cdot (s \cdot \textbf{u}) = (r s) \cdot \textbf{u}$ , associativity of scalars
$\textbf{u} + \textbf{0} = \textbf{u}$ , $\textbf{0}$ is the zero vector
$\textbf{u} + (- \textbf{u}) = \textbf{0}$ , $- \textbf{u}$ is called the negative of $\textbf{u}$
$r \cdot (\textbf{u} + \textbf{v} ) = r \textbf{u} + r \textbf{v}$
$(r + s) \cdot \textbf{u} = r \cdot \textbf{u} + s \cdot \textbf{u}$
$1 \textbf{u} = \textbf{u}$

## Examples with Solutions

Example 2
Evaluate the following and express the answer as a vector
a)       $2 \begin{bmatrix} 2 \\ -3 \\ 0 \\ 8 \\ -9 \end{bmatrix} + 3 \begin{bmatrix} -7 \\ 0 \\ -4 \\ 2\\ 1 \end{bmatrix}$           b)       $-2 \left( \begin{bmatrix} 0 \\ 1 \\ -1 \\ 4 \\ -5 \end{bmatrix} - 2 \begin{bmatrix} -1 \\ 2 \\ 7 \\ 0\\ -1 \end{bmatrix} \right) + 2 \begin{bmatrix} 0 \\ -2 \\ -2 \\ 1\\ -1 \end{bmatrix}$
Solution to Example 2
a)
Apply scalar multiplication of vectors
$= \begin{bmatrix} 4 \\ -6 \\ 0 \\ 16 \\ -18 \end{bmatrix} + \begin{bmatrix} -21 \\ 0 \\ -12 \\ 6\\ 3 \end{bmatrix}$
$= \begin{bmatrix} -17 \\ -6 \\ -12 \\ 22 \\ -15 \end{bmatrix}$

b)
Apply scalar multiplication of vectors to the terms within the parentheses
$= -2 \left( \begin{bmatrix} 0 \\ 1 \\ -1 \\ 4 \\ -5 \end{bmatrix} + \begin{bmatrix} 2 \\ -4 \\ -14 \\ 0\\ 2 \end{bmatrix} \right) + 2 \begin{bmatrix} 0 \\ -2 \\ -2 \\ 1\\ -1 \end{bmatrix}$
Apply addition of vectors within the parentheses
$= -2 \begin{bmatrix} 2 \\ -3 \\ -15 \\ 4 \\ -3 \end{bmatrix} + 2 \begin{bmatrix} 0 \\ -2 \\ -2 \\ 1\\ -1 \end{bmatrix}$
Apply scalar multiplication and addition of vectors and simplify
$= \begin{bmatrix} -4 \\ 2 \\ 26 \\ -6\\ 4 \end{bmatrix}$

Example 3
Find the unknown vectors in the following equations
a)       $2 \textbf {u} = 2 \begin{bmatrix} 4 \\ 6 \\ 0 \\ \end{bmatrix} -3 \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix}$           b)       $-3 \begin{bmatrix} 1 \\ 2 \\ -1 \\ \end{bmatrix} - \textbf {u} = - 2\begin{bmatrix} -1\\ 3\\ 4\\ \end{bmatrix}$
Solution to Example 3
a)
Apply scalar multiplication and addition of vectors to the right side of the equation
$2 \textbf {u} = \begin{bmatrix} 8 \\ 9 \\ 0 \\ \end{bmatrix}$
Scalar multiply both sides of the equation by $\dfrac{1}{2}$
$(\dfrac{1}{2}) 2 \textbf {u} = (\dfrac{1}{2}) \begin{bmatrix} 8 \\ 9 \\ 0 \\ \end{bmatrix}$
Simplify to obtain
$\textbf {u} = \begin{bmatrix} 4 \\ 9/2 \\ 0 \\ \end{bmatrix}$
b)
Scalar multiply and simplify
$\begin{bmatrix} -3 \\ -6 \\ 3\\ \end{bmatrix} - \textbf {u} = \begin{bmatrix} 2\\ -6\\ -8\\ \end{bmatrix}$

Add $- \begin{bmatrix} -3 \\ -6 \\ 3\\ \end{bmatrix}$ to both sides of the equation

$\begin{bmatrix} -3 \\ -6 \\ 3\\ \end{bmatrix} + (- \begin{bmatrix} -3 \\ -6 \\ 3\\ \end{bmatrix} ) - \textbf {u} = \begin{bmatrix} 2\\ -6\\ -8\\ \end{bmatrix} + (- \begin{bmatrix} -3 \\ -6 \\ 3\\ \end{bmatrix} )$
Simplify the right side
$- \textbf {u} = \begin{bmatrix} 5\\ 0\\ -11\\ \end{bmatrix}$
Multiply both sides of the above vector equation by $- 1$ and simplify to obtain
$\textbf {u} = \begin{bmatrix} - 5\\ 0\\ 11\\ \end{bmatrix}$

Example 4
Find the real numbers $r$ and $s$ such that
a)       $r \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + s \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \\ \end{bmatrix}$           b)       $r \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} + s \begin{bmatrix} 2 \\ 2 \\ \end{bmatrix} = \begin{bmatrix} -1\\ 3\\ \end{bmatrix}$
Solution to Example 4
a)
Scalar multiply the vectors on the left side
$\begin{bmatrix} r \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ s \\ \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \\ \end{bmatrix}$
Add the vectors of the left side
$\begin{bmatrix} r \\ s \\ \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \\ \end{bmatrix}$
Two vectors are equal if their components are equal; hence
$r = 2$ and $s = - 3$

b)
Scalar multiply the vectors on the left side
$\begin{bmatrix} r \\ 2 r \\ \end{bmatrix} + \begin{bmatrix} 2 s \\ 2 s \\ \end{bmatrix} = \begin{bmatrix} -1\\ 3\\ \end{bmatrix}$
Add the two vectors on the left side
$\begin{bmatrix} r + 2 s \\ 2 r + 2 s \\ \end{bmatrix} = \begin{bmatrix} -1\\ 3\\ \end{bmatrix}$
Use equality of vectors to write
$r + 2 s = - 1$ and $2 r + 2 s = 3$
Use any method to solve the above system of 2 equations with 2 unknowns to obtain
$r = 4$ and $s = -5/2$