# 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
$$= \begin{bmatrix} 4 \\ -6 \\ 0 \\ 16 \\ -18 \end{bmatrix} + \begin{bmatrix} -21 \\ 0 \\ -12 \\ 6\\ 3 \end{bmatrix}$$
Apply addition of vectors
$$= \begin{bmatrix} -17 \\ -6 \\ -12 \\ 22 \\ -15 \end{bmatrix}$$

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 $$s = -5/2$$

## More References and links

1. Vector Spaces - Questions with Solutions
2. Linear Algebra and its Applications - 5 th Edition - David C. Lay , Steven R. Lay , Judi J. McDonald
3. Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres