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