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 \( \textbf{u} \)
\( r \cdot (\textbf{u} + \textbf{v} ) = r \textbf{u} + r \textbf{v} \)
\( (r + s) \cdot \textbf{u} = r \cdot \textbf{u} + s \cdot \textbf{u} \)
\( 1 \textbf{u} = \textbf{u} \)
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 \( s = -5/2 \)
