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Vector Space
Vectors in ℝ𝒏
Vectors in ℝ𝟐
The set of all ordered pairs of real numbers is called two dimensional real space and is denoted
by ℝ2 .
ℝ2 = {(𝑥, 𝑦): 𝑥, 𝑦 ∈ ℝ}, where ℝ2 = ℝ × ℝ
Example: (1,2); (2, −1); (3,4); (√3, 5); (0, e) ∈ ℝ2
Vectors in ℝ𝟑
The set of all ordered triplets of real numbers is called three dimensional real space and is
denoted by ℝ3
ℝ3 = {(𝑥, 𝑦, 𝑧): 𝑥, 𝑦, 𝑧 ∈ ℝ}, where ℝ3 = ℝ × ℝ × ℝ
Vectors in ℝ𝒏
If 𝑛 is a positive integer then the set of all ordered triplets of real numbers is called real 𝑛-space
and is denoted by ℝ𝑛 and if 𝑢 ∈ ℝn ; 𝑢 = (𝑢1 , 𝑢2 , … … , 𝑢𝑛 ), then 𝑢 is called a 𝑛-dimensional
vector in ℝ𝑛
ℝ3 = {(𝑥, 𝑦, 𝑧): 𝑥, 𝑦, 𝑧 ∈ ℝ}, where ℝ3 = ℝ × ℝ × ℝ
Addition of two vectors in ℝ𝟐
If 𝑢 = (𝑢1 , 𝑢2 ) and 𝑣 = (𝑣1 , 𝑣2 ) be two vectors in ℝ𝟐 then 𝑢 + 𝑣 = (𝑢1 + 𝑣1 , 𝑢2 + 𝑣2 )
Subtraction of two vectors in ℝ𝟐
If 𝒖 = (𝑢1 , 𝑢2 ) and 𝒗 = (𝑣1 , 𝑣2 ) be two vectors in ℝ𝟐 then 𝒖 − 𝒗 = (𝑢1 − 𝑣1 , 𝑢2 − 𝑣2 )
Scalar multiplication of vectors in ℝ𝟐
Let 𝑘 ∈ ℝ be a scalar and 𝑢 ∈ ℝ2 where 𝒖 = (𝑢1 , 𝑢2 ) is a vector in ℝ2 then 𝑘𝒖 = 𝑘(𝑢1 , 𝑢2 ) =
(𝑘𝑢1 , 𝑘𝑢2 )
Zero Vector:
The vector whose components are all zero is called the zero vector and is denoted by 0 and
defined by 𝟎 = (0,0, … . . ,0)
Dot or Inner product of two non-zero vectors:
Let 𝒖(𝑢1 , 𝑢2 ) and 𝒗(𝑣1 , 𝑣2 ) are two non-zero vectors and 𝜃 be the angle between them then the
dot product of 𝑢 and 𝑣 is denoted by 𝑢, 𝑣 and defined by
𝒖. 𝒗 = ‖𝑢‖‖𝑣‖cosθ
= 𝑢1 𝑣1 + 𝑢2 𝑣2
Parallel vectors in ℝ𝟐 :
Let 𝒖, 𝒗𝜖ℝ2 then the vector 𝑢 is called parallel vector of 𝑣 if 𝒖 = 𝑘𝑣; where 𝑘 is any non-zero
scalar. In this case if 𝑘 > 0 then they are same directed and if 𝑘 < 0 then they are opposite
directed.
Perpendicular vectors in ℝ𝟐 :
Let 𝒖, 𝒗 ∈ ℝ2 then 𝒖 and 𝒗 are said to be perpendicular (or orthogonal) if 𝒖. 𝒗 = 0
Example: Let 𝒖, 𝒗 ∈ ℝ2 where 𝒖 = (3, −4), 𝒗 = (8,6);
𝒖 ∙ 𝒗 = 3 ∙ 8 + (−4) ∙ 6 = 0
Hence 𝒖 and 𝒗 are perpendicular vector.
Distance between two vectors in ℝ𝟐 :
Let 𝒖, 𝒗 ∈ ℝ2 where 𝒖(𝑢1 , 𝑢2 ) and 𝒗(𝑣1 , 𝑣2 ), then the distance between 𝒖 and 𝒗, denoted by
𝑑(𝑢, 𝑣) are defined by the following:
𝑑(𝑢, 𝑣) = √(𝑢1 − 𝑣1 )2 + (𝑢2 − 𝑣2 )2
Example: If 𝒖 = (5,4), 𝒗 = (1,1) then 𝑑(𝑢, 𝑣) = √(5 − 1)2 + (4 − 1)2 = √16 + 9 = 5
Norm or Length in ℝ𝟐 :
Let 𝒖 ∈ ℝ2 , where 𝑢 = (𝑢1 , 𝑢2 ) then the norm or length of 𝑢 denoted by ‖𝑢‖ and is defined by
‖𝒖‖ = √𝑢 ∙ 𝑢 = √𝑢12 +𝑢22
Example: Let 𝒖, 𝒗 ∈ ℝ3 ; where 𝑢 = (1,0, −1), 𝑣 = (2, −3,1) then find
(𝑖)2𝒖 + 3𝒗
(𝑖𝑖)𝒖 ∙ 𝒗
(iii) ‖𝒗‖
(𝑖𝑣) ‖𝒖 − 𝒗‖
(𝑣)𝑑(𝑢, 𝑣)
Solution:
(𝑖) 2𝒖 + 3𝒗 = 2(1,0, −1) + 3(2, −3,1) = (2,0, −2) + (6, −9,3) = (8, −9,1)
(𝑖𝑖) 𝒖 ∙ 𝒗 = 1 ∙ 2 + 0 ∙ (−3) + (−1) ∙ 1 = 2 − 1 = 1
(𝑖𝑖𝑖) ‖𝒗‖ = √4 + 9 + 1 = √14
(𝑖𝑣) 𝒖 − 𝒗 = (1 − 2,0 + 3, −1 − 1) = (−1,3, −2)
∴ ‖𝒖 − 𝒗‖ = √(−1)2 + 32 + (−2)2 = √14
(𝑣)𝑑(𝑢, 𝑣) = √(1 − 2)2 + (0 + 3)2 + (−1 − 1)2 = √14 .
Vector Spaces
Let 𝐹 be a field of scalars and 𝑽 be a non-empty set of vectors. If 𝑽 contains the following rules
of vector addition and scalar multiplication, then 𝑽 is called vector space over 𝐹.
In vector addition:
𝐴1 : 𝒖, 𝒗 ∈ 𝑉 ⇒ (𝒖 + 𝒗) ∈ 𝑽
𝐴2 : 𝒖, 𝒗, 𝒘 ∈ 𝑽 ⇒ (𝒖 + 𝒗) + 𝒘 = 𝒖 + (𝒗 + 𝒘)
𝐴3 : 𝒖, 𝟎 ∈ 𝑽 ⇒ (𝒖 + 𝟎) = (𝟎 + 𝒖) = 𝒖,
here 𝟎 is vector
𝐴4 : 𝒖, 𝒗 ∈ 𝑽 ⇒ 𝒖 + 𝒗 = 𝒗 + 𝒖
𝐴5 : 𝒖 ∈ 𝑽 ⇒ −𝒖 ∈ 𝑽 ⇒ 𝒖 + (−𝒖) = (−𝒖) + 𝒖 = 0
In scalar multiplication:
𝑀1 : 𝑎 ∈ 𝐹 𝑎𝑛𝑑 𝒖 ∈ 𝑽 ⇒ 𝑎𝒖 ∈ 𝑽
𝑀2 : 𝑎 ∈ 𝐹 𝑎𝑛𝑑 𝒖, 𝒗 ∈ 𝑽 ⇒ 𝑎(𝒖 + 𝒗) = 𝑎𝒖 + 𝑎𝒗
𝑀3 : 𝑎, 𝑏 ∈ 𝐹 𝑎𝑛𝑑 𝒖 ∈ 𝑽 ⇒ (𝑎 + 𝑏)𝒖 = 𝑎𝒖 + 𝑏𝒖
𝑀4 : 𝑎, 𝑏 ∈ 𝐹 𝑎𝑛𝑑 𝒖 ∈ 𝑽 ⇒ (𝑎𝑏)𝒖 = 𝑎(𝑏𝒖)
𝑀3 : 1 ∈ 𝐹 ⇒ 1. 𝒖 = 𝒖. 1 = 𝒖;
𝒖∈𝐹
Subspace:
Let 𝑊 be a non empty subset of a vector space 𝑽 over the field 𝐹. We call 𝑊 a subspace of 𝑽 if
and only if 𝑊 is a vector space over the field 𝐹 under the laws of vector addition and scalar
multiplication defined on 𝑽, or equivalently, 𝑊 is a subspace of 𝑽 wherever 𝑤1 , 𝑤2 ∈ 𝑊 and
𝛼, 𝛽 ∈ 𝐹 implies that 𝛼𝑤1 + 𝛽𝑤2 ∈ 𝑊.
Example: Show that 𝑆 = {(𝑎, 𝑏, 𝑐) ∣ 𝑎, 𝑏, 𝑐 ∊ ℝ 𝑎𝑛𝑑 𝑎 + 𝑏 + 𝑐 = 0 } is a subspace of the vector
space ℝ𝟑 .
Solution: Here 𝟎 ∊ ℝ𝟑 , 𝟎 = (0,0,0) ∊ 𝑆
∴0+0+0= 0
Hence 𝑆 is non empty set.
Again, let 𝒖 = (𝑎1 , 𝑏1 , 𝑐1 ) ∊ 𝑆 and 𝒗 = (𝑎2 , 𝑏2 , 𝑐2 ) ∊ 𝑆
𝑎1 + 𝑏1 + 𝑐1 = 0 and 𝑎2 + 𝑏2 + 𝑐2 = 0
For any scalars α ,β we get 𝛼𝑢 + 𝛽𝑣 = 𝛼(𝑎1 , 𝑏1 , 𝑐1 ) + 𝛽(𝑎2 , 𝑏2 , 𝑐2 )
= (𝛼𝑎1 + 𝛽𝑎2 , 𝛼𝑏1 + 𝛽𝑏2 , 𝛼𝑐1 + 𝛽𝑐2 )
Now, 𝑎 + 𝑏 + 𝑐 = (𝛼𝑎1 + 𝛽𝑎2 + 𝛼𝑏1 + 𝛽𝑏2 + 𝛼𝑐1 + 𝛽𝑐2 )
= 𝛼(𝑎1 + 𝑏1 + 𝑐1 ) + 𝛽(𝑎2 + 𝑏2 + 𝑐2 ) = 𝛼. 0 + 𝛽. 0 = 0
∴ 𝛼𝒖 + 𝛽𝒗 ∈ 𝑆
Hence 𝑆 is a subspace of ℝ𝟑 .
Linear combination, Dependence and Independence Vectors
Linear combination:
Let 𝑽(𝐹) be a vector space, where 𝑣1 , 𝑣2 , 𝑣3 … 𝑣𝑛 ∊ 𝑽 and 𝛼1 , 𝛼2 , 𝛼3 … 𝛼𝑛 ∊ 𝐹. If 𝛼1 𝑣1 +
𝛼2 𝑣2 + 𝛼3 𝑣3 + ⋯ + 𝛼𝑛 𝑣𝑛 = 𝒖 ∊ 𝑽, then 𝑢 is called linear combination of 𝑣1 , 𝑣2 , 𝑣3 … 𝑣𝑛 .
Linear dependence of vectors:
Let 𝑽(𝐹) be a vector space, where 𝑣1 , 𝑣2 , 𝑣3 … 𝑣𝑛 ∊ 𝑽 and 𝛼1 , 𝛼2 , 𝛼3 … 𝛼𝑛 ∊ 𝐹. If 𝛼1 𝑣1 +
𝛼2 𝑣2 + 𝛼3 𝑣3 + ⋯ + 𝛼𝑛 𝑣𝑛 = 𝟎, and at least one of the element of the set {𝛼1 , 𝛼2 , 𝛼3 … 𝛼𝑛 } is not
zero , then the vectors 𝑣1 , 𝑣2 , 𝑣3 … 𝑣𝑛 are linearly dependence.
Linear independence of vectors:
Let 𝑽(𝐹) be a vector space, where 𝑣1 , 𝑣2 , 𝑣3 … 𝑣𝑛 ∊ 𝑽 and 𝛼1 , 𝛼2 , 𝛼3 … 𝛼𝑛 ∊ 𝐹. If 𝛼1 𝑣1 +
𝛼2 𝑣2 + 𝛼3 𝑣3 + ⋯ + 𝛼𝑛 𝑣𝑛 = 𝟎, and all of the elements of the set {𝛼1 , 𝐶, 𝛼3 … 𝛼𝑛 } are zero , then
the vectors 𝑣1 , 𝑣2 , 𝑣3 … 𝑣𝑛 are linearly independence.
Example: Write the vector 𝒖 = (1, −2,5) as a linear combination of the vectors
𝒖𝟏 = (1,1,1), 𝒖𝟐 = (1,2,3) and 𝒖𝟑 = (2, −1,1).
Solution: Let 𝛼1 𝒖𝟏 + 𝛼2 𝒖𝟐 + 𝛼3 𝒖𝟑 = 𝒖 ; where 𝛼1 , 𝛼2 , 𝛼3 are scalars.
𝛼1 (1,1,1) + 𝛼2 (1,2,3) + 𝛼3 (2, −1,1) = (1, −2,5)
(𝛼1 + 𝛼2 + 2𝛼3 , 𝛼1 + 2𝛼2 − 𝛼3 , 𝛼1 + 3𝛼2 + 𝛼3 ) = (1, −2,5)
We can write from above equation
𝛼1 + 𝛼2 + 2𝛼3 = 1
𝛼1 + 𝛼2 + 2𝛼3 = −2
𝛼1 + 3𝛼2 + 𝛼3 = 5
Solve the equations any methods, we get 𝛼1 = −6, 𝛼2 = 3 and 𝛼3 = 2.
∴−6 𝒖𝟏 + 3𝒖𝟐 + 2𝒖𝟑 = 𝒖
Hence 𝒖 is a linear combination of the vectors 𝒖𝟏 , 𝒖𝟐 𝑎𝑛𝑑 𝒖𝟑 .
Example : Test whether the following vectors 𝒖𝟏 = (1, 0, 1), 𝒖𝟐 = (−3, 2, 6), 𝒖𝟑 = (4, 5, −2)
are linearly dependent or independent.
Solution: Let 𝛼1 𝒖𝟏 + 𝛼2 𝒖𝟐 + 𝛼3 𝒖𝟑 = 𝟎 ; where 𝛼1 , 𝛼2 , 𝛼3 are scalars.
𝛼1 (1, 0, 1) + 𝛼2 (−3, 2, 6) + 𝛼3 (4, 5, −2) = (0, 0, 0)
(𝛼1 − 3𝛼2 + 4𝛼3 , 0 + 2𝛼2 + 5𝛼3 , 𝛼1 + 6𝛼2 − 2𝛼3 ) = (0, 0, 0)
We can write from above equation
𝛼1 − 3𝛼2 + 4𝛼3 = 0
0 + 2𝛼2 + 5𝛼3 = 0
𝛼1 + 6𝛼2 − 2𝛼3 = 0
Solve the equations any methods, we get 𝛼1 = 0, 𝛼2 = 0 and 𝛼3 = 0.
Hence the vectors 𝒖𝟏 , 𝒖𝟐 𝑎𝑛𝑑 𝒖𝟑 are linearly independent.
Exercise:
Write the vector 𝒖 as a linear combination of the vectors 𝒖𝟏 , 𝒖𝟐 and 𝒖𝟑 , where
1.
2.
3.
𝒖 = (2, 3, 4) , 𝒖𝟏 = (1, 0 ,1), 𝒖𝟐 = (0, −1, 1), 𝒖𝟑 = (−1, −1, 1)
𝒖 = (5, −2, 1) , 𝒖𝟏 = (−1, 2, 0), 𝒖𝟐 = (3, 1, 2), 𝒖𝟑 = (0, 1, −1)
𝒖 = (4, 2, 1, 0) , 𝒖𝟏 = (3, 1, 0, 1), 𝒖𝟐 = (1, 2, 3, 1), 𝒖𝟑 = (0, 3, 6, 6).
Test whether the following vectors are linearly dependent or independent.
4.
5.
6.
7.
𝒖𝟏 = (1, 1, −1), 𝒖𝟐 = (1, 0, 2), 𝒖𝟑 = (1, 1, 1)
𝒖𝟏 = (6, 2, 3, 4), 𝒖𝟐 = (0, 5, −3, 1), 𝒖𝟑 = (0, 0, 7, −2)
𝒖𝟏 = (1, −1, 2), 𝒖𝟐 = (3, −5, 1), 𝒖𝟑 = (2, 7,8), 𝒖𝟒 = (−1, 1, 1)
𝒖𝟏 = (1, −2, 3), 𝒖𝟐 = (5, 6, −1), 𝒖𝟑 = (3, 2, 1) .
Textbook/ References :
1.
2.
3.
4.
5.
Linear Algebra & Its Application - D. C. Lay.
Elementary Linear Algebra – H. Anton & C. Rorres.
Advanced Engineering Mathematics – E. Kreyszig.
Vector Analysis- M. R. Spiegel (Schaum's Outline Series)
A course in Vector and Matrix Analysis for Engineers and Physicists - A.K.
Mukhopadhyay & S.Sengupta