Vector spaces

1. VECTOR SPACES

2. LINEAR COMBINATION

3. DEFINITION:
A vector v is called a linear combination of vectors
𝑣1, 𝑣2, … … . . , 𝑣 𝑛if there exist scalars 𝑘1, 𝑘2, … … … , 𝑘 𝑛 such that,
𝑣= 𝑘1 𝑣1+𝑘2 𝑣2+……+𝑘 𝑛 𝑣 𝑛
For an example, Type equation here.a vector (a,b)Є𝑅2
is a linear
combination of (1,0)and (0,1) as it can be expressed as
(a,b)=a(1,0)+b(0,1).

4. WORKING RULES :
To check weather a vector 𝑣 is a linear combination of
vectors 𝑣1, 𝑣2, … … . . , 𝑣 𝑛, we have the following method:
Suppose that 𝑣= 𝑘1 𝑣1+𝑘2 𝑣2+……+𝑘 𝑛 𝑣 𝑛 .
Compare the corresponding components to get a linear
system in 𝑘1, 𝑘2, … … … , 𝑘 𝑛.

5. Solve the linear system by the usual methods(Gauss
elimination or Gauss Jordan elimination.)
If the system is consistent, then v is a linear
combination of 𝑣1, 𝑣2, … … . . , 𝑣 𝑛. If the system is
inconsistent ,then v is not a linear combination of
𝑣1, 𝑣2, … … . . , 𝑣 𝑛.

6. EXAMPLES
 Example 1:
Check whether the vector v = (2,1,3) is a linear
combination of 𝑣1 = (1,0,1) and 𝑣2 = −1,2,4 .
 Suppose that v = 𝑘1 𝑣1 + 𝑘2 𝑣2. This is
(2,1,3) = 𝑘1(1,0,1) + 𝑘2(-1,2,4)
(2,1,3) = (𝑘1 − 𝑘2, 2𝑘2, 𝑘1+4𝑘2).
 Comparing the corresponding components,
𝑘1 - 𝑘2 = 2 , 2𝑘1 = 1 , 𝑘1 + 4𝑘2 = 3

7.  The augmented matrix for the system is
1 −1 2
0 2 1
1 4 3
 Applying the operation 𝑅3→𝑅3 − 𝑅1 , we obtain
1 −1 2
0 2 1
0 5 1
 Applying the operation 𝑅2 →(1 2)𝑅2, we obtain
1 −1 2
0 1 1
2
0 5 1

8.  Applying the operation 𝑅3 → 𝑅3 - 5𝑅2, we obtain
1 −1 2
0 1 1
2
0 0 −3
2
 From the last row of the above matrix, we get 0 =
−3
2
, which is not possible. Thus the system is
inconsistent and hence v is not a linear
combination of 𝑣1 and 𝑣2.

9.  Example 2
Express the polynomial p(x) = -9-7x-15𝑥2
as a
linear combination of 𝑝1(x) = 2 + x + 4𝑥2
, 𝑝2(x) = 1
– x + 3𝑥2
, 𝑝3(x) = 3 + 2x + 5𝑥2
.
 Suppose that p(x) = 𝑘1 𝑝1(x) + 𝑘2 𝑝2(x) + 𝑘3 𝑝3(x).
-9 – 7x -15𝑥2
= 𝑘1(2+x+4𝑥2
) + 𝑘2(1-x+3𝑥2
) +
𝑘3(3+2x+5𝑥2
)
-9 – 7x -15𝑥2
= (2𝑘1+𝑘2+3𝑘3) + (𝑘1 − 𝑘2+2𝑘3)x
+ (4𝑘1+3𝑘2+5𝑘3)𝑥2

10.  Equating the corresponding coefficients of 1,x and
𝑥2
on both sides, we get
2𝑘1 + 𝑘2 + 3𝑘3 = -9
𝑘1 - 𝑘2 + 2𝑘3 = -7
4𝑘1 + 3𝑘2 + 5𝑘3 = -15
 The augmented matrix of the system is

2 1 3 − 9
1 −1 2 − 7
4 3 5 − 15

11.  Applying 𝑅1↔𝑅2, we get
1 −1 2 − 7
2 1 3 − 9
4 3 5 − 15
 Applying 𝑅2→ 𝑅2-2𝑅1 and 𝑅3→ 𝑅3-4𝑅1,we get
1 −1 2 − 7
0 3 −1 5
0 7 −3 13
 Applying 𝑅2 → (1
3
) 𝑅2, we get
1 −1 2 − 7
0 1 − 1
3
5
3
0 7 −3 13

12. Applying 𝑅3→ 𝑅3-7𝑅2, we get
1 −1 2 − 7
0 1 −
1
3
5
3
0 0 −
2
3
4
3
Applying 𝑅3→(−
3
2
) 𝑅3, we get
1 −1 2 − 7
0 1 −
1
3
5
3
0 0 1 − 2

13. Which is row echelon form. The system
corresponding to the last matrix is
𝑘1-𝑘2+2𝑘3= -7
𝑘2- 1
3
𝑘3 =5
3
𝑘3 = -2
Using back substituting, we get
𝑘1=-2, 𝑘2=1, 𝑘3=-2
Hence
P(x)=-2𝑝1(𝑥)+𝑝2(𝑥)-2𝑝3 𝑥 .

14. Example :3
Check weather the vector v=(0,4,5)is a linear
combination of 𝑣1=(0,-2,2) and 𝑣2=(1,3,-1) ?
Solution :
Suppose that 𝑣=𝑘1 𝑣1+𝑘2 𝑣2
that is (0,4,5)=𝑘1(0,-2,2)+𝑘1(1,3,-1)
(0,4,5)=(𝑘2, -2𝑘1+3𝑘2,2𝑘1-𝑘2)
Comparing the corresponding components. We
obtain

15. 𝑘2=0…….1
-2𝑘1+3𝑘2=4…….2
2𝑘1-𝑘2=5……..3
Here 𝑘2=0 so solve the equation 2 and 3 we get
𝑘1=-2 and 𝑘1=5
2
so which is not
possible so 𝑣is not linear combination.

16. SPAN

17. DEFINITION:
Let S= 𝑣1, 𝑣2, … … . . , 𝑣 𝑛 be a set of vectors in a vector space
V. Then the set of all linear combinations of the vectors in S is
called the space spanned by
𝑣1, 𝑣2, … … . . , 𝑣 𝑛. If we denote this set by W then we say
that 𝑣1, 𝑣2, … … . . , 𝑣 𝑛 span W. symbolically,
W=span(S) or W=span 𝑣1, 𝑣2, … … . . , 𝑣 𝑛 .

18. LINEAR
DEPENDENCE&INDEPENDENCE

19. DEFINITION:
The vectors 𝑣1, 𝑣2, … … . . , 𝑣 𝑛 are said to be linear dependent
if there exist scalars 𝑘1, 𝑘2, … … … , 𝑘 𝑛,not all zero such that
𝑘1 𝑣1+𝑘2 𝑣2+……+𝑘 𝑛 𝑣 𝑛 =0
And linearly independent if,
𝑘1 𝑣1+𝑘2 𝑣2+……+𝑘 𝑛 𝑣 𝑛 =0 => 𝑘1=𝑘2=……..=𝑘 𝑛=0.

20. THEOREMS:
1)Let S= 𝑣1, 𝑣2, … … . . , 𝑣 𝑛 be a set of n vectors. Then S is
linearly dependent if and only if one of the vectors in S can be
expressed as a linear combination of the other vectors in S.
2) A set of two vectors is linearly dependent if and only if
one vector is a scalar multiple of the other.
.

21. 3)A set containing zero vector is linearly dependent.
4)If 𝑣1, 𝑣2, … … . . , 𝑣 𝑘 are vectors in 𝑅 𝑛
and k>n , then the
vectors are linearly dependent

22. WRONSKIAN
Let 𝑓1(𝑥) , 𝑓2(𝑥),… 𝑓𝑛(𝑥) be n-1 times
continuously differentiable function of the
interval (-∞,∞). Then
 W 𝑥 =
𝑓1(𝑥) 𝑓2 𝑥 … . . 𝑓𝑛(𝑥)
𝑓′1(𝑥) 𝑓′
2
𝑥 … … 𝑓′ 𝑛(𝑥)
.
.
.
𝑓
(𝑛−1)
1
.
.
.
𝑓
(𝑛−1)
2
.
.
.
𝑓
(𝑛−1)
𝑛
Is called the Wronskian of the functions
𝑓1(𝑥) , 𝑓2(𝑥),… 𝑓𝑛(𝑥).

23. LINEAR INDEPENDENCE OF
FUCTION:
Let 𝑓1 x , 𝑓2 𝑥 , … … … , 𝑓𝑛 𝑥 be n-1 times continuously
differentiable functions on the interval (−∞, ∞).If the
wronskian of these function is nonzero for at least one point
in this interval , then these functions are linearly independent
in 𝐶(𝑛−1)
(−∞, ∞).

24. EXAMPLES
 Examples : 1
Let 𝑣1=(1,-1,0), 𝑣2=(0,1,-1) and 𝑣3=(0,0,1) be element in 𝑅3
.
Show that the set of vector {𝑣1, 𝑣2, 𝑣3} is linearly independent.
 Solution : We consider the vector equation
𝑘1 𝑣1+ 𝑘 𝑣2 + 𝑘 𝑣3 = 0
substituting for 𝑣1, 𝑣2, 𝑣3, we obtain
𝑘1(1,-1,0) + 𝑘2(0,1,-1) + 𝑘3(0,0,1)=0
(𝑘1,-𝑘1+𝑘2,-𝑘2+𝑘3)=0

25. Comparing we obtain
𝑘1=0, -𝑘1+𝑘2=0 and -𝑘2+𝑘3=0
The solution of these equation is 𝑘1=𝑘2=𝑘3=0.
Therefore , the given set of vectors is lineary
independent.
 Alternative
det(𝑣1,𝑣2,𝑣3)=
1 0 0
−1 1 0
0 −1 1
= 1 ≠ 0.
Therefore , the given vectors is linearly
independent.

26. Example : 2
Let 𝑣1=(1,-1,0), 𝑣2=(0,1,-1), 𝑣3=(0,2,1) and 𝑣4=(1,0,3) be
element of 𝑅3. Show that the set of vector {𝑣1, 𝑣2, 𝑣3, 𝑣4} is
linearly dependent.
Solution : The given set of element will be dependent if
there exists scalar 𝑘1, 𝑘2, 𝑘3, 𝑘4 not all zero, such that
𝑘1 𝑣1+ 𝑘2 𝑣2+ 𝑘3 𝑣3+ 𝑘4 𝑣4=0
Substituting for 𝑣1, 𝑣2, 𝑣3, 𝑣4 and comparing, we obtain
𝑘1+ 𝑘4=0, -𝑘1+ 𝑘2+2 𝑘3=0,- 𝑘2+ 𝑘3+3 𝑘4=0

27. The solution of this equation is
𝑘1=- 𝑘4, 𝑘2=
5𝑘4
3
, 𝑘3=−
4𝑘4
3
, 𝑘4 arbitrary.
Substituting equation (1) and cancelling 𝑘4, we
obtain
- 𝑣1+
5
3
𝑣2-
4
3
𝑣3+ 𝑣4=0
Hence there exist scalar not all zero, such that
(1) is satisfied. Therefore, the set of vector is
linearly dependent.

28. Example : 3
Check whether the set S={𝑥 +4𝑥2
,3+6𝑥+2𝑥2
,2+10𝑥-4𝑥2
}
is linearly independent 𝑃2.
Solution : Let 𝑃1(𝑥 )=2-𝑥 +4𝑥2
𝑃2(𝑥 )=3+6𝑥 +2𝑥2
𝑃3 (𝑥 )=2+10𝑥 -4𝑥2
Now, we know that
𝑘1 𝑃1(𝑥)+ 𝑘2 𝑃2(𝑥)+……..+ 𝑘 𝑛 𝑃𝑛(𝑥)=0

29.  𝑘1(2-𝑥 +4𝑥2
)+𝑘2(3+6𝑥 +2𝑥2
)+𝑘3(2+10𝑥 -4𝑥2
)=0+0𝑥 +0𝑥2
(2𝑘1+3𝑘2+2𝑘3) +(-𝑘1+6𝑘2+10𝑘3) 𝑥 +(4𝑘1+2𝑘2-4𝑘3) 𝑥2 = 0+0𝑥 +0𝑥2
Equating the corresponding coefficient of 𝑥2, 𝑥 ,1 on both the sides,
we get
2𝑘1+3𝑘2+2𝑘3=0
-𝑘1+6𝑘2+10𝑘3=0
4𝑘1+2𝑘2-4𝑘3=0

30. The coefficient of the matrix of the system
A=
2 3 2
−1 6 10
4 2 −4
det(A)=2(-24-20)-3(4-40)+2(-2-24)
= -88+108-52=-32≠0
Hence , the 𝑃1(𝑥 ), 𝑃2(𝑥 ) and 𝑃3(𝑥 ) is linearly
independent.