Linear dependence of vectors,Linear independence of
vectors,Linear combinations of vectors
Written by :-Thokchom Sarojkumar Sinha
Define :(a) Linear dependence of vectors
(b) Linear independence of vectors
(c) Linear combination of vectors
Illustrate each of them with examples.
Solution :- (a) Linear dependence of vectors
A set containing the vectors u1,u2,u3,…,ur defined over a field F is
said to be linearly dependent
if there exists scalars a1,a2,a3,…,ar є F (not all zero) such that
a1u1 + a2u2 + a3u3 + …+ arur = 0
(b) Linear independence of vectors
Any set containing the vectors u1,u2,u3,…,ur defined over a field F is
said to be linearly
independent (L.I.) if it is not linearly dependent I.e.,if every equation of
the form
a1u1 + a2u2 + a3u3 + …+ arur = 0 implies ai = 0 for all i such that 1≤i≤r
(d) Linear combination of vectors
Any vector u expressible in the form
u = a1u1 + a2u2 + a3u3 + …+ arur is called a linear combination of
vectors u1,u2,u3,…,ur where a1,a2,,a3,…,ar are any r scalars, not all zero.
If the vectors u1,u2,u3,…,ur are linearly dependent, then one of the
vectors is a linear combination of the remaining vectors.
Let the vectors u1,u2,u3,…,ur be linearly dependent , then
a1u1 + a2u2 + a3 u3 + …+ ar ur = 0 ----------------------------------------- (1)
Where a1,a2,a3,…,ar are scalars, not all zero.
We can write (1) as
a1u1 = - (a2u2 + a3u3 + ….+ arur)
or, u1 = (- a2/a1)u2 + (- a3/a1)u3 + …..+ (- ar/a1)ur where a1 ≠ 0
which implies that u1 is a linear combination of the vectors
u2,u3,…,ur.
Determine whether the following vectors in R3(R) are linearly dependent or
linearly independent.
(a) (1,2,1), (2,1,4), (4,5,6), (1,8,-3)
(b) (1,2,3), (4,1,5), (- 4,6,2)
Solution :- (a) Let u1 = (1,2,1), u2 = (2,1,4), u3 = (4,5,6), u4 = (1,8,-3)
Using u1 to reduce the first component to zero, we get ,
u2 – 2u1 = (2,1,4) – 2(1,2,1) = (2,1,4) – (2,4,2) = (0,-3,2) ---------------(2)
u3 – 4u1 = (4,5,6) – 4(1,2,1) = (4,5,6) – (4,8,4) = (0,-3,2) ---------------(3)
u4 – u1 = (1,8,-3) – (1,2,1) = (0,6,-4) ----------------------------------(4)
From Eqn (2) and Eqn (3), we have u2 – 2u1 and u3 – 4u1 are
equivalent.
Therefore by adding Eqns (2) and (3), we get
(u2 – 2u1) + (u3 – 4u1) = 2(0,-3,2)
or, u2 + u3 – 6u1 = (0,-6,4)
-----------------------------------------------(5)
Adding Eqn (4) and Eqn (5), we get
(u4 – u1) + (u2 + u3 – 6u1) = (0,0,0)
or, - 7u1 + u2 + u3 +u4 = 0 --------------------------------------------------(6)
so the vectors u1,u2,u3,u4 are linearly dependent;are connected by the
relation (6)
(b) Let u1 = (4,5,6), u2 = (4,1,5), u3 = (-4,6,2)
Also let a1u1 + a2u2 + a3u3 = 0 ----------------------------------------------(1)
or, a1(4,5,6) + a2(4,1,5) + a3(-4,6,2) =(0,0,0)
or, (4a1,5a1,6a1) + (4a2,a2,5a2) + (-4a3,6a3,2a3) (0,0,0)
or, (4a1 + 4a2 – 4a3,5a1 + a2 + 6a3,6a1 + 5a2 + 2a3) = (0,0,0)
This => 4a1 + 4a2 – 4a3 = 0
5a1 + a2 + 6a3 = 0
6a1 + 5a2 + 2a3 = 0
=> a1 + a2 – a3 = 0 ----------------------------------------------------(2)
5a1 + a2 + 6a3 = 0 ---------------------------------------------------(3)
6a1 + 5a2 + 2a3 = 0 --------------------------------------------------(4)
From (2) a3 = a1 + a2
Using this in (3), we get
5a1 + a2 +6a1 + 6a2 = 0 and 6a1 + 5a2 + 2a1 + 2a2 = 0
or, 11a1 + 7a2 = 0 ----------------------------------------------------(5)
and 8a1 + 7a2 = 0 ----------------------------------------------------- (6)
Subtracting (6) from (5), we get
3a1 = 0 => a1 = 0
Putting a1 = 0 in (6) , we get
8.0 + 7a2 = 0 or, 7a2 = 0
Therefore a2 = 0
Putting a1 = 0, a2 = 0 in (2) we get
0 + 0 – a3 = 0 or, a3 = 0
Therefore a1 = a2 = a3 = 0
Thus every relation of the form (1) implies that ar = 0 for r = 1,2,3
Hence the given vectors are Linearly Independent.
In the vector space R3, express the vector (1,-2,5) as a linear combination
of the vectors (1,1,1),(1,2,3) and (2,-1,1).
Solution:-Let u1 = (1,-2,5), u2 = (1,1,1), u3 = (1,2,3), u4 = (2,-1,1)
If u1 is a linear combination of the vectors u2,u3 and u4, we get
u1 = a2u2 + a3u3 + a4u4 -------------------------------------------(1)
where a2,a3,a4 are scalars.
i.e., (1,-2,5) = a2(1,1,1) + a3(1,2,3) + a4(2,-1,1)
= (a2,a2,a2) + (a3,2a3,3a3) + (2a4,-a4,a4)
= (a2 + a3 + 2a4, a2 + 2a3 –a4, a2 + 3a3 +a4)
This => 1 = a2 + a3 +2a4
-2 = a2 +2a3 – a4
5 = a2 + 3a3 + a4
 a2 + a3 + 2a4 = 1 ----------------------------------------------------(2)
a2 + 2a3 - a4 = -2 ----------------------------------------------------(3)
a2 + 3a3 + a4 = 5 ---------------------------------------------------(4)
Subtracting the Eqn (3) from Eqn (4),we get
a3 + 2a4 = 7 -----------------------------------------------------------(5)
From Eqn (2) and Eqn (5),we get
a2 + 7 = 1 or, a2 = - 6
From Eqn (5) a3 = 7 – 2a4 ----------------------------------------------------(6)
Putting the value of Eqn (6) with a2 = - 6 in Eqn (3), we get,
- 6 + 2(7 – 2a4) – a4 = - 2
or, - 6 + 14 – 4a4 – a4 = -2
or, 8 – 5a4 = - 2
or, - 5a4 = - 10
Therefore a4 = ( - 10)/( - 5) = 2
Therefore from (6),
a3 = 7 – 2.2 = 7 – 4 = 3
Therefore a2 = - 6, a3 = 3, a4 = 2
We have from (1)
u1 = - 6u2 + 3u3 + 2u4
i.e., (1, -2 ,5) = - 6(1,1,1) + 3(1,2,3) + 2(2,- 1,1)
Hence the result.
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