LINEAR COMBINATIONS

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Linear Combination
Let V be a vector space and S be a non empty subset of V. A vector
v ∈ V is called a linear combination of vectors of S if there exists a finite
number of vectors u1 , u2 , u3 , · · · , un ∈ S and scalars
a1 , a2 , a3 , · · · , an ∈ F such that

v = a1 u1 + a2 u2 + a3 u3 + · · · + an un

Note
In any vector space V, 0 · v = 0 ∀ v ∈ V.
i.e., Zero vector is a linear combination of any non empty subset of
V.
Consider (2, 1), which is the linear combination of (1, 0)&(0, 1).
i.e., (2, 1) = 2(1, 0) + 1(0, 1)

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Example 1
Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear
combination of S or not?
Solution:
Let a, b ∈ R. Then

(2, 2) = a(1, 2) + b(2, 1)

= (a, 2a) + (2b, b)
= (a + 2b, 2a + b)
a + 2b = 2 (1)
2a + b = 2 (2)
2 2
Solve (1) and (2), we get a= , b= .
3 3
∴ (2, 2) = 23 (1, 2) + 32 (2, 1).

Hence linear combination exist.

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Example 2
Check whether the vector v = (2, −5, 3) is a linear combination of the
vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not?

Solution: Let a, b, c ∈ R. Then

(2, −5, 3) = ae1 + be2 + ce3
= a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7)
= (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c)
= (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c)
a + 2b + c = 2 (3)
−3a − 4b − 5c = −5 (4)
2a − b + 7c = 3 (5)
The above equations are written as follows
    
1 2 1 a 2
 −3 −4 −5   b  =  −5 
2 −1 7 c 3
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Solve above system of equations by Cramer’s rule.
Let
1 2 1
∆ = −3 −4 −5
2 −1 7
= 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8)
= −33 + 22 + 11
= 0
⇒ ∆ is inconsistent.

∴ There is no solution.

Hence linear combination does not exist.

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Example 3
For which values of k will the vector v = (1, −2, k) in R3 be a linear
combination of u = (3, 0, −2) and w = (2, −1, 5).

Solution: Let a, b ∈ R. Then

(1, −2, k) = au + bw
= a(3, 0, −2) + b(2, −1, 5)
= (3a, 0, −2a) + (2b, −b, 5b)
= (3a + 2b, −b, −2a + 5b)
3a + 2b = 1 (5)
−b = −2 (6)
−2a + 5b = k (7)
From (5) and (6), we get a = −1, b = 2.
∴ a, b exists.
From (7), k = 12.
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Example 4
 
3 1
Write down the matrix E = as a linear combination of the
1 −1
     
1 1 0 0 0 2
matrices A = ,B= and C =
1 0 1 1 0 −1

Solution: Let a, b, c ∈ R.
Then

E = aA + bB + cC
       
3 1 1 1 0 0 0 2
= a +b +c
1 −1 1 0 1 1 0 −1
     
a a 0 0 0 2c
= + +
a 0 b b 0 −c
 
a a + 2c
=
a+b b−c

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 a = 3
a + 2c = 1
a+b = 1
b − c = −1

Solve above equations, we get

a = 3
b = −2
c = −1

∴ a, b, c exists.

Hence the linear combination of E is 3A − 2B − C

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Span
Let S be a nonempty subset of a vector space V. The span of S, denoted
span(S), is the set consisting of all linear combinations of the vectors in
S.

Note
For convenience, we define span(φ) = {0}.

Generate
A subset S of a vector space V generates or spans V, if span(S)=V.

We also say that the vectors of S generate or span V.

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Example 1
Determine whether the vector v = (x, 1) belong to span(S), where
S = {(1, 1), (1, 0)}.

Solution:
Let a, b ∈ F. Then

a(1, 1) + b(1, 0) = (x, 1)

(a, a) + (b, 0) = (x, 1)
(a + b, a) = (x, 1)
⇒ a+b = x
a = 1

Solving the above equations, we get a = 1, b=x−1

∴ v = 1(1, 1) + (x − 1)(1, 0)
Hence v is spanned by S.

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Example 2
Determine whether the vector v = 2x3 − x2 + x + 3 belong to span(S),
where S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}.

Solution: Let a, b, c ∈ F. Then

a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v
(ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3
ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3
Comparing the coefficients of x3 , x2 , x, constant, we get
a = 2 (8)
a + b = −1 (9)
a+b+c = 1 (10)
a+b+c = 3 (11)
Solving above equations, we get a = 2, b = −3, c = 2
Substitute in (11), we get a + b + c = 2 − 3 + 2 = 1 6= 3
∴ v is not a linear combination of S. Hence S is doesn’t span v.
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Example 3
Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F 3 .

Solution: Let a, b, c ∈ F.
Then

a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (x, y, z)

(a, a, 0) + (b, 0, b) + (0, c, c) = (x, y, z)
(a + b, a + c, b + c) = (x, y, z)
⇒ a+b = x
a+c = y
b+c = z

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Solving the above equations, we get

1
a = (x + y − z)
2
1
b = (x − y + z)
2
1
c = (−x + y + z)
2

1 1 1
(x, y, z) = (x+y−z)(1, 1, 0)+ (x−y+z)(1, 0, 1)+ (−x+y+z)(0, 1, 1)
2 2 2
Hence every vector in F 3 is spanned by S.
i.e., F 3 = Span(S)

∴ The vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F 3

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Example 4
       
1 0 0 1 0 0 0 0
Show that the matrices , , ,
0 0 0 0 1 0 0 1
generate M2×2 (F)

Solution: Let a, b, c, d ∈ F.Then

         
1 0 0 1 0 0 0 0 x y
a +b +c +d =
0 0 0 0 1 0 0 1 z w
         
a 0 0 b 0 0 0 0 x y
+ + + =
0 0 0 0 c 0 0 d z w
   
a b x y
=
c d z w
         
1 0 0 1 0 0 0 0 x y
x +y +y +w =
0 0 0 0 1 0 0 1 z w
       
1 0 0 1 0 0 0 0
∴ M2×2 (F) = span , , ,
0 0 0 0 1 0 0 1
∴ Given vectors generates M2×2 (F).
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Theorem
The span of any subset S of a vector space V is a subspace of V.
Moreover, any subspace of V that contains S must also contain the span
of S.
Proof:
Let S = φ, then span(φ) = {0}.
∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in
any subspace of V)

Let S 6= φ, then S contains a vector z.

Now, 0z = 0 ∈ span(S).
Let x, y ∈ span(S).
Then there exist vectors u1 , u2 , · · · , um , v1 , v2 , · · · , vn ∈ S and
scalars a1 , a2 , · · · , am , b1 , b2 , · · · , bn such that

x = a1 u1 + a2 u2 + · · · + am um
y = b1 v1 + b2 v2 + · · · + bn vn
.
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Then

x + y = a1 u1 + a2 u2 + · · · + am um + b1 v1 + b2 v2 + · · · + bn v
∈ span(S)

For any scalar c,

cx = c(a1 u1 + a2 u2 + · · · + am um )
= (ca1 )u1 + (ca2 )u2 + · · · + (cam )um
∈ span(S)

∴ x + y and cx are in span(S).

Thus span(S) is a subspace of V.

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Let W denote any subspace of V that contains S. i.e., S ⊆ W
To prove: span(S) ⊆ W

Let w ∈ span(S).
Then w = c1 w1 + c2 w2 + · · · + ck wk for some vectors w1 , w2 , · · · , wk ∈ S
and some scalars c1 , c2 , · · · , ck .

Since S ⊆ W, we have w1 , w2 , · · · , wk ∈ W.

We know that "If W is a subspace of a vector space V and

w1 , w2 , · · · , wn ∈ W, then a1 w1 + a2 w2 + · · · + an wn ∈ W for any scalars
a1 , a2 , · · · , an ".

By the above fact, we have

c1 w1 + c2 w2 + · · · + ck wk ∈ W
⇒w∈W
∴ span(S) ⊆ W
Hence proved.
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