3 Dimension&basis

Span of any set S Definition : Span of set S is set of all finite linear combinations of elements of S.

Thm : [S] is a subspace of V for any nonempty set S of V.

Rajiv Kumar

Math II

Trivial linear combination If all scalars α1 , α 2 ,...,α n are zeros, then linear combination

α u + α u +,...,+α u 1

1

2

2

n

n

is called trivial linear combination.

Rajiv Kumar

Math II

Nontrivial linear combination

If at least one of αi’s is not zero,

α u + α u + ,..., + α u then 1

1

2

2

n

is called non-trivial linear combination Rajiv Kumar

Math II

n

Linearly Dependent and Independent Set

Defn : A finite set S = { u , u ,..., u Is said to linearly independent if 1

2

n

α1u1 + α 2u2 +,...,+α nun = 0V ⇒ α1 = α 2 ......... = α n = 0

}

When is the set S linearly dependent ? How to test if a finite set S is linearly independent or dependent ?

Rajiv Kumar

Math II

Take arbitrary linear combination of vectors of a set

S = { u1 ,..., u n }

and equate

it to zero vector,

α1u1 + α 2u2 +,..., +α nu n = 0V now solve for

(i)

α1 , α 2 ,..., α n

We can prove all αi are 0, then S is linearly independent If

If there exists a nontrivial solution of equation (i) i.e. at least one of α ’s is i

not zero, then set

S = { u , u ,..., u 1

2

n

}

Rajiv Kumar

is called LD. Math II

Corr : Any set S = { 0 , u , u ,..., u } V

1

2

n

containing zero vector in it is always LD. Consider α ≠ 0 0

α 0 + 0u +,...,+0u = 0 0

V

1

n

V

in this allscalars

α , 0, 0,…,0 are not zeros since 0

α ≠ 0. 0

Therefore S is LD.

Q.10 : Given that set { u, v, ω } is Linearly

Independent,

check

whether { u + v, v + ω , ω + u} is LI.

α ( u + v ) + β ( v + ω ) + γ ( ω + u ) = 0V

(α + γ ) u + ( β + α ) v + ( γ + β )ω = 0V

Since { u, v, ω} is L.I.

α +γ = β +α = γ + β = 0

⇒α = β =γ = 0

Example : Determine whether the Set

{

}

S = 1, x + x ,− x + x,3 x 2

2

of P vector space of all polynomials of is LI or LD. Rajiv Kumar

Math II

Take the linear combination of the vectors of set S equal to oP

α (1) + β ( x + x ) + γ ( − x + x ) + δ (3x) = 0 P 2

2

On rearranging in powers of x, we get,

α + ( β + γ + 3δ ) x + ( β − γ ) x =0 P 2

Rajiv Kumar

Math II

0+0x+0x = α + ( β + γ + 3δ ) x + ( β − γ ) x 2

2

comparing coefficient of like powers of x we get equations α=0, β+γ +3δ=0, β- γ =0 Which has infinitely many solutions , a possible solution is α=0, δ=-2/3, β= γ =1 Hence S is linearly dependent. Rajiv Kumar

Math II

Example : Determine whether the Set

{

}

S = x,1 + x, x − x + 1 2

of P vector space of polynomials of degree ≤ 2 is LI or LD. 2

Take the linear combination of the vectors of set S equal to 0V 2 α ( x ) + β (1 + x ) + γ x − x + 1 = 0

(

)

On rearranging in powers of x, we get, ( β + γ ) + (α + β − γ ) x + ( γ ) x

2

=0 P 2

Basis and Dimension Definition: Let V be a vector space. A subset B of V is called a basis of V, if (i) B is Linearly Independent (ii) [B] = V

Definition : If a vector space V has a basis consisting of a finite number of elements then space is said to be Definition : Number of elements in a finite dimensional. Basis of a finite dimensional vector

space V is called dimension of V. Rajiv Kumar

Math II

Thm:If set

S = { u1 , u2 ,..., um }

a vector space V is LD, then every superset

of

S = { u1 , u 2 ,..., u m , u m +1 ,....... u n } of S is also L.D.

S= { u , u ,..., u } .As set S is LD there exist a nontrivial linear combination of elements of S equal to 0V , i.e there exist αi not all zero such that α u + α u + ... + α u = 0 ---(i) 1

1

1

2

2

m

2

m

m

V

Now consider superset

S ′ = { u1 , u2 ,..., um , um +1 ,...un } of S. Now (i) ⇒

α 1u1 + α 2u2 + ... + α mum + 0um+ 1 + ... + 0un = 0V

Hence there exist a non trivial linear combination of elements of S' equal to 0v i.e set S' is linearly dependent.

Theorem: If a set S = { u , u ,..., u 1

2

n

}

o

a vector space V is linearly

independent, then every subset of S is is also linearly independent.

Rajiv Kumar

Math II

Theorem: Suppose S = { u1 , u2 ,..., uk } is an ordered set of a vector space V. If u ≠ 0 , then set S is LD, iff one of the vectors um of { u ,..., u } belongs to the span of { u ,..., u } 1

V

1

1

k

m −1

Fact

If { u1, u2 ,..., um } is Linearly independen t

and u∈ [{ u1, u2 ,..., um } ], u ≠ ui then { u1, u2 ,..., um , u}

is Lineardependent. Rajiv Kumar

Math II

Fact

If { u1, u2 ,..., um } is Linearly independen t and u∉ [{ u1, u2 ,..., um } ] then { u1, u2 ,..., um , u}

is Linearindependen t. Rajiv Kumar

Math II

Example : Determine whether the Set x x x S= {xe ,(1+x) e ,(1-x) e }, of vector space ζ(0,∞) LI or LD. Rajiv Kumar

Math II

x

x

αxe +β(1+x) e x +γ(1-x) e =0V x as e ≠ 0 for any x, -x multiply with e , we get ⇒ αx +β(1+x) +γ(1-x) =0V Rajiv Kumar

Math II

⇒(β+γ) + (α+β-γ)x =0V Then there exist infinitely many Solutions , for example α= -2 β=1, γ=-1 is a solution hence Set is linearly dependent Rajiv Kumar

Math II

Example : Determine whether the Set x x 3 x S={xe ,(1+x) e ,(1-x ) e }, of vector space ζ(0,∞) LI or LD. Rajiv Kumar

Math II

x

x

αxe +β(1+x) e 3 x +γ(1-x ) e =0V ⇒ αx +β(1+x) 3 +γ(1-x ) =0V x as e ≠ 0 for any x

Rajiv Kumar

Math II

3

⇒(β+γ) + (α+β)x-γx =0V ⇒ γ=0 β + γ=0 α+ β =0 ⇒

γ=0 =α= β

Hence set is linearly independent. Rajiv Kumar

Math II

Q. Find a basis for a subspace U of V4 in the following 3x1+x2-x3+x4 = U={(x1,x2,x3,x4)∈V4 : 0 x1+3x2-x3 = 0 Essentially U is solution set of two Homogenous equations . Rajiv Kumar

Math II

Answer : {(1/4,1/4,1,0),(-3/8,1/8,0,1)}

Rajiv Kumar

Math II

Fact

If { u1, u2 ,..., um }

is Linearly independen t and u∈ [{ u1, u2 ,..., um } ]

then { u1, u2 ,..., um , u} , u ≠ ui

is Lineardependentand [{ u1, u2 ,..., um } ] = [{ u1, u2 ,..., um , u} ] Rajiv Kumar

Math II

Corollary : A finite subset S = { u1 , u2 ,..., uk } S= of a vector space V containing a nonzero vector contains a linearly independent subset A of S such that [A]=[S]. Rajiv Kumar

Math II

Q.2 page 104 Q. Is the subset S of V4 a basis of V4 ,? If not find a basis of [S] S={(1,-1,0,1), (0,0,0,1), (2,1,0,1),(3,2,1,0)}

Rajiv Kumar

Math II

Example Q. Is the subset S of V4 a basis of V4 ,? If not find a basis of [S] S={(1,-1,0,1), (1,0,0,0), (2,={u 1,0,1),(3,2,1,0)} 1, u 2 , u3 , u 4 }

Solution : set {u1} is ≠(0,0,0,0) Linearly independent as u1 Rajiv Kumar

Math II

{u1, u2 } is linearly independent As u2 ≠ αu1 for any α real Now number to check, if set { u1, u2 , u3 } is linearly independen

or dependent. For that we need to Check whether u3 is in span of

{ u , u } or not?

Rajiv Kumar

Math II

(2,-1,0,1) =α (1,-1,0,1)+ β (1,0,0,0), We get from inspection that α=1,β =1 Hence

u3 is in span of

{ u1, u2 } i.e. , set { u1, u2 , u3 } Is linearly dependent. But [{ u1, u2 } ]= [{ u1, u2 , u3 }] Rajiv Kumar

Math II

Now we check

u4 is in span of

{ u1, u2 } or not? For that (3,2,1,0) =c (1,-1,0,1)+ d (1,0,0,0),

And we have equations c+d=3 -c=2, 1=0c+0d &0=c+0d Which would mean 0=1, which is Not true , hence c&d does not exis Rajiv Kumar

Math II

u4 is not in span of { u1, u2 } i.e { u1, u2 , u4 } is Linearly independent and [S]= [{ u1, u2 , u4 }] As u3 is in [{ u1, u2}] , hence In [{ u1, u2 , u4 }] i.e. A= { u1, u2 , u4 } i.e.

Rajiv Kumar

Math II

Q. Find a basis for a subspace U of V in the following U={p(x)∈P3 : p(1)=0 =p'(0) }

Answer : {x2-1, x3-1} Rajiv Kumar

Math II

Q. Find a basis for a subspace U of V in the following U={p(x)∈P4 : p"(1) =p'(1) }

Rajiv Kumar

Math II

Definition of sum of two sets

A +C = {u+w : u∈A, w∈C

? What is sum of x-axis and y- axis In XY plane. Thm : If U &W are two subspaces of vector space V then U+W =[U∪W] Rajiv Kumar

Math II

Q. 9 If U &W are two distinct (n-1) dimensional subspaces of an n dimensional vector space V ,n >1, then prove that dim (U∩W)=n-2

Rajiv Kumar

Math II

As U & W are distinct there exist an element u in U such u is not in W Now U & W are n-dimensional Let S’ = { u1, u2 ,…………. un-1 } And S’’ ={ v1, v2 ,…………. vn-1 } Be basis of U & W respectively Rajiv Kumar

Math II

u∉ W = [S’’] u∉ [S’’ ] and S’’ is linearly independent i.e. Set ={ v1, v2 ,…………. vn1,u} is linearly independent and has n elements hence it is But U+W= a basis of V [U∪W] would imply that U+W=V Rajiv Kumar

Math II

dim (U+W) =dim (U) + dim(W) -dim(U∩ W) Would give the result.

Rajiv Kumar

Math II

Q. Find a basis for a subspace U of V in the following U={p(x)∈P3 : p"(1) =p'(0) =0 } p(x)∈U ⇒ 2

3

q0 + q1x+ q2x + q3x =p(x) &p"(1) =p'(0) =0 Rajiv Kumar

Math II

2

p'(x)= q1+2q2x+ 3 q3x p’’(x)= 2q2+ 6 q3x

Now p’(0)=0 ⇒q1=0 p’’(1)=0⇒2q2+ 6 q3=0 Hence U={q0 + q1x+ q2x2+ q3x3: q1=0 & q2= -3q3 } Rajiv Kumar

Math II

U={q0-3q3x2+ q3x3: q1, q3 ∈R} U={q0+q3 ( -3x2 +x3): q1, q3 ∈R} U=[{1, -3x2 +x3} ]

Now as set {1, -3x2 +x3} is linearly independent, {1, -3x2 +x3} is a basis of U. Rajiv Kumar

Math II

Q.4 page 104 Q. Determine the dimension of [S]. S={(1,-1,0,1), (0,0,0,1), (2,1,0,1),(3,2,1,0)}

? If we just want to find dim([S]) Can we do it in better manner If elementary row operations are performed on rows of a matrix A Does the span of row vectors of Matrix A change? It does not. Rajiv Kumar

Math II

Hence if we write vectors as rows of the matrix and reduce the matrix to row echelon form, then row rank is nothing but dim[S].

Rajiv Kumar

Math II

Thm : If V has a basis of n elements then every set of p vectors with p>n is linearly dependent. Proof not included .

Rajiv Kumar

Math II

Definition: Let B = {u1 ,u2,…….. un,} be an ordered basis for V. Then a vector v ∈ V can be written as: v = α 1u1+ α 2u2 +…….+ α nun. The vector (α 1, α 2,……., α n) is called the coordinate vector of v relative to the Rajiv Kumar

Math II

It is denoted by [v]B. α 1, α 2,……., α n are called the coordinates of vector v relative to the ordered basis B. The coordinates of a vector relative to the standard basis are simply called the coordinates of the vector. It should be clear that [v]B is unique in view of following Rajiv Kumar

Math II

Theorem: In a vector space V let B = {u1,u2,…….. ,un} span V. Then the following two conditions are equivalent: (i) {u1 ,u2,…….. un,} is a linearly independent set. ( ii) If v ∈ V, then the expression Rajiv Kumar

Math II

Example: Find the coordinates of 3x+4x2 with respect to the ordered basis B={1,x+x2, x2-1} of P2 Let α+β(x+x2)+γ(x2-1) = 3x+4x2 (α-γ)+βx+(β+ γ)x2= 3x +4x2 Rajiv Kumar

Math II

Equating coefficients of powers of x we get α-γ=0, β= 3, β+ γ=4 ⇒ γ =1= α, β=3 (α,β,γ) = (1,3,1) are coordinates of the polynomial 3x+4x2 with respect to the given basis., i.e. [3x+4x2]B= (1,3,1) Rajiv Kumar

Math II

Thm : In an n-dimensional vector space V any set of n linearly independent vectors is a basis. Proof : If

S = { u , u ,..., u 1

2

m

}

Is set of n linearly independent Vectors, for proving S is a basis of V we shall prove [S]=V Rajiv Kumar

Math II

Let u∈V & u∉S , we shall prove u∈ [S], Now set

S ∪ {u} = { u , u ,..., u , u} 1

2

m

Is a set containing more than n elements, hence as per theorem Stated earlier set S∪{u} is LD i.e there exist a nontrivial linear Combination of elements of S∪{u} Equal to 0V. Rajiv Kumar

Math II

i.e there exist αi not all zero such that α1u1 +α2u2 + ... +αmum +αm +1u = 0V

---(i)

Now first we prove αm+1≠0 Because if αm+1=0, then (i) ⇒ α1u1 + α2u2 + ... + α mum = 0V

And linear independence of set S Would mean all αi 1 ≤i≤m are o Rajiv Kumar

Math II

Which contradicts the fact that

α1u1 + α 2u2 + ... + α mum + α m +1u = 0V

Was a nontrivial linear combination Hence αm+1≠0 , i.e. u=-(1/ αm+1) (α1u1 + α 2u2 + ... + α mum ) i.e. u∈[S] Hence V is a subset of [S] And we already have [S] is a subset V hence [S]=V, i.e S is a basis of V Rajiv Kumar

Math II