Vector Subspace

 Vector Subspace

A non empty subset W of a vector space V over a field F is said to be a subspace of V if W itself forms a vector space over F w.r.t addition and scalar multiplication in V.

Let V be a vector space, then the zero space {0} and the whole space V are the trivial or improper subspace of V. All other subspace of V are called proper subspace of V.

Theorem 1:

A non empty set W of a vector space V is subspace of V if
(i) W is closed under addition
i.e x,y ∈ W ⟹ x + y ∈ W
(ii) W is closed under scalar multiplication
i.e α ∈ F, x ∈ W ⟹ αx ∈ W

Proof:

Suppose that W is a subspace of V, then

(i) W is closed under addition

i.e x,y ∈ W ⟹ x + y ∈ W

(ii) W is also closed under scalar multiplication

i.e α ∈ F, x ∈ W ⟹ αx ∈ W

Conversely suppose that W is a non empty subset of V such that conditions (i) and (ii) are satisfied.

Let x,y ∈ W

Now y ∈ W ⟹  (-1)y ∈ W by (ii)

⟹ -y ∈  W

Therefore x,y  ∈  W ⟹ x, -y  ∈  W

⟹ x + (-y)  ∈  W by (i)

⟹ x - y ∈ W.

Thus W is subgroup of V under addition.

Since W is subset of V, axiom V₂) and V₃) of Vector space V will be satisfied.

Hence W is a vector space and so it is a subspace of V.

Theorem 2.

A non empty subset W of a vector space V over a field F is a subspace of V if α, β  ∈  F and x,y  ∈  W ⟹ αx +  βy  ∈  W.

Proof:

Suppose W is a subspace if V , then

α, β ∈ F and x,y  ∈  W⟹ αx  ∈ W , βy  ∈ W

( since W is closed under scalar multiplication)

⟹ αx +  βy  ∈  W. (W is closed under addition)

Conversely suppose that

α, β ∈ F and x,y  ∈  W  ⟹ αx +  βy  ∈  W

Taking α = β = 1, we get

x,y  ∈  W ⟹ 1x + 1y ∈ W

⟹ x + y ∈ W

Therefore it is closed under addition.

α ∈ F and x ∈  W ⟹ αx + 0y ∈ W, taking β = 0

Thus α ∈ F and x ∈  W ⟹ αx ∈ W

Therefore it is closed under scalar multiplication.

Hence W is a subspace of V by theorem 1.

Example 1:

Consider the following subset of R³ (R)

(i) W₁ = {(x₁ , x₂, 0) : x₁, x₂ ∈ R}.
(ii) W₂ = {(x₁ , x₂, x₃) : x₁ + x₂ +x₃ = 0}.
(iii) W₃ = {(x₁ , x₂, x₃) : x₁ + x₂ +x₃ = 1}.

Then W₁, W₂ are subspaces of R³, but W₃ is not a subspace of R³.

Solution:

(i) Let x = (x₁ , x₂, 0) , y = ( y₁, y₂, 0) ∈ W₁ and α, β ∈ R, then αx + βy = (αx₁ , αx₂, 0) + (βy₁, βy₂, 0)

= (αx₁ + βy₁,  αx₂ + βy₂, 0) ∈ W₁

Hence W₁ is a subspace of R³

(ii) Let x = (x₁ , x₂, x₃), y = ( y₁, y₂, y₃) ∈ W₂

such that x₁ + x₂ + x₃ = 0 ,  y₁ + y₂ + y₃ = 0

Then αx + βy = (αx₁ + βy₁,  αx₂ + βy₂, αx₃ + βy₃)

where α, β ∈ R

Now α(x₁ + x₂ + x₃) + β( y₁ + y₂ + y₃) = 0

or (αx₁ + βy₁) +  (αx₂ + βy₂) + ( αx₃ + βy₃) = 0

Therefore αx + βy = ∈ W₂

Hence W₂ is a subspace of R³.

(iii) Let x = (x₁ , x₂, x₃), y = ( y₁, y₂, y₃) ∈ W₃

such that x₁ + x₂ + x₃ = 1 ,  y₁ + y₂ + y₃ = 1

If α, β ∈ R and  α + β ≠ 1then

αx + βy = (αx₁ + βy₁,  αx₂ + βy₂, αx₃ + βy₃)

but αx₁ + βy₁ +  αx₂ + βy₂ + αx₃ + βy₃

= α(x₁ + x₂ + x₃) + β( y₁ + y₂ + y₃) 

= α + β ≠ 1

Therefore αx + βy = ∉ W₃

Hence W₃ is not a subspace of R³.

Example 2:

In the vector space V(R) of all real valued functions, let W = {f ∈ V: f(-x) = f(x) ∀ x ∈ R}

be the set of even functions. Then W is a subspace of V.

Solution:

(af + bg)(-x) = (af)(-x) + (bg)(-x)

= af(-x) + bg(-x)

= af(x) + bg(x)  (since f,g are even functions)

=  (af)(x) + (bg)(x)

=  (af + bg)(x)

Therefore af + bg is also even functions

Thus ∀ a, b ∈ R, f, g ∈ W

⟹  (af + bg) ∈ W

Hence W is a subspace of V.

Theorem 3:

The intersection of any two subspaces of a vector space V is also a subspace of V.

Proof:

Let W₁ and W₂ be two subspaces of a vector space V, Then clearly 0 ∈ W₁, 0 ∈ W₂

so that 0 ∈ W₁ ∩ W₂

Therefore W₁ ∩ W₂ is non empty.

Now let α, β ∈ F and x, y ∈ W₁ ∩ W₂

⟹ α, β ∈ F and (x, y ∈ W₁ and x, y ∈ W₂)

⟹ αx + βy ∈ W₁ and αx + βy ∈ W₂

since W₁ and W₂ are subspace of V

⟹ αx + βy ∈ W₁ ∩ W₂

Hence W₁ ∩ W₂ is a subspace V.

Theorem 4:

If { Wᵢ}ᵢ∈ᵢ be as arbitrary family of subspace of V, then ∩ᵢ∈ I Wᵢ is a subspace of V.

Proof:

Left as an exercise.

Note: If W₁ and W₂ are two subspaces of Vector space V, then W₁ ∪ W₂ need not be a subspace of V.

For example, if R be the field of real numbers

Then W₁ = {(x₁, x₂, 0): x₁, x₂ ∈ R} and

W₂ =  {(0, x₂, x₃): x₂, x₃ ∈ R} are two subspaces of R³(R).

Now we consider the element x = (1, 2, 0) and y = (0, 3, 4) of W₁ ∪ W₂, then x + y = (1, 5, 4) ∉ W₁ ∪ W₂

Hence W₁ ∪ W₂ is not a subspace of R³(R).

Theorem 5:

Let W₁ and W₂ be two subspace of a vector space V, then W₁ ∪ W₂ is a subspace of V iff W₁ ⊆ W₂ or W₂  ⊆ W₁.

Proof:

Let W₁ and W₂ be two subspace of a vector space V.

also let W₁ ⊆ W₂ or W₂  ⊆ W₁ then either

W₁ ∪ W₂ = W₂  or W₁ ∪ W₂ = W₁

Hence W₁ ∪ W₂ is a subspace of W.

Conversely suppose W₁ and W₂ are two subspaces of V in such a way that W₁ ∪ W₂ is also a subspace.

If possible, let W₁ ⊆ W₂ and W₂  ⊆ W₁

Then ∃ x, y such that

x ∈ W₁ but x ∉ W₂ and y ∈ W₂ but y ∉ W₁ 

Now x ∈ W₁ ⟹ x ∈ W₁ ∪ W₂

y ∈ W₂ ⟹ y ∈ W₁ ∪ W₂

Since W₁ ∪ W₂ is a subspace of V and so

x + y = W₁ ∪ W₂ ⟹ x + y ∈ W₁ or x + y ∈ W₂

Now if x + y = W₁ , x ∈ W₁ ⟹ (x + y) - x ∈ W₁

⟹ y ∈ W₁

which is a contradiction

Again if x + y ∈ W₂, then

x + y ∈ W₂,  y ∈ W₂ ⟹ (x + y) - y ∈ W₂

⟹ x ∈ W₂ 

which is again contradiction

Hence either W₁ ⊆ W₂ or W₂  ⊆ W₁

Example 1:

If V₃(F) = {(a, b, c): a, b, c ∈ F} is a vector space. Then a set W₃ = {(a, b, c): α₁a + α₂b + α₃c = 0 for fixed α, α, α ∈ F} I'd a subspace of V₃(F).

Solution:

Here W₃ is clearly a subset of V₃(F).

Also for x = (a₁, b₁, c₁), y = (a₁, b₂, c₂) ∈ W₃ such that

α₁a₁ + α₂b₁ + α₃c₁ = 0 ........(1)

α₁a₂ + α₂b₂ + α₃c₂ = 0 .........(2)

and α, β ∈ F, we have

αx +  βy = α(a₁, b₁, c₁) + β(a₂, b₂, c₂ )

= (αa₁, αb₁, αc₁) + (βa₂, βb₂, βc₂ )

= (αa₁ + βa₂;  αb₁ + βb₂; αc₁ +  βc₂ )

Now α₁(αa₁ + βa₂) +  α₂(αb₁ + βb₂) + α₃(αc₁ +  βc₂ )

= α(α₁a₁, α₂b₁, α₃c₁) + β(α₁a₁, α₂b₂, α₃c₂)

= α0 +  β0  [ by 1 and 2]

= 0

Hence αx +  βy = W₃ and W₃ is subspace of V₃(F).

Example 2:

If α₁, α₂, α₃ are fixed elements of a field F, then the set W of all ordered triads (x₁, x₂, x₃) of elements of F such that 

α₁x₁, α₂x₂, α₃x₃ = 0

is a subspace of V₃(F).

Solution:

Let x= ( x₁, x₂, x₃) and y = ( y₁, y₂ ,y₃) be any two elements of W.

Then x₁, x₂, x₃, y₁, y₂, y₃ are elements of F and are such that

α₁x₁, α₂x₂, α₃x₃ = 0 ........(1)

α₁y₁, α₂y₂, α₃y₃ = 0 .........(2)

Let α, β be any two elements of F, we have

αx +  βy = α( x₁, x₂, x₃) + β( y₁, y₂ ,y₃)

= ( αx₁, αx₂, αx₃) + ( βy₁, βy₂ ,βy₃)

= ( αx₁ +βy₁, αx₂ + βy₂ , αx₃ + βy₃)

Now α₁( αx₁ +βy₁) + α₂(αx₂ + βy₂) + α₃( αx₃ + βy₃)

=  α( α₁x₁, α₂x₂, α₃x₃) + β( α₁x₁, α₂x₂, α₃x₃)

= α0 +  β0  [ by 1 and 2]

= 0

αx +  βy = ( αx₁ +βy₁ , αx₂ + βy₂, αx₃ + βy₃)

Hence W is a subspace of V₃(F).

  

Example 3:

Let V(F) be any vector space. Then V itself and the subset of V consisting of zero vector only are always subspace of V.

Solution:

Left as exercise.

These two are called improper subspace. If V has any other subspace, then it is called a proper subspace. The subspace of V consisting of zero vector only is called the zero subspace.

Example 4:

Prove that U = {(a, b, c) : a, b, c are rationals} is not a subspace of R³(R).

Solution:

U will be a subspace of R³(R) iff :

(i) x, y ∈ U ⟹ x + y ∈ U and

(ii) α ∈ R, x ∈ U ⟹  αx ∈ U

Now let √2 ∈ R and x = (a, b, c) ∈ U where a, b, c are rational numbers. Then

 αx = √2(a, b, c) = (√2a, √2b , √2c ) ∉ U.

Since √2a, √2b , √2c are not rational numbers.

Hence condition (ii) is not satisfied for U.

As such U is not subspace of R³(R).

Example 5:

If V = R³ , W ⊆ V. Show that W is not a subspace of V, if

W  = {(a, b, c) : a² + b² + c² ≤ 1}.

Solution:

Let α is a any real number such that α ≻ 1 and x ∈ W such that x = (a, b, c). Then a² + b² + c² ≤ 1

Now αx = α(a, b, c) = (αa, αb, αc)

Since  (αa)² + ( αb )² + (αc)² = a²( a² + b² + c² )  ≤ a ≻ 1

Therefore α ∈ R, x ∈ W but αx ∉ W

Hence W is not a subspace of V.