Subspaces (Vector Spaces): Example Solved Problems

Subspaces (Vector Spaces)

Example Problems

WORKED EXAMPLES

Example 1

Prove that a subset W of a vector space V is a subspace of V if and only if  αw₁+βw₂ ∈ W for all w₁, w₂ ∈ W and α, β ∈ F.

Solution:

Necessary part:

Assume that W is a subspace of V.

To prove that αw₁+βw₂ ∈ W for all w₁, w₂ ∈ W; α, β ∈ F

Let w₁, w₂ ∈ W; α, β ∈ F

αw₁ ∈ W and βw₂ ∈ W        (.W is a subspace).

⇒ αw₁ + βw₂ ∈ W.

Sufficient part:

Assume: αw₁+ βw₂ ∈ W for all w₁, w₂ ∈ W.

To prove: W is a subspace of V.

By Theorem (5) it is enough to prove that

(a) 0 ∈ W

(b) x+y ∈ W whenever x ∈ W, y ∈ W.

(c) cx ∈ W, x ∈ W, c ∈ F.

(a) Let w₁ = w₂ = w ∈ W

Take α=−1 and β=1

−w+w=0 ∈ W.

(b) Let w₁, w₂ ∈ W then by assumption

αw₁ + βw₂ = W

Take α= 1, β=1 ∈ F

W₁ + W₂ ∈ W

(c) Let w₁ = w, w₂ = 0 ∈ W

  α ∈ F

⇒ αw₁+ β(0) ∈ W

  Aw₁ ∈ W.

Example 2

Determine whether the following subsets of the given vector spaces are subspaces or not.

Solution:

 (a) Let V=R² and S be the set of all points in x and y axis (only) in

R²

(ie)

S= {(x, 0), (0, y)}

Consider u = (2, 0) ∈ S and v = (0, 3) ∈ S.

 u+v = (2, 3) ∉ S.

It is not closed under the operation addition (+).

Hence S is not a subspace of R².

 (b) Let V=M_2×3 (R) and S be the set of all 2×3 matrices with the 2^nd row all zero.

Thus S is closed under addition and scalar multiplication and so it is a subspace of M_2x3 (R).

 (c) Let V=R³ and S be the set of all points in xy−plane in R³.

 (ie). S={(x, y, 0) | x and y are real numbers }

 Consider u= (x₁,y₁, 0) and v = (x₂, x₂, 0) ∈ S

 u + v = (x₁ + x_2, y₁ + x², 0) ∈ S

Let r ∈ R (scalar)

 ru = (rx, ry, 0) ∈ S

 S is closed under + and scalar multiplication.

  It is a subspace of R³.

(d) Let c (R) denote the set of all continuous real−valued functions defined on R and S be the set of all functions f in [0, 3] with f(2) = 0.

Consider f, g→S and α∈R then ƒ (2) = 0, g (2)=0.

(f+g) (2) = ƒ(2) + g(2) = 0+0 = 0∈ S.

(α ƒ) (2) = αƒ (2) = α (0) = 0 ∈ S.

Thus S is closed under + and scalar multiplication.

  S is a subspace of c (R).

(e) Let the vector space be c (R), S be the set of all functions "f ' defined

by f(2) = 3 then

 (f+g) (2) = ƒ(2) + g(2) = 3 + 3 = 6 ≠ 3 ∉ S.

  S is not a subspace of c (R).

 (f) Let W be the set of vectors in R³ of the form form (a, b, 1/2a−2b)

Consider (a, b, a/2 – 2b ) + ( c, d, c/2 − 2d)

= ( a + c, b + d, 1/2 (a + c) − 2(b + d) )

k (a, b, a/2 – 2b ) = ( ka, kb, k( a/2 − 2b))

W is a subspace of R³.

Example 3

Let Fⁿ = {(a₁, a₂,... a_n) / α_i ∈ F} be a vector space. Let W = { (a₁, a₂, 0, 0 ..... 0)/a₁, a₂ ∈ F } is a subset of Fⁿ. prove that W is a subspace of Fⁿ.

Solution:

Let w₁ = (a₁, a₂, 0, 0, ..... 0)

 w₂ = (b₁, b₂, 0, 0, ... 0), w₁, w₂ ∈ W.

Let α, β ∈ F

 αw₁+ βw₂ = α (a₁, a₂, 0, 0, ... 0) + β(b₁, b₂, 0, 0, ... 0)

= (αa₁, αa₂, 0, 0, ...) + (βb₁, βb₂, 0, 0, 0)

= (αa₁ + βb₁, αa₂ + βb₂, 0, 0, ... 0) ∈ W.

 W is a subspace of Fⁿ.

Example 4

 Let V=R³; W={(a₁, a₂, a₃)/2α₁−7a₂+a₃=0} verify whether it is a subspace or not.

Solution:

Let w₁ = (a₁, a₂, α₃); w₂ = (b₁, b₂, b₃); w₁, w₂ ∈ W.

αw₁+βw₂ = [(αa₁, αa₂, αa₃) + (βb₁, βb₂, βb₃)]

= (αa₁ + βb₁, αа₂ + βb₂, αa₃ + βb₃)

= 2 (αa₁ + βb₁)−7 (αa₂ + βb₂) + (αa₃ + βb₃)

= 2αa₁ + 2βb₁−7αa₂−7βb₂+αa₃ + βb₃

= α(2α₁−7α₂+ a₃) + β(2b₁ −7b₂+b₃)

= α(0) + β(0)

= 0 ∈ W

 W is a subspace of V.

Example 5

Let W = {(a₁, a₂, a₃)/a₁=a₃+2} prove that it is not a subspace of V.

Solution:

Let W= {(a₁, a₂, a₃)/a₁ =a₃ +2}.

Let w₁ = {(a₁, a₂, a₃)} and w₂ = { b₁, b₂, b₃ }

α w₁ + βw₂ = (α a₁ +βb₁, α a₂+ βb₂, α a₃ + βb₃)

 = (α a₁ + βb₁) − αа₃−βb₃−2      (a₁−a₃−2)

= α (a₁−a₃) + β(b₁−b₃) − 2

= α (a₁‒a₁+2)+ β(b₁−b₁+2)−2

= 2α+2β−2

 αw₁ + βw₂ ∉ W.

  W is not a subspace.