Vector SubSpaces: Theorems Part 2 - Example Solved Problems

Vector SubSpaces: Theorems Part 2

Example Problems

 

Example 7

Show that Fⁿ is the direct sum of the subspaces

W₁ = { (a₁, a₂ ..... ‚ a_n) ∈ Fⁿ, a_n = 0 } and

W₂ = {(a₁, a₂,..., a_n) = Fⁿ, a₁ = a₂ = ... = a_n−1=0}.

Solution:

Let us consider p, q ∈ W₁.

 p = (a₁, a₂,... a_n);         a_n=0

q= (b₁, b₂, ... b_n);          b_n = 0

 αp+βq = α(a₁, a₂, ... 0) + β(b₁, b₂, ..... 0)

= (α a₁ + βb₁, αa₂ + βb₂ ,... 0) ∈ W₁.

Hence W₁ is subspace of V=Fⁿ.

Let us consider r, s ∈ W₂.

where r = (0, 0, ... a_n)

and

 s = (0, 0, ..... b_n)

 ar+βs = (α0+β0, α0+β0, ... , αa_n+ βb_n)

= (0,0,..., αa_n + Bb_n)

 W₂ is a subspace of V=Fⁿ

Clearly W₁ ∩ W₂ = {0}

W₁+ W₂ = (a₁, a₂,... a_{n− 1},0)+(0, 0, ..., 0, a_n)

= (a₁, a₂, ... a_n) ∈ Fⁿ

 Fⁿ = W₁  W₂

 (ie). Fⁿ is the direct sum of the subspaces W₁ and W₂.

 

Example 8

Let V be the vector space of all functions from the real field R into R. Show that W is not a subspace of V where

(i) W={f; f(7) = 2+ƒ (1) }

(ii) W consists of all non−negative functions, (ie) All functions "f" for which f(x) ≥0 for all x ∈ R.

Solution:

(i) Suppose f, g ∈ W

 (ie) ƒ(7)=2+ƒ(1) and g (7) = 2 + g (1)

(f+g) (7)=ƒ(7)+g(7) = 2+f(1)+2+g(1)

= 4+f(1)+g(1) = 4+ (f+g) (1)

 ≠ 2+(f+g) (1)

Hence f+g ∉ W and so W is not a subspace of V.

(ii) Let k = −2 and f ∈ V be defined by f(x) = x² then f ∈ W.

Since f(x) = x²≥0 for all x ∈ R.

 kƒ (5) = (− 2) (5²) = −50 <0 (k=−2)

Hence kf ∉ W and so W is not a subspace of W.

 

Example 9

In any vector space V, show that (a+b) (x+y)=ax+ay + bx+by for any x,y ∈ V and any a, b ∈ F.

Solution:

Noting that (a+b) ∈ F, we have

(∴ α (u+v) = αu+αv)

 (a+b)(x+y)=(a+b)x+(a+b)y

= ax + bx + ay + by = ax + ay + bx + by

as required.

 

Example 10

How many matrices are there in the vector space M_m×n (z₂)?

Solution:

There are 2^mn vectors in this vector space.

 

Example 11

Prove that (aA +bB)^t= aA^t +bB^t for any A, B ∈ M_mxn (F) for any a, b ∈ F.

Solution:

We have (aA)_ij = a A_ij, (bB)_ij = b B_ij

So (aA+bB)_ij = a•A_ij + b•B_ij

[(aA +bB)^t]_ij = a•A_ji + b•B_ji

Now

 (A^t)_ij = A_ji  and (B^t)_ij = B_ji

(aA^t)_ij = a • A_ji and (bB^t)_ij=b•B_ji

  [aA^t + bB^t]_ij = a•A_ji + b.B_ji

  (aA + bB)^t = aA^t + bB^t as required.

 

Example 12:

Prove that (A^t)^t = A for each A ∈ M_m×n (F).

Solution:

We have (A^t)_ij=A_ji

Thus [(A^t)^t]_ij = (A^t)_ji = A_ij

So that (A^t)^t = A as required.

 

Example 13:

Prove that A+ A^t is symmetric for any square matrix A.

Solution:

We know that

 A+A^t = A^t+A

= A^t+(A^t)^t

= (A + A^t)^t

 

Example 14:

Prove that tr (aA +bB) = a tr (A) + b tr (B)

Solution:

tr (aA + bB) =  (aA + bB)_ii

= aA_ii + bB_ii

= aA_ii + bB_ii

= a tr (A) + b tr (B)

 

Example 15:

What are the possible subspace of R²?

Solution:

(i) {0} is the subspace of R².

(ii) R² is the subspapce of R² and

(iii) Lines through the origin are subspace of R².

 

Example 16:

Prove that if U is a subspace of a vector space V and u₁, u₂, u₃, …. u_n ∈ U then a₁u₁ + а₂u₂ +  а₃u₃ + ... + а_nu_n ∈ U for any scalars (a₁, a₂, a₃ ... а_n) ∈ F.

Solution:

Given that U is a subspace of a vector space V and u₁, u_2, u₃ ….. u_n ∈ U then we have to prove that а₁u₁ + a₂u₂ + a₃u₃ + …..  … + a_nu_n ∈ U where a₁, a₂, a₃ ... a_n ∈ F.

Let us prove this by using mathematical induction.

Let S (n) = a₁u₁ + а₂u₂ + a₃u₃ + ... + а_nu_n

                   ...(1)

To prove S(1) is true, put n = 1 in (1)

 S (1) = a₁u₁ ∈ U

(a₁ ∈ F, u₁ ∈ U ⇒ α₁ u₁ ∈ U)

Now assume that S (k) is true.

а₁u₁ + a₂u₂ + a₃u₃ + …..  … + a_ku_k ∈ U

To prove S(k+1) is true

а₁u₁ + a₂u₂ + a₃u₃ + …..  … + a_ku_k + a_k+1u_k+1 ∈ U

  S(k) + a_k+1 u_k+1 ∈ U.

  S(k+1) is true whenever S (k) is true.

   а₁u₁ + a₂u₂ + a₃u₃ + …..  … + a_nu_n ∈ U

  U is a subspace of a vector space V.

 

Example 17

Prove that the intersection of the subspaces W₁ & W₂ of a vector space V is a subspace.

Solution:

Let us consider x, y ∈ W₁ ∩ W₂ and a ∈ F.

since x ∈ W₁ ∩ W₂, there must be x ∈ W₁ and x ∈ W₂.

Similarly for y ∈ W₁ ∩ W₂ there must be y ∈ W₁ and y ∈ W₂.

Since x, y ∈ W₁ ∩ W₂ is a subspace of V, x+y ∈ W₁ & x+y= W₂.

..x + y ∈ W₁ ∩ W₂.

Here W₁ ∩ W₂ is a subspace of V.

 

Example 18

Let V=R³. Prove that W={(a₁, a₂, a₃) ∈ R³/a₁ = 3a₂ & a₃=−a₂} is a subspace over R³.

Solution:

Let W = { (a₁, a₂, a₃)/a₁ =3a₂ and a₃ = −ɑ₂ }

Let w₁ = (a₁, a₂, a₃} and w₂ = {b₁, b₂, b₃], w₁, w₂ ∈ W.

 αw₁+βw₂ = α (a₁, a₂, a₃) + β (b₁, b₂, b₃)

= [(α a₁, α α₂, α α₃) + (βb₁, βb₂, βb₃)]

= (αa₁ + βb₁, αa₂ + βb₂, αa₃ + βb₃)

= 3 (α a₂+ βb₂) + (αa₂ + βb₂) − (αa₂ + βb₂)

= 3αa₂+3βb₂+αa₂+βb₂−αa₂−βb₂

= 3αa₂+3βb₂

= 3 (αa₂+ βb₂)

= 3a₂

⇒ αw₁ + βw₂ ∈ W

.. W is a subspace over R³.

 

Example 19

Let V=R³ prove that W={(a₁, a₂, a₃) ∈ R³/a₁+2a₂−3a₃=1} is not a subspace of V.

Solution:

W = {(a₁, a₂, a₃) ∈ R³/a₁ +2a₂−3a₃ = 1}

Let w₁ = (a₁, a₂, a₃) & w₂ = (b₁, b₂, b₃), where w₁, w₂ ∈ W.

αw₁+βw₂ =α (a₁, a₂, a₃) + β(b₁, b₂, b₃)

= (α a₁, α α₂, α α₃) + (βb₁, Bb₂, βb₃)

= (α a₁ + βb₁, αα₂ + βb₂, αa₃ + βb₃)

= (αa₁ + βb₁) + 2(α a₂+ βb₂) ‒ 3(αa₃ + βb₃).

= αa₁ + βb₁ +2α a₂ + 2 βb₂ − 3αa₃ − 3βb₃

= α (a₁+2a₂−3a₃) + β(b₁+2b₂−3b₃)

= α+B   (because  a₁+2a₂ − 3a₃ = 1 & b₁+2b₂−3b₃ = 1)

 αw₁+βw₂ = α+β ∉ W

  W is not a subspace of V.

 

Example 20

Let V=R³. Prove that W={(a, b, c)/a²+b²+c²=1} is not a subspace of V.

Solution:

Given that W={(a, b, c) / a² + b² + c² = 1}.

Let w₁ = (a₁, a₂, a₃) & w₂ = (b₁, b₂, b₃), where w₁, w₂ ∈ W.

 αw₁+ βw₂ = α(a₁, a₂, a₃) + β(b₁, b₂, b₃)

= (αа₁, αα₂, αα₃) + (βb₁, βb₂, βb₃)

= (αa₁ + βb₁, αa₂+ βb₂, αa₃ + βb₃)

= (αa₁ + βb₁)² + (αа₂ + βb₂)² + (αa₃ + βb₃)²

= α²a₁² + β²b₁² + 2αβα₁b₁ + α²a₂² + β²b₂² + 2αβa₂b₂ + α²a₃² + β²b₃² + 2αβa₃b₃

= α² (a₁² + a₂² + a₃²) + β²(b₁² + b₂² + b₃²) + 2αβ(a₁b₁ + a₂b₂+ a₃b₃)

= α² + β²+2αβ(a₁b₁+a₂b₂+a₃b₃)

  αw₁ + βw₂ ∉ W.

Hence W is not a subspace of V.

 

Example 21

Prove that a subset W of a vector space V is a sub−space of V if and only if 0 ∈ W and ax+y ∈ W where a ∈ F and x, y ∈ W.

Solution:

It is given that V is a vector space and W is a subset of V.

Necessary part: If W is a subspace of V then

 (i) 0 ∈ W and (ii) ax+y ∈ W, where a ∈ F and x, y ∈ W.

W is a subspace of V ⇒ (i) 0 ∈ W.

(ii) x+y ∈ W where x, y ∈ W.

(iii) ax ∈ W where a ∈ F and x ∈ W.

These conditions hold good.

Converse part: If (i) 0 ∈ W

 (ii) ax + y ∈ W where a ∈ F and x, y ∈ W,

then W is a subspace of V.

(i) x+y ∈ W where x, y ∈ W.

(ii) ax ∈ W where a ∈ F and x ∈ W.

(iii) 0 ∈ W.

(iv) Each element in W has an additive inverse in W.

(i) 1 ∈ F, if x, y ∈ W then 1.x+y = x+y ∈ W.

(ii) a ∈ F, x ∈ W, 0 ∈ W then ax+0 = ax ∈ W

(iii) 0 ∈ W by (i)

(iv) Each vector in W has an additive inverse in W.

If x ∈ W, then (−1) x ∈ W ⇒  − x = (− 1) x.

Here ‒1 ∈ F and x ∈ W. Hence the additive inverse of each element of W is also in W.

Hence W is a subspace of V.

 

Example 22

Let V be a vector space of all 2×2 matrices over R. Verify whether W is a subspace of V or not for the following

(i) W consists of all matrices with non−zero determinant.

(ii) W consists of all matrices A such at A²=A.

Solution:

Hence A+ A ∉ W.

  W is not a subspace of V because W is not closed under addition.

 

Example 23

Let V be the vector space of all square n×n matrices over reals. Verify whether the following are sub−spaces of V or not.

(i) Collection of all symmetric matrices.

(ii) Collection of all skew−symmetric matrices.

(iii) Collection of all scalar matrices.

(iv) Collection of all singular matrices.

Solution:

It is given that V = { A : A = [a_ij]_{n ×n} : α_ij ∈ R }

(i) Let W be the collection of all symmetric matrices.

Then it is obvious that W  V.

Let x, y ∈ W where x = [b_ij] for b_ij = b_ij and

 y= [c_ij] for c_ij=c_ji

Now αx + βy = α[b_ij] + β[c_ij] where α, β ∈ R

= [αb_ij + βc_ij]

= [d_ij]

    where αb_ij + βc_ij = d_ij.

Now d_ji = αb_ji + βc_ji

= αb_ij + βc_ij

= d_ij

Hence d_ij=d_ji

Here [d_ij] is a symmetric matrix and d_ij ∈ W.

Hence W is a subspace of V.

(ii) Let W = Collection of all skew − symmetric matrices

W = { [a_ij]_mxn/ a_ij = ‒aj for all a_ij ∈ R)

Here clearly it says that W  V.

Let αx+βy = α [a_ij] + β[b_ij] for α, β ∈ R.

= [α a_ij + β b_ij]

= c_ij where c_ij= αa_ij + βb_ij

Now c_ij = αa_ji + βb_ji

= −αaj−βb_ij

= − [c_ij]

 [c_ij] is also skew symmetric.

Hence W is a subspace of V.

(iii) Let W= Collection of all scalar matrices.

W = { [α_ij]_n×n/a_ij= k for i = j, k ∈ R }

= 0 for i≠j. Then WV.

  W is a subspace of V.

 (iv) Let W = Collection of all singular matrices.

Here W is not a subspace of V because W is not closed under the operation addition.

W is not a subspace of V.

 

Example 24

Prove that the union of two subspaces of a vector space is a subspace.

Solution:

Let A= {(a, 0, 0)/a ∈ R} and B={(0, b, 0)/b ∈ R}

Then A + B = (a, 0, 0) + (0, b, 0) = (a, b, 0) ∈ W.

Hence W is a subspace of R³.

 

Example 25

Prove that the set W = {(a₁, a₂, a₃) ∈ R³, 5a₁²−3a₂²+6a₃²=0} is not a subspace of V.

Solution:

Let W= {(a₁, a₂, a₃) ∈ R³, 5a₁²−3a₂²+6a₃² = 0}

Let W₁ = (a₁, a₂, a₃) and W₂ = (b₁, b₂, b₃) where w₁, w₂ ∈ W

 αw₁+βw₂ = α (a₁, a₂, a₃) + β (b₁, b₂, b₃)

= (α α₁, α α₂, α α₃) + (βb₁, βb₂, βb₃)

= (α a₁+βb₁, α a₂+ βb₂, αa₃ + βb₃)

= 5 (α a₁ + βb₁)² − 3 (α a₂ + βb₂)² + 6 (α a₃ + β b₃)²

= 5 [a² a₁² + β²b₁² + 2α βa₁ b₁] − 3 [α² a₂²+β²b₂² + 2αβa₂b₂] + 6 [a² a₃² +β² b₃² +2αβa₃b₃]

= 5α²a₁² +5β²b₁²+10αβa₁b₁ − 3a²a₂² − 3β² b₂² − 6αβa₂b₂ + 6α²a₃²+6β²b₃² + 12αβa₃b₃.

= α² [5a₁² − 3a₂² + 6a₃²] + β²[5b₁² – 3b₂² + 6b₃²] + αβ[10a₁b₁ − 6a₂b₂ + 12a₃b₃]

= α²[0] + β²[0] + 2αβ[5a₁b₁− 3a₂b₂ + 6a₃b₃]

 ≠ 0

 αx + βy ∉ W.

Hence W is not a subspace of V

 

Example 26

Let W₁ = {(a₁, a₂, α₃) ∈ R³, a₁ = 3a₂ and a₃ =  ‒a₂} and W₂ = {(a₁, a₂, α₃) = R³, 2a₁−7a₂+a₃=0}. Prove that W₁∩W₂ is a subspace of R³.

Solution:

It is given that 2a₁−7a₂+ a₃ = 0.

 By using W₁ in W₂, we have

2(3a₂) − 7(a₂) + (− a₂) = 0

 6a₂−7a₂−a₂=0

 −2a₂=0

 a₂=0

From W₁:

Since a₂ =0, a₁ = 3 (0) = 0 and a₃ = − (0) = 0

  a₁ =0, a₂ = 0 and a₃ = 0

W₁ ∩ W₂ = {0}

The null set is a subspace of itself and W₁ ∩ W₂ = 0.

The set W₁ ∩ W₂ is a subspace of R³.

 

Example 27

Let S be any non−empty set and F be a field. Prove that for any s₀ ∈ S {ƒ ∈ F (S, F) : ƒ (s₀) = 0 } is a subspace of F (S,F).

Solution:

It is given that S be any non−empty set.

 (f+g) (s₀) =ƒ(s₀) + g (s₀) = 0+0 = 0.

Thus under addition it is satisfied.

 cf(s₀) = c(0) = 0. It is satisfied under multiplication.

Then zero function is in the set.

 It is a subspace of F (S, F).

 

Example 28

Let S be a non−empty set and F be a field. Let C (S, F) be the set of all functions f ∈ F (S, F) such that ƒ (s) = 0 for all but a finite number of elements. Then prove that C (S, F) is a subspace of F (S, F).

Solution:

From the given statement, the number of non−zero points of (f+g) is lesser than the number of non−zero points of ƒ and g.

Then it is closed under addition operation.

Similarly, the number of non−zero points of Cf(s) is equal to the number of f.

Thus it is closed under multiplication.

Here zero function is in the set.

It is a subspace of F (S, F).

 

Example 29

Is the set of all differentiable real valued functions defined on R a subspace of C (R).

Solution:

The sum of two differentiable functions is again differentiable.

The product of one scalar and one differentiable function is again differentiable and zero function is also differentiable.

 The set of all differentiable real valued functions defined on R is again a subspace of C (R).

 

Example 30

Let Cⁿ (R) be the set of all real valued functions defined on R has a continuous n^th derivative. Then prove that Cⁿ(R) is a subspace of F (R, R).

Solution:

Let f ⁿ and gⁿ are the n^th derivative of f and g.

 f ⁿ + gⁿ is the n^th derivative of ƒ+ g.

If f ⁿ and gⁿ are continuous then f ⁿ+gⁿ is also continuous.

Cf ⁿ is the n derivative of C and it is continuous.

Here zero function is the null vector.

  Cⁿ (R) is a subspace of F (R, R).

 

Example 31

Let F₁ and F₂ be two fields. Prove that the set of all even functions in F (F₁, F₂) and the set of all odd functions in F(F₁, F₂) are the subspace of F (F₁, F₂).

Solution:

Let V be the vector space of all functions.

Let Ve and Vo be the subsets of even and odd functions.

Now we have to prove that Ve and Vo are subspace of V.

For even functions

Let f_e and g_e ∈V_e and c is any scalar.

(cf_e + g_e)(−x) = cf_e(−x) + g_e(− x)

= cf_e (x) + g_e(x)         (since it is an even function)

= (cfe + g_e) (x)

Thus cf_e + g_e is an even function.

Hence f_e , g_e ∈ V_e implies cf_e + g_e ∈ V_e

Hence V_e is a subspace of V.

For odd functions

Let f₀ and g₀ ∈ V₀ and c is any scalar.

(c f₀ + g₀) (−x) = cf₀ (− x) + g₀ (−x)

= c[−f₀ (x)] − g₀(x)             (.it is an odd function)

= −[c f₀ (x) + g₀ (x)]

= −[c f₀ + g₀)] (x).

Thus c f₀ + g₀ is an odd function.

Hence f₀, g₀ ∈ V₀ which implies that c f₀ + g₀ ∈ V₀.

Hence V₀ is a subspace of V.

 

Example 32

Is the set W= {f(x) ∈ P(F) ; f (x) = 0 or f (x) has degree n } a subspace of P (F) if n≥1.

Solution:

The given set is not a subspace for n ≥1.

When n=2; the given set W is not closed under addition.

 

Example 33

Let W₁ and W₂ be sub spaces of V. (i) Show that W₁+ W₂ is a subspace of V that contains W₁ and W₂. (ii) Show that any subspace of V that contains W₁ and W₂ and W₁+ W₂.

Solution:

(i) We know that by the definition of subspace

 0 ∈ W₁ and 0 ∈ W₂ ⇒ 0+0 ∈ W₁ + W₂.

 W₁+ W₂ is a non−empty subset of V.

Let x, y ∈ W₁+ W₂ and a, b are any scalars.

 x=x₁+x² and y=y₁+x² where x₁, y₁ ∈ W₁ and x², x² ∈ W₂.

Since W₁ is a subspace, x₁,y₁ ∈ W₁ and a, b are scalars.

Then ax₁+by₁ ∈ W₁.

Since W₂ is a subspace, x², x² ∈ W₂ and a, b are scalars then ax² + bx² ∈ W₂.

 (ax₁+by₁)+(ax₂+ bx₂) ∈ W₁+ W₂

 (ax₁ + ax₂) + (by₁ + bx₂) ∈ W₁ + W₂

 a (x₁+x₂) + b (y₁+ y₂) ∈ W₁+ W₂

⇒ a(x) +b(y) ∈  W₁+ W₂

  W₁+ W₂ is a subspace of V.

Here w₁ ∈ W₁, means that w₁ = w₁ + 0

w₁ = w₁+0 ∈ W₁+ W₂ as 0 ∈ W₂.

That means W₁  W₁+ W₂

Now w₂ ∈ W₂ means that w₂=0+ w₂

 w₂ =0+w₂ ∈ W₁ + W₂ as 0 ∈ W₁

That means W₂  W₁+ W₂

From (1) and (2) W₁  W₁ + W₂ and W₂  W₁ + W₂

(ii) Let us consider U V be the subspace of V.

(ie) W₁U and W₂  U

Now we have to prove that W₁ + W₂  U.

 ax + y ∈ W₁+ W₂ (ax ∈ W₁ and y∈ W₂)

  ax ∈ U and y ∈ U

Since U is a subspace, ax+ye ∈ U

⇒ W₁+ W₂  U.

 

Example 34

Let W₁ and W₂ be two subspaces of V. Prove that V is the direct sum of W₁ and W₂ iff each vector in V can be written as x₁+x² where x₁ ∈ W₁ and x² ∈ W₂.

Solution:

Let W₁ and W₂ be two subspaces of V.

 (i) V=W₁ W₂.

 (ii) For Z ∈ V, z=x₁+x² when x₁ ∈ W₁ and x² ∈ W₂.

Let us consider V=W₁  W₂  ⇒ x₁ + x² = z;

x₁ ∈ W₁ and x² ∈ W₂

Suppose that the above is not unique. Then let

 Z=x₁'+x²'; x₁' ∈ W₁ and x²' ∈ W₂

  x₁ + x₂ = x₁′ + x₂′ ⇒ x₁−x₁'=x₂−x₂

As W₁ and W₂ are subspaces x−x₁' ∈ W₁ and x₂−x₂' ∈ W₂.

 ⇒ x₁x₁' = x₂−x₂' ∈ W₁ ∩ W₂

Hence W₁∩W₂={0} x₁=x₁' and x₂=x₂'.

Now let us consider z=x₁+x₂ for x₁ ∈ W₁ and x₂ ∈ W₂.

To prove W₁ ∩ W₂ = {0}

Suppose W₁ ∩ W₂ has a non−zero vector x, then

x=x+0:  x ∈ W₁ and 0 ∈ W₂

y=0+y;  0 ∈ W₁ and y ∈ W₂

This is a contradiction to our hypothesis.

  W₁ ∩ W₂ = {0}

 

Example 35

A matrix M is called skew−symmetric if M^t = −M. A skew symmetric matrix is square. Let F be a field. Prove that the set W₁ of all skew symmetric n×n matrices with entries from F is a subspace of M_nxn (F). Now assume that F is not characteristic 2 (two) and let W₂ be the subspace of M_nxn (F), consisting of all symmetric n×n matrices. Prove that M_n×n (F) = W₁  W_2.

Proof:

We know that (M₁ + M₂)^t = M₁^t + M₂^t  = − (M₁ + M₂)

and (CM)^t = CM^t = – CM.

Here zero matrix is skew symmetric.

The set of all skew−symmetric matrices is a space.

Define M_n×n (F) = {A:A ∈ M_n×n(F)}

= { (A + A′) + (A − A'); A ∈ M_n×n(F)) }

= W₁+W₂

and also W₁ ∩ W₂ = {0}.

Then A+A^t is symmetric and A−A^t is skew symmetric.

It is given that F is of characteristic of 2.

Then W₁ = W₂

M_n×n (F) =W₁  W₂