Vector Spaces: 2 Marks Important Questions with Answer

Linear Algebra:

UNIT I: VECTOR SPACES

TWO MARKS QUESTIONS AND ANSWERS

1. Prove that the identity vector is unique in vector space

The vector 0 is unique. (ie) The identity is unique.

Proof:

Let x ∈ V and e, e' ∈ V

 x+e = V and x+e' = V

x+e = x+e'

 e = e'           (By cancellation law)

2. Prove that the inverse vector is unique in vector space.

The vector y is unique (ie) The inverse is unique

Proof:

Let x ∈ V and y, y' ∈ V

 x+y = 0 and x+y' = 0

x+y = x+y'

y=y'                (By cancellation law)

3. In a vector space V, show that

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

Solution:

We know that for each element a in F and each pair of elements x, y in V, then

 a (x + y) = ax + ay and also for each pair of elements (a, b) in F and each element x in V then (a+b) x = ax + bx.

Based on these two conditions, (a+b) (x + y) can be written as

 (a+b) (x + y) = a (x + y) + b (x+y) (or) (a+b)x + (a + b)y

 = ax + ay + bx+ by (or) ax + bx + ay + by

4. Define a vector space V over a field F.

A vector space (or linear space) V over a field "F" consists of a set on which two operations (called addition & scalar multiplication) are defined so that for each pair of elements x, y ∈ V there is a unique element x+ye ∈ V and for each element "a" in F and each element x in V there is a unique element ax in V, such that the following conditions hold.

1. For all x, y ∈ V, x+y=y+x (Commutative law)

2. For all x, y, z ∈ V, (x + y) + z = x + (y + z). (Associative law)

3. There exists an element in V denoted by 0 such that x+0= x_for each x in V.

4. For each element x in V there exists an element y in V such that x+y=0.

5. For each element x in V, 1.x = x.

6. For each pair of elements a, b ∈ F and each element x in V (a b) x = a (bx).

7. For each element "a" in F and each pairs of elements x, y in V, a (x + y) = ax + ay.

8. For each pair of elements a, b in F and each element x in V, (a + b) x = ax + bx.

5. Define a subspace of a vector space over F.

A subset W of a vector space V over a field F is called a subspace of V if W is a vector space over F with the operations of addition and scalar multiplication defined on V.

In any vector space V, {0} and V are subspaces.

6. Let Fⁿ = {(a₁, a₂, а₃ ... а_n)/a_i ∈ F} be a vector space. Let W = { (a₁, a₂, 0, 0, ..... 0)/a₁, a₂ ∈ F} is a subset of Fⁿ. Then 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)

= (α a₁, α a₂, 0, 0, ...) + (β b₁, βb₂, 0, 0, 0)

= (α a₁ + βb₁, α a₂ + β b₂, 0, 0, ... 0) ∈ W.

 W is a subspace of Fⁿ.

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

Solution:

There are 2^mn vectors in this vector space.

8. 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.

9. 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

10. What are the possible subspaces 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.

11. Prove that the union of two subspaces of a vector space is not 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.

12. Let V=R³; S={(1, 2, 0), (0, −5,−7)} and V= (2,−5,7) ∈ V. Verify V is a linear combination of S or not.

Solution:

Let us consider v = αv₁ + βv₂

 (2,−5,7) = α(1, 2, 0) + β(0, −5,−7) = (α, 2α, 0) + (0, −5β, −7 β)

 (2,−5,7) = (α, 2α−5β, −7B)

Comparing on both sides we get α = 2, 2α−5β = −5 and −7β=7

From the first and last terms, we have

α=2 and β=− 1.

Substitute these two values in the middle term, we get

2(2)−5(−1)=−5

⇒ 9 ≠ −5

v is not a linear combination of S.

13. Prove that a subset W of a vector space V is a subspace of V if and only if L[W] = W.

Solution:

Given that W is a subspace of V.

To prove L (W) = W, let x ∈ L(W), such that x_i ∈ W, α_i ∈ F.

 x=Σα_ix_i for  i = 1, 2, 3 ... n

 x=Σα_ix_i ∈ W.

Since W is a subspace of V, it is closed under addition and multiplication.

 L(W)  W and W  L (W)

  L (W) = W.

Hence the proof.

14. Verify whether the vectors u = (1, 2, 3); v = (0, 1, 2) and W = (0, 0, 1) generates R³ or not.

Solution:

 Let (x, y, z) ∈ R³, and a, b, c ∈ F.

 (x, y, z) = au + by + cw

 = a(1, 2, 3) + b(0, 1, 2) + c(0, 0, 1)

 = (a, 2a, 3a) + (0, b, 2b) + (0, 0, c)

(x, y, z) = (a, 2a + b, 3a+2b+c)

Comparing the corresponding terms on both sides, we get

a = x;

2a + b = y;

b=y−2x

3a+2b+c=z

c=z−2(y−2x)−3x.

c=z−2y+x

Hence any vector in R³ can be generated by u, v, w.

15. Show that the matrices  generate M_2×2(R).

Solution:

By comparing on both sides a = a₁₁, b=a₁₂, c=a₂₁ and d=a₂₂.

  The given matrices generate M_2×2(R).

16. Show that P_n(F) is generated by {1, x, x², x³ ... xⁿ }

Solution:

Let S = {1, x, x², x³, x², ... x_n }.

Let α=a₀+a₁x+a₂x² + a₃t³…  +a_nxⁿ be any arbitrary number of P_n where a₀, a₁, a₂ …a_n ∈ F.

Then α is the linear combination of polynomials 1, x, x², x³... xⁿ over the field F.

S generates P_n(F).

17. Prove that the set

S = { (1, 0, 0, − 1), (0, 1, 0, − 1), (0, 0, 1, −1), (0, 0, 0, 1)} is linearly independent.

Solution:

Now we must show that the only linear combinations of vectors in S that equals the zero vector is the one in which all the coefficients are zero.

Suppose that a₁, a₂, a₃ and a₄ are scalars such that

 a₁(1, 0, 0, −1)+ a₂(0, 1, 0, −1) + a₃(0, 0, 1, −1) + a₄(0, 0, 0, 1) = (0, 0, 0, 0)

(a₁, 0, 0, −a₁) + (0, a₂, 0, − a₂) + (0, 0, a₃, −a₃) + (0, 0, 0, a₄) = (0, 0, 0, 0)

⇒ a₁=0; a₂=0; a₃=0

 −a₁−a₂−a₃ + a₄ =0

⇒ а₄=0.

Clearly the only solution to this system is

 a₁ =0, a₂ = 0, a₃=0 and a₄ = 0 and so S is linearly independent.

18. Let V=R⁴. Prove that v₁ = (1, 0, 1, 0); v₂ = (0, 1, 0, 1); v₃=(0,0,0,2) are linearly independent?

Solution:

We know that av₁+bv₂+cv₃=0

a(1, 0, 1, 0) + b(0, 1, 0, 1) + c(0, 0, 0, 2) = 0

(a, 0, a, 0) + (0, b, 0, b) + (0, 0, 0, 2c) = (0, 0, 0, 0)

(a, b, a, b + 2c) = (0, 0, 0, 0)

a=0; b=0; b+2c=0.

⇒ c = 0

 a=0=b=c

The vectors v₁, v₂, v₃ are linearly independent.

19. Define minimal generating set.

Minimal generating set: Let V be a vector space over F and S contained in V then S is called minimal generating set for V if

(i) L(S) = V.

(ii) No proper subset of S will generate.

20. Define finite and infinite dimensional vector spaces over F.

Finite Dimensional: A vector space is called finite dimensional if it has a basis consisting of a finite number of vectors. The unique number of vectors in each basis for V is called the dimension of V and is denoted by dim (V).

A vector space that is not finite−dimensional is called infinite−dimensional.

Examples:

1. The vector space {0} has dimension zero.

2. The vector space Fⁿ has dimension n.

3. The vector space M_m×n (F) has dimension mn.

4. The vector space P_n(F) has dimension n + 1.

5. Over the field of complex numbers, the vector space of complex numbers has dimension 1. (A basis is { 1 })

6. Over the field of real numbers, the vector space of complex numbers has dimension 2. {A basis is { 1, i } }

21. Verify the following set forms a basis or not for P₂(R) given that { −1 − x + 2x², 2 + x−2x², 1−2x+4x² }

Solution:

To verify these vector are linearly independent or not let us choose a, b, c as scalars such that

 a(−1−x+2x²) + b(2+x−2x²) + c(1−2x+4x²) = 0

(−a−ax +2ax²) + (2b+ bx − 2bx²) + (c−2cx+4cx²)=0

Then −a+2b+c=0         ...(1)

−a+b−2c=0         ...(2)

2a−2b+4c=0         ...(3)

= −1[4 −4] − 2[− 4 + 4] + 1[2 − 2] = 0

 The given set of vectors are linearly dependent.

Hence these set of vectors does not form a basis for P₂(R).

22. Find the dimension of W, where W = { (a₁, a₂, a₃, a₄, α₅) ∈ F⁵/a₁+a₃+a₅ = 0, a₂ = a₄ }.

Solution:

 a₁, a₂, a₃ are choosen by us to be they are independent.

  (a₁, a₂, a₃, а₂, − а₁ − а₃) = α₁ (1, 0, 0, 0, − 1) + a₂ (0, 1, 0, 1, 0) + a₃(0, 0, 1, 0, −1)

{(1, 0, 0, 0, −1), (0, 1, 0, 1, 0), (0, 0, 1, 0, −1)) is a basis of W.

  dim (W) = 3.

23. Find the dimension of W for W = { (x₁, x₂, x₃)/x₁+x₂+x₃=0}

Solution:

  (x₁, x₂, x₁−x₂) = x₁(1, 0, − 1) + x₂(0, 1, − 1)

  B = {(1, 0, −1), (0, 1, −1)} is a basis of W

 dim (W) = 2.

24. Find the dimensions of W, where

W= {(a₁, a₂, a₃, a₄, а₅) ∈ F⁵/α₁−α₃−α₄=0}

Solution:

 (a₁, a₂, a₃, а₁ ‒ a₃, a₅) = a₁(1, 0, 0, 1, 0) + a₂(0, 1, 0, 0, 0) + a₃(0, 0, 1, − 1, 0) + a₅(0, 0, 0, 0, 1).

 B = { (1, 0, 0, 1, 0), (0, 1, 0, 0, 0), (0, 0, 1, − 1, 0), (0, 0, 0, 0, 1)} is a basis of W

 . dim (W) = 4.

25. Find the dimensions of W for

W = { (a₁, a₂, a₃, a₄, α₅) ∈ F⁵/a₂ = a₃ = a₄ and a₁+a₅ = 0 }

Solution:

 (a₁, a₂, a₂, a₂,−a₁) = a₁ (1, 0, 0, 0, −1) + a₂ (0, 1, 1, 1, 0)

  B = {(1, 0, 0, 0, ‒1), (0, 1, 1, 1, 0)}

  dim (W) = 2.