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QUESTION: 1

If {v_{1}, v_{2}, v_{3}} is a linearly independent set of vectors in a vector space over which one of the following sets is also linearly independent?

Solution:

QUESTION: 2

Consdierthe subspace W = {(x_{1}, x_{2},.... ,x_{10}) x_{n} = x_{n-1} + x_{n - 2} for 3 __<__ n __<__ 10) of the vector space the dimension of W is

Solution:

QUESTION: 3

Which one of the following is a subspace of the vector space

Solution:

QUESTION: 4

If A and B are 3 × 3 real matrices such that rank (AB) = 1, then rank (BA) cannot be

Solution:

Here A & B a re 3 × 2 real matrices such that rank (AB) = 1

So, |AB| = 0 ⇒ |A| |B| = 0 (∴ |AB| = |A| |B|)

⇒ either |A| or |B| should be zero

So, |BA| = |B||A| = 0

⇒ BA is singular

Hence rank (BA) cannot be 3. (Because BA is 3 × 3 matrix)

QUESTION: 5

Which of the following sets of functions from is a vector space over

Solution:

QUESTION: 6

Let T : R^{3} → R^{3} be the linear transformation define by T(x, y, z) = (x + y, + z, z + x) for all (x, y, z) ∈ 3. Then

Solution:

Let T : R^{2} → R^{3} be the L.T. defined as

T (x, y, z) = (x + y, y + z, z + x) ∀ (x, y, z) ∈ R^{3}

N (T ) = {( x, y, z)|T ( x, y, z) = 0}

= {(x, y, z)|x + y = 0,y + z = 0, z + x = 0}

= {(0, 0, 0)| x, y, z ∈R}

⇒ dim N (T) = 0

Rank (N) + Nullity T = dim R^{3 }= 3

Rank (T) + 0 = 3 ⇒ Rank of T = 3

QUESTION: 7

If the matrixthe A

is

Solution:

Here,

So, Answer is skew-Hermitian matrix.

QUESTION: 8

Let V be a 3 dimensional vector space over the field of 3 elements. The number of distinct 1 dimensional subspaces of V is

Solution:

QUESTION: 9

Let M = {(a_{1},a_{2},a_{3}): a_{i} ∈ (1,2,3,4); a_{1} + a_{2} + a_{3} = 6} then the number of elements in M is

Solution:

QUESTION: 10

The dimension of the vector space of all symmetric matrices A = (a_{ij}) of order n x n (n __>__ 2) with real entries, a_{ii} = 0 and trace zero is

Solution:

The dimension of the space of *n*×*n* symmetric matrices with diagonal equal to zero is . Now, in your case the diagonal is not zero but the sum of its elements is zero, that means that you have *n*−1 elements which can vary. SO you get as expected.

QUESTION: 11

Let n be a positive integer and let H_{n} be the space of all n x n matrices A = (a_{ij}) with entries in satisfying a_{ij} = a_{rs} whenever i + j = r + s (i, j, r, s = 1, 2,..., n) then the dimention of H_{n}, as a vector space over

Solution:

QUESTION: 12

The dimension of the vector space of all symmetric matrices of order n x n (n __>__ 2) with real entries and trace equal to zero is

Solution:

QUESTION: 13

Let {X,Y,Z) be a basis of Consider the following statements P and Q

P : {X + Y,Y + Z, X - Z) is a basis of

Q : {X + Y + Z,X + 2Y - Z, X - 3Z} is a basis of

Which of the above statements hold true?

Solution:

QUESTION: 14

Let V denote the vector space C^{5}[a, b] over R and

Solution:

QUESTION: 15

Let M be the space of all 4 x 3 matrices with entries in the finite field of three elements. Then the number of matrices of Rank three In M is

Solution:

QUESTION: 16

Consider the subspace W = {[a_{ij}] : a_{ij} = 0, if i is even) of all 10 x 10 matrices(real) then the dimension of W is

Solution:

QUESTION: 17

The dimension of the vector space over the field

Solution:

QUESTION: 18

Then the dim V is

Solution:

QUESTION: 19

and let V = {(x, y, z) : |A| = 0}.Then the dimension of V equals

Solution:

QUESTION: 20

A basis of V = {(x, y, z, w) : x + y - z = 0, y + z + w = 0, 2x + y - 3z = 0)

Solution:

The system of 3 equations in 4 unknowns is

1x + 1y +(-1)z +0w =0……..(1)

0x + 1y + 1z + 1w = 0……….(2)

2x + 1y +(-3)z +1w = 0……..(3). The row operation (3) - 2×(1) reduces (3) to

0x +(-1)y+(-1)z+(-1)w =0…….(4), which is just (-1)× (3). Hence the given system is equivalent to the pair of equations:

x = -y+z and y= -z -w. This pair is equivalent to x=(z+w)+z = 2z+w and y = -z -w, where z and w are completely arbitrary. Thus

[x,y,z,w] =[2z+w,-z-w,z,w] =

z[2,-1,1,0] + w[1,-1,0,1].

We had already observed that the rank of the echelon matrix equivalent to the given coefficient matrix is 2, and hence its nullity = 4–2 = 2, which is the same as the dimension of the solution space V. Thus a basis of V may be taken as

B={[2,-1,1,0], [1,-1,0,1]}.

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