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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?
Consdierthe subspace W = {(x_{1}, x_{2},.... ,x_{10}) x_{n} = x_{n1} + x_{n  2} for 3 < n < 10) of the vector space the dimension of W is
Which one of the following is a subspace of the vector space
If A and B are 3 × 3 real matrices such that rank (AB) = 1, then rank (BA) cannot be
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 = BA = 0
⇒ BA is singular
Hence rank (BA) cannot be 3. (Because BA is 3 × 3 matrix)
Which of the following sets of functions from is a vector space over
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
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
Here,
So, Answer is skewHermitian matrix.
Let V be a 3 dimensional vector space over the field of 3 elements. The number of distinct 1 dimensional subspaces of V is
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
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
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.
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
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
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?
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
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
and let V = {(x, y, z) : A = 0}.Then the dimension of V equals
A basis of V = {(x, y, z, w) : x + y  z = 0, y + z + w = 0, 2x + y  3z = 0)
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,zw,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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