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Let {v_{1}, v_{2}, ........,v_{16}} be an ordered basis for V= C^{16}. If T is a linear transformation on V defined by T(v) = v_{i+1 }for 1 ≤ i ≤ 15 and T(v_{16}) = (v_{1} + v_{2} + ... + v_{16}) Then,
Let B = {v_{1}, v_{2}, ........,v_{16}} be an ordered basis for V= C^{16}.
If T is a linear transformation on V defined by T(v) = v_{i+1} for 1 ≤ i ≤ 15 and T(v_{16}) = (v_{1 + }v_{2} + .......+ v_{16}).
Thus, the matrix of T relative to the ordered basis B = {v_{1}, v_{2}, ........,v_{16}} is
Then Rank [T_{B}]= 16
Hence, T is invertible and it has rational eigen values.
Let T : R^{3} > R^{3} be a linear transformation defined by T(x, y, z) =(x + y  z, x + y + z, y  z)
Then, the matrix of the linear transformation T with respect to the ordered basis B = {(0,1,0), (0,0,1), (1,0, 0) of R^{3} is
Let T : R^{3} > R^{3} be a linear transformation defined by
T(x,y,z) = (x + y  z , x + y + z, y  z)
We need to find the matrix of T with respect to the ordered basis B = {(0, 1,0), (0,0, 1), (1, 0,0)} of R^{3}.
Now
T(0,1,0) = (1,1,1)
= 1(0, 1, 0) + 1(0, 0, 1) + 1(0, 0, 0)
T(0, 0, 1) = (1, 1, 1)
= 1(0, 1, 0 )  1(0, 0 , 1)  1( 1, 0 , 0)
T(1, 0, 0) =(1,1,0)
= 1(0,1,0) + 0(0,0,1) + 1(1,0,0)
Therefore, the matrix of T with respect to the ordered basis
B = {(0, 1, 0), (0, 0, 1), (1, 0, 0)} is
Let T : R^{4} > R^{4} be the linear map satisfying
T{e_{1}) =e_{2}, T(e_{2}) = e_{3}, T(e_{3}) = 0, T(e_{4}) = e_{3},
where {e_{1} e_{2}, e_{3} e_{4}} is the standard basis of R^{4}. Then,
We are given that a linear map
T : R^{4} > R^{4} satisfying
T(e_{1}) = e_{2}, T(e_{2}) = e_{3}, T(e_{3})  0, T(e_{4}) = e_{3}
Where {e_{1} e_{2}, e_{3}, e_{4}} is the standard basis of R^{4}.
Now, let (x,y,z,t) t ∈ R^{4}.
Then (x,y,z,t) = [xe_{1} + ye_{2} + ze_{3} + te_{4}]
Taking the image under linear transformation T, we get
T(x, y, z, t) = T[xe_{1} + ye_{2} + ze_{3} + te_{4}] Using the linearity of T, we get
T(x, y, z, t) = xT(e_{1}) + yT(e_{2}) + zT(e_{3}) + tT(e_{4})
Substituting the value of T(e_{1}), T(e_{2}), T(e_{3}) and T(e_{4}),
we get
T(x, y, z, t) = xe_{2} + ye_{3} + z • 0 + t • e_{3}
or equivalently
T(x, y, z,t) = (0 ,x ,y + t,0)
Let (x, y, z, t) ∈ ker T.
Then T(x, y, z, t) = (0, 0, 0, 0)
Using the definition of linear transformation T, we get
(0, x , y + t, 0) = (0, 0, 0, 0)
Comparing the components, we get
x =0, y + t = 0, z is arbitrary.
Therefore, ker T = {(0, y, z, y ) : yz ∈ R}.
Hence,dim ker T = 2.
Using rank nullity theorem
Rank T = 4  dim ker T
= 4  2 = 2.
Now T^{2}(x,y,z,t) = T[T(x, y, z, t)]
= T(0, x , y + t, 0) = (0, 0, x, 0)
and T^{3}(x, y, z, t) = T[T^{2}(x, y, z, t)]
= T(0,0,x,0)
= (0,0, 0 ,0 )
Hence, T is nilpotent linear transformation of index 3.
Consider the basis {u_{1} u_{2}, u_{3}} of R^{3}, where u_{1} = (1, 0, 0), u_{2} = (1,1,0) , u_{3 }= (1,1,1) . Let {f_{1}, f_{2}, f_{3}} be the dual basis of {u_{1}, u_{2}, u_{3}} and f be a linear functional defined by f(a,b,c) = a + b + c, (a, b, c) ∈ R^{3}. If f = a_{1}f_{1} + a_{2}f_{2} + a_{3}f_{3} then (α_{1}, α_{2}, α_{3}) is
Let {u_{1}, u_{2}, u_{3}) = {(1, 0, 0), (1, 1, 0), (1,1, 1)}
be the basis o f R^{3} and {f_{1}, f_{ 2}, f_{3}} be the dual basis of {u_{1}, u_{2}, u_{3}} a n d f is a linear functional defined by f(a,b,c) = a + b + c for all (a, b, c) ∈ R^{3}.
Also given that
Since {f_{1} + f_{2} + f_{3}} is dual basis of { u_{1 }+ u_{2} + u_{3}}.
Therefore,
f_{1}(u_{1}) = 1, f_{1}(u_{2}) = 0, f_{1}(u_{3}) = 0
f_{2}(u_{1}) = 0 , f_{3}(u_{2}) = l , f_{2}(u_{3}) = 0
f_{3}(u_{1}) = 0 , f_{3}(u_{2}) = 0 , f_{3}(u_{3}) = 1
Let(a,b,c) = xu_{1} +yu_{2} + zu_{3}
Substituting th e values of u_{1}, u_{2} and u_{3,} we get
(a, b, c) = x( 1 ,0 , 0) + y (1,1, 0 ) + z (l, 1 ,1 ) = ( x + y + z , y + z ,z)
Comparing the components of the coordinates, we get
a = x + y + z, b = y + z, c = z
Solving for x, y, z, we get
x = a  b , y = b  c , z = c
= z = c.
Hence, {a  b , b  c , c} is the dual basis o f {u_{1,}u_{2}, u_{3}).
Now since we are given that
we get,
or equivalently or equivalently
C om paring th e coefficients o f a, b, c, we get
Solving for α_{1}, α_{2}, α_{3} ,we get
α_{1}=1 α_{2 }=2_{ }α_{3 }=3_{ }
For a matrix [M] = , the transpose of the matrix is equal to the inverse of the matrix [M]' = M 1 The value of x is
Correct Answer : b
Explanation : Given a matrix is equal to its conjugate transpose hence it is hermitian. also, it is skew also because the transpose of a given matrix is equal to negative of the given matrix.
Hence it is a skewHermitian matrix
If A is a nonzero column vector (n x 1), then the rank of matrix ,AA^{'} is
If P and Q are nonsingular matrices, then for matrix M, which of the following is correct?
If the rank of an n x n matrix A is (n  1), then the system of equations Ax= b has
Let A be a matrix of order m x n and R is nonsingular matrix of order n, then
Let A be a square matrix of order n, then nullity of A is
A is a 3 x 4 real matrix and AX = b is an inconsistent system of equations. The highest possible rank of A is
The system of linear equations 4x + 2y =7, 2x + y = 6 has
If x + 2y  2u = 0, 2x  y  u = 0, x + 2z  u = 0, 4x y + 3 z  u= 0 is a system of equations, then it is
Let T : P_{3}[0 ,1] > P_{2}[0 , 1] be defined by (T_{p}) (x) = P"(x) + P'(x). Then the matrix representation of T with respect to the basis {1, x, x^{2}, x^{3}} and {1, x, x^{2}} of P_{3}[0, 1] and P_{2}[0, 1], respectively is
Let T : P_{3}[0, 1] > P_{2}[0,1] be a linear transformation defined by (T_{p}) (x) = p"(x) + p'(x ). We need to find the matrix of T with respect to the basis { l , x , x^{2}, x^{3}} and {1, x, x^{2}} of P_{3}[0, 1] and P_{2}[0, 1] respectively.
Now,
and
Therefore, the matrix of T with respect to the basis {1, x, x^{2}, x^{3}} and { 1,x,x^{2}} of P_{3}[0, 1] and P_{2}[0, 1] respectively is
Let T : R^{4 }—> R^{4} be defined by
T(x, y , z, w) = (x + y + 5w, x + 2 y + w,  y  z + 2w, 5x + y+ 2z).
Then dimension of the eigen space of T is
We are given th at a linear transform ation T : R^{4} —> R^{4} defined by
T(x, y, z, w) = x + y + 5w, x + 2y + w, w  z + 2w, 5x + y + 2z)
Let B = {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0 , 1 , 0), (0, 0, 0 ,1 ) } be standard basis of R^{4}. Then
T(l, 0 ,0 ,0 ) = (1,1, 0,5) = 1( 1, 0, 0, 0) + 1(0, 1, 0, 0) + 0(0, 0, 1, 0) + 5(0, 0, 0, 1)
T(0, 1 ,0 ,0 ) = ( 1 ,2 ,0 , 1)
= 1( 1, 0, 0, 0) + 2(0 , 1, 0, 0) + 0(0,
0, 1, 0) + 1(0, 0, 0, 1)
T(0, 0, 1, 0) = (0, 0,  1 , 2 )
= 0 (1 ,0 , 0, 0) + 0 ( 0 , 1 ,0 , 0 )  l (0, 0, 1, 0) + 2(0, 0, 0, 1)
T(0, 0, 1) = (5, 1, 2, 0) = 5(1, 0,0 ,0 )+ 1 (0 , 1,0,0)+ 2 (0, 0, 1, 0) + 0(0,0,0,1)
Therefore, the matrix of linear transformation T with respect to the basis B is which is a
symmetric matrix and hence diagonalisable.
Hence, dimension of eigen space is 4
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