Test: Linear Algebra - 10 - Mathematics MCQ

Let B = {v₁, v₂, ........,v₁₆} be an ordered basis for V= C¹⁶.
If T is a linear transformation on V defined by T(v) = v_i+1 for 1 ≤ i ≤ 15 and T(v₁₆) = -(v_{1 +} v₂ + .......+ v₁₆).
Thus, the matrix of T relative to the ordered basis B = {v₁, v₂, ........,v₁₆} is

Then Rank [T_B]= 16
Hence, T is invertible and it has rational eigen values.

Let T : R³ ---> R³ 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 R³.
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

We are given that a linear map
T : R⁴ --> R⁴ satisfying
T(e₁) = e₂, T(e₂) = e₃, T(e₃) - 0, T(e₄) = e₃
Where {e₁ e₂, e₃, e₄} is the standard basis of R⁴.
Now, let (x,y,z,t) t ∈ R⁴.
Then (x,y,z,t) = [xe₁ + ye₂ + ze₃ + te₄]
Taking the image under linear transformation T, we get
T(x, y, z, t) = T[xe₁ + ye₂ + ze₃ + te₄] Using the linearity of T, we get
T(x, y, z, t) = xT(e₁) + yT(e₂) + zT(e₃) + tT(e₄)
Substituting the value of T(e₁), T(e₂), T(e₃) and T(e₄),
we get
T(x, y, z, t) = xe₂ + ye₃ + z • 0 + t • e₃
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 T²(x,y,z,t) = T[T(x, y, z, t)]
= T(0, x , y + t, 0) = (0, 0, x, 0)
and T³(x, y, z, t) = T[T²(x, y, z, t)]
= T(0,0,x,0)
= (0,0, 0 ,0 )
Hence, T is nilpotent linear transformation of index 3. 

Let {u₁, u₂, u₃) = {(1, 0, 0), (1, 1, 0), (1,1, 1)}
be the basis o f R³ and {f₁, f ₂, f₃} be the dual basis of {u₁, u₂, u₃} a n d f is a linear functional defined by f(a,b,c) = a + b + c for all (a, b, c) ∈ R³.
Also given that

Since {f₁ + f₂ + f₃} is dual basis of { u₁ + u₂ + u₃}.
Therefore,
f₁(u₁) = 1, f₁(u₂) = 0, f₁(u₃) = 0 
f₂(u₁) = 0 , f₃(u₂) = l , f₂(u₃) = 0
f₃(u₁) = 0 , f₃(u₂) = 0 , f₃(u₃) = 1
Let(a,b,c) = xu₁ +yu₂ + zu₃
Substituting th e values of u₁, u₂ 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,u₂, u₃).
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 α₁, α₂, α₃ ,we get
α₁=1 α₂ =2     α₃ =3   

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 skew-Hermitian matrix

Let T : P₃[0, 1] --> P₂[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 , x², x³} and {1, x, x²} of P₃[0, 1] and P₂[0, 1] respectively.
Now,

and 
Therefore, the matrix of T with respect to the basis {1, x, x², x³} and { 1,x,x²} of P₃[0, 1] and P₂[0, 1] respectively is

We are given th at a linear transform ation T : R⁴ —> R⁴ 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 R⁴. 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