Test: Linear Algebra - 8 - Mathematics MCQ

We are given that a linear transformation T : R⁷ —> R⁷ defined by

Let
be an ordered basis of R⁷
then
So, the matrix corresponding to linear transformation T with respect to basis B is

Implies T_B = I_7x7
Now = 1
Thus, determinant of T is 1.
Now,
T(x₁, x₂, . . . , x₆, x₇) = (x₇, x₆, . . . , x₂, x₁)
Therefore, ,T(x₁, x₂, . . . , x₆, x₇)= T(T(x₁, x₂, . . . , x₆, x₇)) = T(x₇, x₆, . . . , x₂, x₁)
= (x₁, x₂, . . . , x₆, x₇)
Implies that T² = 1
Hence, the smallest n such that Tⁿ = I is 2 which is an even number.
Also ,T³(x₁, x₂, . . . , x₆, x₇)
     = T²(T(x₁, x₂, . . . , x₆, x₇)
     = (x₇, x₆, . . . , x₂, x₁)
     = T(x₁, x₂, . . . , x₆, x₇)
Implies that T³ = T
Implies T⁶ = (T³)² = T² = I Therefore,T⁷(x₁, x₂, . . . , x₆, x₇)
= T⁶(x₇, x₆, . . . , x₂, x₁) = T(x₁, x₂, . . . , x₆, x₇)
Implies T⁷ = T ≠ I

We are given here that T is a linear transformation from R² ---> R² such that T(y) = αc and T(x) = 0, where x and y are linearly independent in R².
Therefore,
T(T(y)) = T(αx) = αT(x) = α . 0 = 0
Implies T²(y) = 0
Hence,T is a Nilpotent linear transformation of index 2.
Thus T= 
 

We are given that the characteristic polynomial o f 3 x 3 matrix, A is
λ³ + 3λ² + 4λ - 3 = 0
and x = trace (A)
       y = det(A)
The general characteristic polynomial of 3 x 3 matrix is given by,
λ³ - trace (A) λ² + λ + |A| = 0
Comparing this equation by the given equation, we get
trace = -3 
or x = -3
and |A| = -3
or y = -3
Hence, x = y = -3
 

We need to find the characteristic vector of the matrix corresponding to the characteristic root 1.
Put λ = 1 in the equation (A - λl)X = 0

or x₁ + 3x₂ = 0
Let x₂ = k
then x₁ = - 3k
The eigen vector corresponding the eigen value λ = 1 is,
=  or 

We are given that the system of equation
x + 2y - z =11
3 x + y - 2z = 10
x - 3 y = 5
The given system of equation can be written in the augmented matrix as,

Applying the operations "R² ---> R² - 3R₁" and R₃ —> R₃ - R₁ these operations yield

and also "R₃ ---> R₃ - R₂ ", which gives,

Here, rank of (A) ≠ rank of aug (A).
Hence, the given system of equation has no solution.

We need to find the minimal polynomial of the 3 x 3 real matrix
A = 
First we need to find the characteristic polynomial of the given matrix, that is
ch_A(x) = |xI - A| = (x - a) (x - a) (x - b)
Further, the minimal polynomial of the given matrix 
= l.c.m. of (x - a), (x - a), (x - b)
or m_A (x) =(x - a) (x - b)

We need to find the number of onto linear transformation from R³ to R⁴.
Since dim R⁴ = 4 > 3 = dim R³. Therefore, there is no onto linear transformation from R³ to R⁴.
Suppose T : U ---> V be an onto linear transformation.
Then dim V ≤ dim U

Let (1,2) and (3,4) be two vectors of R².
Then T[(1, 2) + (3,4)] = T(4, 6)]
= (4 x 6, 4) = (24, 6)
and T(l, 2) = (1 x 2,1) = (2, 1)
T(3, 4) = (3 x 4,.3) = (12, 3) Thus,T(l, 2) +  T(3, 4) = (2, 1) + (12, 3) = (14, 4)
Therefore, T[(1, 2) + (3 ,4)] ≠ T(1 ,2) + T(3,4)
Hence, T is non linear.
To show that a given map T is a linear transformation we must show that
 T(αx + βy) = αT(x) + βT(y)
where x,y are vectors and a, P are scalars while to show that the given map is non-linear conveniently we try to find a related contrast example.

Since T₁(V)  we also have T₂(T₁(V)) T₂(U) and
so, dim (T₂(T₁(V))) - dim (T₂(U)) then Rank T₂oT₂ = dim ((T₂oT₁) (v)) = dim [T₂ (T₁(V))]
≤ dim T₂(U)
= Rank T₂
 

Let V be the vector space of polynomial functions of degree three or less. Let the ordered basis for V consisting of the functions
x^j : j = 0 ,1 ,2 , 3
and let D be the differentiation operator. We need to find the matrix of differentiation operator D in the basis x^j, j = 0,1, 2, 3, that is {1, x, x², x³}.
Therefore, D(1) = d/dx(1) = 0 
= 0.1 + 0.x + 0.x² + 0.x³
D(x) = d/dx (x) = 1 = 1.1 + 0 . x + 0 .x² + 0.x³ ax
D(x²) =d/dx(x²) = 2x = 0.1 + 2.x + 0.x² + 0.x³ ax
D(x³) = d/dx (x³) = 3x² = 0.1 + 0.x + 3.x² + 0.x⁴
Thus, the matrix of differentiation operator is

We are given that a linear transformation T : R² -> R² defined by T(x, y) = (x - y, 2x + y)
We need to find T^-1
Now, let (x, y) € ker T
then T(x, y) = (0, 0)
Using the definition of linear transformation, we get
( x - y , 2x+y) = (0, 0)
Comparing the components of the co-ordinates, we get
x - y = 0, 2x+ y = 0
Implies, x = 0 and y = 0
Therefore, ker T = {(0,0)}
Hence, T is one-one.
Since T is a linear transformation from R² to R² but R² is a two dimensional (finite dimensional) vector space.
Hence, T is onto and therefore, T^-1 must exist.
Let (x,y) be the image of (a, b) under T^-1. Therefore, 
T^-1(a, b) = (x,y)
Implies, T(x, y) = (a, b)
Using the definition of linear transformation, we get
(x - y, 2x + y) = (a, b)
Comparing the components of the co-ordinates, we get
x - y = a and 2x + y = b
Solving for x and y we get,
 and 
Hence, T^-1(a, b) = 
That is T^-1(x, y) = 

We are given that the matrix
λ
We need to find the eigen values of A. The characteristic equations o f the given matrix is, | A - λI | = 0
or 

Since, λ = 2 satisfies this equation. Therefore, λ = 2 is one of the root of the given matrix,
(λ - 2) (λ² + 2λ + 1) = 0 
or λ = 2, (λ + 1)² = 0
λ = - 1 , - 1

We are given that a matrix A =  has real eigen values. The characteristic equation of the given matrix is,
|A - λI| = 0
(cos θ - λ)² + sin²θ =0
or cos²θ + λ² - 2λcosθ + sin²θ =0
or λ² - 2λcosθ + 1 = 0
or λ = cos θ ± i sin θ
Since, the given matrix has real eigen values, then
sinθ = 0
or θ = nπ for some integer n.

We are given that the n x n matrix A = [a_ij] such that a_ij = 3and _j.
Then , we need to find the nullity o f the given matrix.
Let the matrix,

Reduce the matrix to echelon form using the operations "R₂ --> R₂ - R₁ ", "R₃ ---> R₃ - R₁ " .......
"R_n --> R_n - R₁".
These operations yield--

Here, rank (A) =1. By Sylwester’s law,
dim (A) =rank (A) + nullity (A)
or n = 1 + nullity (A)
or nullity(A) = n - 1

We are given that the system of simultaneous linear equations,
x + y + z = 0
x - y - z = 0
The coefficient matrix is given by,

rank of coefficient matrix = 2. Here, the rank of matrix < no. of unknowns therefore, the system of equation has infinitely many solution in R³. 

We need to find the minimal polynomial m(x) of A_{n x n}, each of whose element is 1.
Let
 
or
 
or A² - nA = 0
Hence m_A(x) = x² - nx

We are given that a linear transformation T :R³ --> R³ defined by
T(x₁, x₂, x₃) = (x₁ + x₂, x₂ + x₃, x₃ + x₄)
We need  to find T^-1
Let (x, y, z) ∈ ker T
then T(x, y, z) =(0, 0, 0)
Using the definition of linear transformation, we get 
(x₁ + x₂, x₂ + x₃, x₃ + x₄) = (0, 0, 0)
Comparing the components of the coordinates, we get
x₁ + x₂ = 0
x₂ + x₃ = 0
x₃ + x₁ = 0
Implies, x₁ = 0, x₂ = 0 and x₃ = 0
Therefore, ker T = {(0,0,0)} , hence, T is one-one.
Since T is a linear transformation from R³ to R³ but R³ is a three dimensional (finite dimensional) vector space.
Hence, T is onto and therefore, T^-1 must exist.
Let (x, y, z) be the image of (a,b,c) under T^-1,
Therefore T^-1(a ,b ,c ) = (x,y,z) Implies, T{x, y, z) = (a, b , c ) Using the definition of linear transformation, we get
(x + y , y + z ,z + x) = (a , b, c) Comparing the components of the co-ordinates, we get
x + y = a
y + z = b
z + x - c
Solving for x, y and z, we get 


and 
Hence,T^-1(a, b, c)

That is 
T^-1(x,y,z) = Hence,T^-1(x₁, x₂, x₃)

 

Let T be a linear operator on R³ defined by
T(1,0, 0) = (1, 2, 1)
T(0, 1, 0) = (3, 1, 5)
T(0,0, 1) = ( 3,- 4 , 7)
We need to find the image of (x, y, z) under T.
Let there exist scalars α, β and γ such that
(x,y, z) = α(l, 0, 0) + β(0, 1, 0) + γ(0, 0, 1)
or equivalently (x, y, z) =(α, β, y) Implies a = x, p = y andy = z Therefore,(x, y, z) = x ( l , 0, 0) + γ( 0, 1, 0) + z (0, 0, 1)
Taking the image under linear transformation T, we get
T(x, y, z) =xT’( l , 0, 0) + y T ( 0, 1 , 0 ) + zT(0, 0, 1) = x ( l , 2 , 1 ) + y ( 3 , 1, 5) + z(3, - 4 , 7)
=(x + 3y + 3z, 2x + y — 4z, x + 5y + Iz)
Now let (x, y, z) ∈ ker T
Then T(x, y, z) = (0, 0, 0)
Using the definition of linear transformation, we get
(x + 3y + 3z, 2x + y - 4z, x + 5y + 7z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
x + 3y + 3z = 0
2x + y - 4z = 0
x + 5y + 7z = 0
Now solving for x, y, z, we get 
Therefore, ker T = 
Hence,dim (ker T) = 1
Therefore, T is not one-one and hence T is not invertible.

Let F be a field and T be a linear operator on F² defined by
T(x₁, x₂) = (x₁+x₂, x₁)
We need to find T^-1.
Let (x₁,x₂) ∈ ker T.
Then T(x₁, x₂) = (0,0)
Using definition of T, we get
(x₁ + x₂, x₁) = (0, 0)
Implies, x₁ + x₂ = 0 and x₁,= 0.
Solving for x₁ and x₂,
we get x₁ = 0 and x₂ = 0.
Hence, ker T = { 0 } and hence, T is one-one.
Since F² is a two dimensional vector space and T is a linear operator on F².
Hence, T is onto.
Therefore, T^-1 exist.
Let (a, b) be the image of (x₁, x₂) under T^-1.
Then T^-1(x₁,x₂) = (a, b)
Implies, T(a, b) = (x₁, x₂)
Using the definition of linear transformation T, we get
(a + b, a) = (x₁, x₂)
Comparing the component of the coordinates, we get
a + b = x₁, a = x₂
Solving for a and b, we get a = x₂,
b = x₁ - x₂
Therefore,T^-1(x₁,  x₂) = (x₂, x₁ - x₂)

Let T be a linear transformation defined by T(x, y, z) = (2x, 2x -5y,2y + z)
We need to find T^-1
Let (x, y, z) ∈ ker T,
Then T(x,y,z) = (0,0,0)
Using definition of T, we get
(2x, 2x - 5y, 2y + z) - (0, 0, 0) Implies, 2x = 0, 2x- 5y = 0 and 2y + z = 0
Solving for x, y and z, we get
x = 0, y = 0 and z =0
Hence, ker T = {(0, 0, 0)}
and thus T is one-one.
Since R³ is a three dimensional (finite dimensional) vector space and T is a linear transformation from R³ to R³.
Hence, T is onto.
Therefore, T^-1 exist.
Let (a, b, c) be the image of (x, y, z) under T^-1 then T^-1(x, y, z) = (a, b, c) Implies, T(a, b, c) = (x,y, z)
Using the definition of linear transformation T,
we get
(2a, 2a - 5b, 2b + c) = (x, y, z) Comparing the components of the co-ordinates,
we get 2a= x ,2a-5b = y and 2b + c = z
Solving for a, b and c we get,

Therefore