Test: Linear Algebra - 11 - Mathematics MCQ

Here we will discuss some properties of skew-symmetric matrix:
When we add two skew symmetric matrices then the resultant matrix is also skew-symmetric.
The determinant of skew symmetric matrix is non-negative.
When the identity matrix is added to the skew symmetric matrix then the resultant matrix is invertible.
The diagonal of skew symmetric matrix consists of zero elements and therefore the sum of elements in the main diagonals is equal to zero.
Scalar product of skew symmetric matrix is also a skew symmetric matrix
Given, A be a square matrix.

We are given that a linear transformation T : C³ --> C³ defined by

We need to find the adjoint T* of T.
Let B = {(1,0,0), (0,1,0), (0,0,1)} be standard basis of C³(C).
Then



Therefore, the matrix of T with respect to standard basis B is

But the matrix corresponding to the adjoint T* of T is defined by

Therefore, T* 

Let T : P₃ —> P₃ be the linear transformation defined by 

If the matrix of T relative to the standard basis B₁ = B₂ ={ 1, x, x², x³} is M and M' denote the transpose of M.
Then we need to find the matrix M + M'.
Now 


Therefore, the matrix of linear transformation T related to the basis is {1, x, x², x³} is 
 
and M '= 
therefore, M+M' = 

Analyzing the properties:

1. T is invertible:

   • A linear operator $T$ is invertible if its determinant is non-zero.
   • The determinant of $T$ is the product of its eigenvalues. Since all the eigenvalues are distinct and non-zero (implied by the fact that they are distinct), the determinant is non-zero.
   • Therefore, $T$ is invertible.

2. T is diagonalizable:

   • A linear operator is diagonalizable if there is a basis of eigenvectors for the vector space.
   • Since $T$ has n distinct eigenvalues, it has n linearly independent eigenvectors (this is a standard result for operators with distinct eigenvalues).
   • Therefore, $T$ is diagonalizable.

Conclusion:

• Since $T$ is both invertible (because it has no zero eigenvalues) and diagonalizable (because it has $n$ distinct eigenvalues), the correct answer is:

$$\boxed{\text{b) T is invertible as well as diagonalizable}}$$

Let T be a linear transformation from R³ —> R³ defined by
T(x,y,z) = ( x + y , y - z )
We need to find the matrix of T with respect to the basis
{(1, 1,1), (1, -1, 0), (0,1, 0)} and {(1,1), (1, 0)}.
Now
T{ 1, 1, 1) = (2, 0) = 0(1, 1) + 2(1, 0)
T(l, -1 ,0 ) = (0,-1) = -1(1,1) +1(1,0) 7(0, 1,0) = (1,1) =1(1, 1) + 0(1,0)
Therefore, the matrix of T with respect to the basis
{(1, 1 , 1 ) , (1, - 1 , 0), (0, 1, 0)} and {(1, 1), (1, 0)} is

Since V is a set of non-zero polynomials in x, so additive identity does not exist.

Let T be a linear transformation given by the matrix (occurring in a typical transportation problem) A linear transformation T is unimodular iff det (T) = ±1

Therefore, Rank T = 3. Also det (T) = 0. Hence, T is unimodular.

We are given that two linear transformations , g : R³ --> R³ defined by
F(x ,y ,z ) = (x ,|y|,z)
and g(x, y, z) = (x + 1, y - 1, z)
Let (0 ,1,2) and (0, - 1 , 2) be two vectors of R³ then ( 0 ,1 , 2) = (0 ,1 ,2 ) 
and (0,-1 ,2 ) = (0 ,1 ,2 )
therefore f(0,1 ,2) + f(0 ,-1, 2) = (0,1,2) + (0,1,2) = (0, 2, 4) 
but f [(0,1,2) + (0,-1,2)] = f(0 0,4) = (0 ,0 ,4)
Hence, f(0,1,2) + f(0, -1, 2) ≠ [(0 ,1, 2) + (0,- 1, 2)]
Thus , f is non linear
Next, g(0, 0, 0) = (1, -1, 0)
but(1,- 1, 0) ≠ (0, 0, 0)
hence, g is also non linear.

We are given that two linear transformations S and T :
R³ --> R³ defined by
S(x, y, z) = (2x, 4x - y, 2x + 3y - z) and T(x, y, z) = (x cos θ - y sin θ, x sin θ - y cos θ, z)
Let (x,y,z) ∈ ker.S
then S(x, y, z) =(0, 0, 0)
Using the definition o f linear transformation, we get
(2x, 4x - y, 2x + 3y - z) = (0, 0, 0) Comparing the components of the co-ordinates, we get
2x = 0
4x - y = 0
2x + 3y - z = 0
On solving these equation for x, y, z, we get
x = 0, y = 0 and z = 0
Hence,ker S = {0, 0, 0}
Thus, S is one-one.
Now, since A is a one-one linear transformation from R³ to R³, which is a finite dimensional vector space therefore S is onto.
Next Let (x, y, z) ∈ ker T
then T(x, y, z) = (0, 0,0)
Using the definition of linear transformation, we get (x cosθ - y sinθ, x sinθ + y cosθ, z) = (0, 0, 0)
Comparing the components of the co-ordinates, we get
x cosθ - y sinθ = 0
x sinθ + y cosθ = 0
z = 0
Now, solving for x, y and z, we get
x = 0 , y = 0 and z = 0
hence, ker T = {0 , 0, 0 }
Therefore, T is one-one
Now, since T is a one-one linear transformation from R³ to R³, which is a finite dimensional vector space therefore T is onto.

We are given that a linear transformation  T : F² ---> F³ defined by T(x₁, x₂) = (x₁, x₁ + x₂, x₂)
We need to find the Nullity of T
let (x₁, x₂) ∈ ker T
Then, T(x₁, x₂) = (0, 0, 0)
Now, using the definition o f Linear transformation we get,
(x₁,x₁ + x₂, x₂) = (0, 0, 0)
Comparing the components of the coordinates, we get
x₁ = 0
x₁ + x₂ = 0
x₂ = 0
Solving for x₁, x₂, we get
x₁ = 0, x₂ = 0
Thus, ker T = {(0,0)}
Hence,Nullity T = dim (ker T) = 0