Mathematics Exam > Mathematics Questions > Let T : R4 --> R4 be the linear map satisf...
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Let T : R4 --> R4 be the linear map satisfying
T{e1) =e2, T(e2) = e3, T(e3) = 0, T(e4) = e3,
where {e1 e2, e3 e4} is the standard basis of R4. Then,
T{e1) =e2, T(e2) = e3, T(e3) = 0, T(e4) = e3,
where {e1 e2, e3 e4} is the standard basis of R4. Then,
- a)T is idempotent*
- b)T is invertible
- c)Rank T= 3
- d)T is nilpotent
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Let T : R4 --> R4 be the linear map satisfyingT{e1) =e2, T(e2) = e3...
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.
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.
Most Upvoted Answer
Let T : R4 --> R4 be the linear map satisfyingT{e1) =e2, T(e2) = e3...
R3 be a linear transformation defined by the equation T(x) = Ax, where A is a 3x4 matrix.
The linear transformation T maps a vector x in R4 to a vector T(x) in R3. The transformation is defined by the matrix A, where each column of A represents the image of one of the standard basis vectors in R4.
Since A is a 3x4 matrix, it means that T maps a 4-dimensional vector x to a 3-dimensional vector T(x). This means that the transformation T reduces the dimensionality of the vector space.
The matrix A can be used to represent the linear transformation T by multiplying it with the vector x. The resulting vector T(x) will have three components, which can be interpreted as the coordinates of the image of x in R3.
Overall, the linear transformation T : R4 -> R3 is defined by the equation T(x) = Ax, where A is a 3x4 matrix.
The linear transformation T maps a vector x in R4 to a vector T(x) in R3. The transformation is defined by the matrix A, where each column of A represents the image of one of the standard basis vectors in R4.
Since A is a 3x4 matrix, it means that T maps a 4-dimensional vector x to a 3-dimensional vector T(x). This means that the transformation T reduces the dimensionality of the vector space.
The matrix A can be used to represent the linear transformation T by multiplying it with the vector x. The resulting vector T(x) will have three components, which can be interpreted as the coordinates of the image of x in R3.
Overall, the linear transformation T : R4 -> R3 is defined by the equation T(x) = Ax, where A is a 3x4 matrix.
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Let T : R4 --> R4 be the linear map satisfyingT{e1) =e2, T(e2) = e3, T(e3) = 0, T(e4) = e3,where {e1 e2, e3e4} is the standard basis of R4.Then,a)T is idempotent*b)T is invertiblec)Rank T= 3d)T is nilpotentCorrect answer is option 'D'. Can you explain this answer?
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Let T : R4 --> R4 be the linear map satisfyingT{e1) =e2, T(e2) = e3, T(e3) = 0, T(e4) = e3,where {e1 e2, e3e4} is the standard basis of R4.Then,a)T is idempotent*b)T is invertiblec)Rank T= 3d)T is nilpotentCorrect answer is option 'D'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let T : R4 --> R4 be the linear map satisfyingT{e1) =e2, T(e2) = e3, T(e3) = 0, T(e4) = e3,where {e1 e2, e3e4} is the standard basis of R4.Then,a)T is idempotent*b)T is invertiblec)Rank T= 3d)T is nilpotentCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let T : R4 --> R4 be the linear map satisfyingT{e1) =e2, T(e2) = e3, T(e3) = 0, T(e4) = e3,where {e1 e2, e3e4} is the standard basis of R4.Then,a)T is idempotent*b)T is invertiblec)Rank T= 3d)T is nilpotentCorrect answer is option 'D'. Can you explain this answer?.
Let T : R4 --> R4 be the linear map satisfyingT{e1) =e2, T(e2) = e3, T(e3) = 0, T(e4) = e3,where {e1 e2, e3e4} is the standard basis of R4.Then,a)T is idempotent*b)T is invertiblec)Rank T= 3d)T is nilpotentCorrect answer is option 'D'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let T : R4 --> R4 be the linear map satisfyingT{e1) =e2, T(e2) = e3, T(e3) = 0, T(e4) = e3,where {e1 e2, e3e4} is the standard basis of R4.Then,a)T is idempotent*b)T is invertiblec)Rank T= 3d)T is nilpotentCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let T : R4 --> R4 be the linear map satisfyingT{e1) =e2, T(e2) = e3, T(e3) = 0, T(e4) = e3,where {e1 e2, e3e4} is the standard basis of R4.Then,a)T is idempotent*b)T is invertiblec)Rank T= 3d)T is nilpotentCorrect answer is option 'D'. Can you explain this answer?.
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