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Let T : R3 ---> R3 be a linear transformation and I be the identity transformation of R3. If there is a scalar c and a non-zero vector x ∈ R3 such that T(x) = cx, then rank (T - cl)
- a)cannot be 0
- b)cannot be 1
- c)cannot be 2
- d)cannot be 3
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Let T : R3 --->R3be a linear transformation and I be the identity t...
Let T : R3 —» R3 be a linear transformation and I be the identity transformation of R3. If there exist a scalar c and a non zero vector x ∈ R3, such that T(x) = cx. Then
(T - c l) (x) = T(x) - (cl) (x)
= cx - cx = 0.
Thus, for non zero x, (T - cl) (x) is zero. Hence, nullity of ( T - cl) cannot be 0. Using Rank nullity theorem.
Rank (T - cI) = 3 - Nullity of (T - cl) Therefore, Rank of (T- cl) can not be 3.
(T - c l) (x) = T(x) - (cl) (x)
= cx - cx = 0.
Thus, for non zero x, (T - cl) (x) is zero. Hence, nullity of ( T - cl) cannot be 0. Using Rank nullity theorem.
Rank (T - cI) = 3 - Nullity of (T - cl) Therefore, Rank of (T- cl) can not be 3.
Most Upvoted Answer
Let T : R3 --->R3be a linear transformation and I be the identity t...
R2 be a linear transformation defined by T(x, y, z) = (2x + y, 3y - z).
To find the standard matrix for T, we need to determine the images of the standard basis vectors for R3.
The standard basis vectors for R3 are:
e1 = (1, 0, 0)
e2 = (0, 1, 0)
e3 = (0, 0, 1)
Let's find the images of these vectors under T:
T(e1) = T(1, 0, 0) = (2(1) + 0, 3(0) - 0) = (2, 0)
T(e2) = T(0, 1, 0) = (2(0) + 1, 3(1) - 0) = (1, 3)
T(e3) = T(0, 0, 1) = (2(0) + 0, 3(0) - 1) = (0, -1)
The images of the standard basis vectors are:
T(e1) = (2, 0)
T(e2) = (1, 3)
T(e3) = (0, -1)
The standard matrix for T is formed by arranging these images as columns:
[ 2 1 0 ]
[ 0 3 -1 ]
Therefore, the standard matrix for the linear transformation T is:
[ 2 1 0 ]
[ 0 3 -1 ]
To find the standard matrix for T, we need to determine the images of the standard basis vectors for R3.
The standard basis vectors for R3 are:
e1 = (1, 0, 0)
e2 = (0, 1, 0)
e3 = (0, 0, 1)
Let's find the images of these vectors under T:
T(e1) = T(1, 0, 0) = (2(1) + 0, 3(0) - 0) = (2, 0)
T(e2) = T(0, 1, 0) = (2(0) + 1, 3(1) - 0) = (1, 3)
T(e3) = T(0, 0, 1) = (2(0) + 0, 3(0) - 1) = (0, -1)
The images of the standard basis vectors are:
T(e1) = (2, 0)
T(e2) = (1, 3)
T(e3) = (0, -1)
The standard matrix for T is formed by arranging these images as columns:
[ 2 1 0 ]
[ 0 3 -1 ]
Therefore, the standard matrix for the linear transformation T is:
[ 2 1 0 ]
[ 0 3 -1 ]
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Let T : R3 --->R3be a linear transformation and I be the identity transformation of R3. If there is a scalar c and a non-zero vector x ∈ R3 such that T(x) = cx, then rank (T - cl)a)cannot be 0b)cannot be 1c)cannot be 2d)cannot be 3Correct answer is option 'D'. Can you explain this answer?
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Let T : R3 --->R3be a linear transformation and I be the identity transformation of R3. If there is a scalar c and a non-zero vector x ∈ R3 such that T(x) = cx, then rank (T - cl)a)cannot be 0b)cannot be 1c)cannot be 2d)cannot be 3Correct 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 : R3 --->R3be a linear transformation and I be the identity transformation of R3. If there is a scalar c and a non-zero vector x ∈ R3 such that T(x) = cx, then rank (T - cl)a)cannot be 0b)cannot be 1c)cannot be 2d)cannot be 3Correct 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 : R3 --->R3be a linear transformation and I be the identity transformation of R3. If there is a scalar c and a non-zero vector x ∈ R3 such that T(x) = cx, then rank (T - cl)a)cannot be 0b)cannot be 1c)cannot be 2d)cannot be 3Correct answer is option 'D'. Can you explain this answer?.
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