IIT JAM Question > Let S and T be linear transformations from a ...
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Let S and T be linear transformations from a finite dimensional vector space V to itself such that S(T(v)) = 0 for all v ∈ V. Then
- a)rank(T) ≥ nullity(S)
- b)rank(S) ≥ nullity(T)
- c)rank(T) ≤ nullity(S)
- d)rank(S) ≤ nullity(T)
Correct answer is option 'A,B,C'. Can you explain this answer?
Most Upvoted Answer
Let S and T be linear transformations from a finite dimensional vector...
In V. Then either S = 0 or T is not injective.
Proof:
Suppose S is not equal to 0. Then there exists a nonzero vector u in V such that S(u) is nonzero. Since T maps V to itself, we can apply T to S(u) to obtain T(S(u)). By the assumption that S(T(v)) = 0 for all v in V, we have S(T(S(u))) = 0. But S(u) is nonzero, so T(S(u)) must be zero, since otherwise S(T(S(u))) would not be zero. This means that T is not injective, since it maps two distinct vectors (S(u) and 0) to the same vector (0).
On the other hand, if S = 0, then S(T(v)) = 0 for all v in V is trivially true. Therefore, either S = 0 or T is not injective, as claimed.
Proof:
Suppose S is not equal to 0. Then there exists a nonzero vector u in V such that S(u) is nonzero. Since T maps V to itself, we can apply T to S(u) to obtain T(S(u)). By the assumption that S(T(v)) = 0 for all v in V, we have S(T(S(u))) = 0. But S(u) is nonzero, so T(S(u)) must be zero, since otherwise S(T(S(u))) would not be zero. This means that T is not injective, since it maps two distinct vectors (S(u) and 0) to the same vector (0).
On the other hand, if S = 0, then S(T(v)) = 0 for all v in V is trivially true. Therefore, either S = 0 or T is not injective, as claimed.
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Let S and T be linear transformations from a finite dimensional vector space V to itself such that S(T(v)) = 0 for all v ∈ V. Thena)rank(T) ≥ nullity(S)b)rank(S) ≥ nullity(T)c)rank(T) ≤ nullity(S)d)rank(S) ≤ nullity(T)Correct answer is option 'A,B,C'. Can you explain this answer?
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Let S and T be linear transformations from a finite dimensional vector space V to itself such that S(T(v)) = 0 for all v ∈ V. Thena)rank(T) ≥ nullity(S)b)rank(S) ≥ nullity(T)c)rank(T) ≤ nullity(S)d)rank(S) ≤ nullity(T)Correct answer is option 'A,B,C'. Can you explain this answer? for IIT JAM 2023 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let S and T be linear transformations from a finite dimensional vector space V to itself such that S(T(v)) = 0 for all v ∈ V. Thena)rank(T) ≥ nullity(S)b)rank(S) ≥ nullity(T)c)rank(T) ≤ nullity(S)d)rank(S) ≤ nullity(T)Correct answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2023 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let S and T be linear transformations from a finite dimensional vector space V to itself such that S(T(v)) = 0 for all v ∈ V. Thena)rank(T) ≥ nullity(S)b)rank(S) ≥ nullity(T)c)rank(T) ≤ nullity(S)d)rank(S) ≤ nullity(T)Correct answer is option 'A,B,C'. Can you explain this answer?.
Let S and T be linear transformations from a finite dimensional vector space V to itself such that S(T(v)) = 0 for all v ∈ V. Thena)rank(T) ≥ nullity(S)b)rank(S) ≥ nullity(T)c)rank(T) ≤ nullity(S)d)rank(S) ≤ nullity(T)Correct answer is option 'A,B,C'. Can you explain this answer? for IIT JAM 2023 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let S and T be linear transformations from a finite dimensional vector space V to itself such that S(T(v)) = 0 for all v ∈ V. Thena)rank(T) ≥ nullity(S)b)rank(S) ≥ nullity(T)c)rank(T) ≤ nullity(S)d)rank(S) ≤ nullity(T)Correct answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for IIT JAM 2023 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let S and T be linear transformations from a finite dimensional vector space V to itself such that S(T(v)) = 0 for all v ∈ V. Thena)rank(T) ≥ nullity(S)b)rank(S) ≥ nullity(T)c)rank(T) ≤ nullity(S)d)rank(S) ≤ nullity(T)Correct answer is option 'A,B,C'. Can you explain this answer?.
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