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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 'C,D'. Can you explain this answer?
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Let S and T be linear transformations from a finite dimensional vector...
To show that S and T are both nilpotent, we need to show that there exists a positive integer n such that S^n = 0 and T^n = 0.
Since S(T(v)) = 0 for all v, this means that the image of T is contained in the null space of S. Let's denote the null space of S as N(S).
Now, consider the image of T applied to itself: T(T(v)). Since the image of T is contained in N(S), this means that T(T(v)) is also in N(S).
Similarly, we can consider T(T(T(v))), which is in N(S) since the image of T is contained in N(S).
We can continue this process, and for any positive integer k, we have T^k(v) in N(S).
Since V is a finite dimensional vector space, this process must eventually terminate. Let k = dim(V)+1. Then, we have T^(dim(V)+1)(v) = 0 for all v in V.
Therefore, T^(dim(V)+1) = 0, which means that T is nilpotent.
Similarly, we can show that S is nilpotent. Since S(T(v)) = 0 for all v, this means that the image of T is contained in the null space of S.
Again, we can consider S(S(v)) and continue this process. Since V is finite dimensional, this process must eventually terminate. Let k = dim(V)+1. Then, we have S^(dim(V)+1)(v) = 0 for all v in V.
Therefore, S^(dim(V)+1) = 0, which means that S is nilpotent.
Therefore, both S and T are nilpotent linear transformations.
Since S(T(v)) = 0 for all v, this means that the image of T is contained in the null space of S. Let's denote the null space of S as N(S).
Now, consider the image of T applied to itself: T(T(v)). Since the image of T is contained in N(S), this means that T(T(v)) is also in N(S).
Similarly, we can consider T(T(T(v))), which is in N(S) since the image of T is contained in N(S).
We can continue this process, and for any positive integer k, we have T^k(v) in N(S).
Since V is a finite dimensional vector space, this process must eventually terminate. Let k = dim(V)+1. Then, we have T^(dim(V)+1)(v) = 0 for all v in V.
Therefore, T^(dim(V)+1) = 0, which means that T is nilpotent.
Similarly, we can show that S is nilpotent. Since S(T(v)) = 0 for all v, this means that the image of T is contained in the null space of S.
Again, we can consider S(S(v)) and continue this process. Since V is finite dimensional, this process must eventually terminate. Let k = dim(V)+1. Then, we have S^(dim(V)+1)(v) = 0 for all v in V.
Therefore, S^(dim(V)+1) = 0, which means that S is nilpotent.
Therefore, both S and T are nilpotent linear transformations.
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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 'C,D'. 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 'C,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 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 'C,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 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 'C,D'. 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 'C,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 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 'C,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 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 'C,D'. Can you explain this answer?.
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